Yes, and thanks for all the fabulous questions. I do need to get to the point landing though So Helmut can take off next week So I would ask that if you have questions that are technical technical or academic academic you ask me after the talk so which means What you should ask me about is how in the world do I apply this to my favorite j-curve space, you know That's what we're really here for right so The why don't you just or is this algebraic geometry questions are good for coffee Good all right, so here's a theorem and today the goal is to actually explain every single word so We want to regularize a modular space which we think of as being cut out as The compact zero set of some section We know by now what an M poly fold is We know essentially what scale smooth is I'm gonna need to tell you what Fred home as we know what a section is Right so the reason we want to model everything on scale hybrid spaces is that we want cut off functions Right exactly Right, but we don't want to worry about that right now So then Really the key is that something is non-empty And what is that so it's a space of other sections that are sc plus in some sense Which I'm gonna need to so if you're seeing sc plus what you should think is compact perturbation that are transverse General position to boundary if there is boundary And I'm gonna be allowing myself to fix a neighborhood of the zero set and the norm And I'm allowing myself to force the section the perturbation to only happen in that neighborhood And to be bounded by that norm since I can scale I'm gonna be able to scale that norm down So I can have a one here So this pretty much means I can make my perturbations as small as I want and that is the reason that I think you Now with these two Riders, I think you don't need any kind of generosity or commiga. You know this gets you to Make your perturbed zero set as close to the unperturbed as you want There is also the Obamacare writer If you like your solutions, you can keep your solutions So What I need to explain at some point is also what an auxiliary norm is what Fred Holm is What else sc plus this is right, so what do I mean by controlling compactness so Right if you just add a perturbation in this infinite-dimensional setting, it's totally not clear that the perturbed zero set Right, which we're going to want here. I Mean it's going to be smooth if it's cut out transversely, but why is it compact? And so we're just going to build this into the definition And then somebody has to you know prove that there are norms and neighborhoods that control compactness so controlling compactness means that if I have any SC plus section that satisfies these two things not necessarily transversality than this set However, a regular it is is compact so Check what's what? Why prime? Oh, right. Why good. Yes. Why one is sort of the quality one fibers And I should say that again at some point that's gonna happen So right so now the implicit function theorem that Joel already told us about gives us Yeah, cuts out nice smooth manifolds and Since I'm asking for general position actually, I know exactly what my boundary in Kona strata are That's important if you sort of do Fleur theory or something Right, you do a one-dimensional moduli space You regularize it and then you're going to get you know that the sum over the Boundary terms is zero because it's a one manifold now You'd like to know that those are actually the once broken trajectories and not the triply broken trajectories or something right, so the actual boundary would be degeneracy one in my perturbed Fleur space and this regularization theorem so says that that's actually exactly where my perturbed zero said hits the degenerate index one Locus in my ambient space and in Fleur spaces degeneracy exactly come from number of breakings So that's the only boundary you get So this is actually has applications Right, so you might ask why in the world did we talk about good enough So the good position is something that happens when you're trying to force more Coherence of your perturbations. You might not be able to make general position But that's because you've already prescribed the section on boundaries, which is something I haven't done here so That's where good position comes from Let's see What else do I need to say right so in order to regularize now I should say that somehow Whatever count I get here is invariant under choices And so I need to particularly well, I really I need to think about what happens when I vary J or whatever in my Setup, but mainly in this theorem. I need to think what it what happens if I take two different perturbations and the claim well Certainly most of these put most of these conditions except for transversality sort of a convex set so I can make a One parameter family and then I wiggle that one parameter family again a little bit to get transversality so Right so I take a pt that goes from p0 to p1 and I built this into Fredholm section over zero one times x So the neat thing is this is all now Fredholm So I don't really have to sort of go and prove again that this is Fredholm What I'm doing here is I'm adding One dimension in the domain That should you know add one to my Fredholm index And then I'm adding a smooth section. That's also compact So that shouldn't nearby sort of there should be Fredholm stability So this shouldn't change my Fredholm index at all. And so, you know all those sort of basic Fredholm Facts I survived in poly folds so This is going to be poly fold Fredholm It's going to be transverse after I wiggle enough by using that theorem again, and then I get a smooth cubordism out Which goes exactly well the only if there was no boundary before then the only boundary I have here comes from the interval And so then it's a cubordism from whatever happens at zero and whatever happens at one So All right, I wanted to write a box except some somehow statements of theorems nowadays are harder than proving theorems Good. Yes Well, and then you have to wiggle a little to get transversality Good so any questions about the statement any non-academic questions good so Right, I want to what I should say what a strong tame M-polyphold bundle it actually is and in particular need to fix notations. I can actually tell you what a Fredholm section is But I'm going to just tell you about this special case that so far has sufficed in all in All applications, so I'm going to say and the nice thing is that this special case will be things that are Automatically strong and tame so I want to not just talk about general Tractions, I'm just going to talk about splicings So an M-polyphold bundle of splicing type is what? Well, so I start with a something that should be a bundle. So I need some kind of suggestion between two polyfolds That are well that are modeled on on splicings Which you may not really know yet, but it's going to become evident once I write down the local tribulations So Together with So it like this to be a vector bundles. I'd better put a real vector bundle Structure on each fiber vector space structure So that's going to be oh goodness Right why sub X you put a superscript T here If you were wondering these if you perturb you might have to actually go into a different no this is that's totally a lie Bundle doesn't change when I change my perturbations. Sorry It just does if I change J, right? Okay, good. So these are actual fibers here, right and I need local trivializations So you see the exercise in writing the poly fold book is to really take your differential geometry book and Just sort of write SC or poly fold in front of everything So Right, yeah at some point You have to worry a little so right, so there are splicings that are just sort of the M-poly fold models and now for a bundle. I want a bundle splicing. So what is that? Well, it happens over an open set in the base I Would like the local bundle to Be SC defiomorphic on fibers To a certain model R, which is again a retract, but it's a specific retract. So I'm gonna at some point probably forget to specify open subsets of Retracts of splicing. So if I do that you can fill them in or you can just ignore this so So whatever this retract is it's gonna have a very nice structure. So there's gonna be one parameter that parameterizes families of projections Which give you the base and the fiber so this whole thing so my My parameter is allowed to That's where all the boundary comes from. So this is where V is some finite dimensional parameter, and that's where all the boundary comes from and then I have a family of projections on a Scale Hilbert space E and one on a scale Hilbert space F so These two are So in particular linear projections Which means also I don't have to worry about scale smoothness of them by themselves I just have to worry about scale smoothness with respect to that parameter. These are usually the gluing parameters so And now I'd only have to write this once because they don't know whether this is a small pie or a large pie so but there are two different families here and on this Right, and this is well, this would be the retraction actually let me write the retraction here Drag and drop So the parameter space of V is here So then the retraction just goes to V Projection on E capital Pi projection on F This is not algebraic geometry. No, sorry First you take the projection and then you build a bundle out of things you projected and then you get a section Why do you do that instead of taking an honest SC infinity vector of one or whatever then taking a section and then taking a projection at the end? Because that's what you have Really if you think about it well ask me that again once I've given you the real example I think that's Or even the toy example, so So you're saying it's an SC splicing because these projections are linear, right? Yeah, and so the nice thing is right So so the all the weirdness now comes from jumps in dimension of the fibers right of these images of the projections so Right. Yes. So the thing is that what I'm asking to be scale smooth is this map I'm not asking Pi V as an operator to depend continuously on V Hmm, it will not in fact well it might but then everything is boring So if it if it varies continuously with V then sorry if the operator Topology is continuous then the dimension of the fibers doesn't change But in infinite dimensions you can let the dimension of the fibers change That was exactly the example that Nate worked out yesterday And so you have this sort of bundle With varying dimensions of fibers and and really what happens is see these retracts. That is your space right, that's where you have the sections.  So the the bigger stuff here Sort of ambient spaces just exist locally and nobody says that they fit together in a nice way So I don't usually have an extension of the section to the bigger stuff. This is not like What I'm talking about here s is not just a restriction of some section in a banach bundle to some weird subspace Right. So the local sort of banach ambient spaces are just local Does that make sense? Good All right So Yeah, ah, yes, I should okay Did I say I didn't did not even say this so Right, so this is the this is the base and this is the fiber. So Really right. So what I should say search that Fibers go to Images and in F. So really right. So this is the base That's where boundary and corners happen and F for fiber And here now it's just it's an N gluing parameters here So that's yeah, it's just some gluing parameters. Give me boundary and some don't Yeah Random choice of words I'm not even going to say why this is tame, but it's automatically strong because Here's the why one bundle so strong means I have a meaningful bundle of Better quality fibers and in this case the fibers are simply given by right. This is my local model and I take Everything in the base but in the fiber. I only take the fiber of quality one so this is This this one is that one. So good. So example, right toy example, right? So this is a an M poll of poly fold So what you see here is an open ball that I've attached sort of two-dimensional Saturn ring to Again with open boundary here, and then I've attached just one interval So that's sort of one-dimensional two-dimensional three-dimensional pieces of my x and in order to for a you know Transverse Section to cut out something smooth here, right? So what I would like to be able is to say Okay, I have a Fredholm index one section over this You know that goes into this two-dimensional domain smoothly and then maybe even into the three-dimensional domain to cut out something Smooth, right? So the goal is to not get a sub poly fold, but it's actual manifold So if that is to be the case I'd better have fibers of different dimensions and The fibers better jump sort of at the same rate as The base dimension jumps and that's sort of nice to explain with this placing. I find right so here in this case Y over x right if I want one dimension here, I should have sort of No vector space over that everything is zero over these points sort of yx should have One and over these points I should have a two-dimensional fiber and so really at some point The condition is going to be really that in this bundle splicing somehow the fibers jump at the same rate But we're going to build that into our definition of Fredholm operator So that's not in here yet. No, right now, right? These could be these ill two things that this jumps at zero and that jumps at one Right, that would not be good news Homework right imagine how Nate's example gave you a little bit of Transition map or a little bit of a chart for a piece here Right and then you might have to might want to put you know another chart That's just two-dimensional and then you'd have sort of an overlap But it just goes from this two-dimensional piece to that two-dimensional pacing here You then need somehow our definition of scale smoothness to say in what sense that's scale smooth What I do right now I do want that Yeah When you say so Fredholm section is gonna mean something that imposes Jumps at the same time. Yeah, actually, maybe let me write this down and then right so So here's a little chart that I would like to so my chart map is gonna be to feel more fake to some open subset of This ambient Splicing so now right I better have a topology in which this thing is open but That's how we built that and What is this right so this here this ambient splicing was the union over? Also are is this direction here? and Each fiber here was some line in L2 so the funny thing is just that these lines in L2 are not constant they sort of Turn into all the infinite dimensions because It's this family of Bump functions that somehow weakly converges to zero because it just gets pushed out to infinity as You as you get closer to that point here so Right so here so what I'm actually going to take here Right, so I'm going to take the projections to be the same Projection on L2 to the same little bump function. That's just zero for V Lesser equal to zero and it's a bump centered at Something like each of the one over V When V is positive So right so so far this is just here. This is just the image of the little pie Because that's just my base Right, but now My fiber over this is going to be the same thing except it uses the same base gluing parameter again so You know over this I'm going to have the same fibers and here right at every point here the fiber Is zero Right, so that's the same thing. I'm going to use maybe I should say here right over V the fiber of My base splicing is zero So the fiber of my bundle splicing is also going to be zero and at a point here this fiber is One-dimensional and so over that point Here my fiber is also going to be our beta V So the local trivialization or the bundle Why over you is? Simply going to be well again this open subset. What is it? Right I could take the points V comma X or V E That lie in this thing here open subset oh right and Over each of them the fiber is again our beta V So these are my fibers What about right so what happens at the interesting point right so the Fiber of my splicing is still just zero Right, and so the fiber of my bundles also just going to be zero It's just that then because the fiber of the base jumps I'm going to have the fiber of the Bundle although that I'm going to have the fibers also jump So that way the jump dimension jumps at you know equally instead of my fret home index actually stays constant Which is a good thing? It's not built in here. No, it's going to be built into the definition of a I should but one could say okay, so Let's actually do this here. So why is fillable if for all V I think the kernel of Pi V This isomorphic to the kernel of capital Pi V So that's the complementary And we're going to have to fill at some point Strong just means that this this nice bundle why one is defined Which in this case is sort of automatically defined because we have these nice fibers So we I could just take the fibers of quality one in the in the ambient Buna space It's a property of a general and poly fold bundle it is automatic for and poly fold bundles of splicing type Right and what you want to think of is when you have this right? Where's my right the real example, right? It is time I think too much both infinite in need and therefore I mean, what do I mean by this right? In some sense, I'm going to need if can you just trust me that this is going to be built into the definition of Fred homefilling Great. Yes, please complain. I guess you out of all have the right to complain Anyway, so the way you should really think about this strong and why one is because you have fibers age three over age three maps So Really what you should think about strong is that? age three in a Pullback tangent bundle makes sense When you was in age three But so far usually our fiber over a Point of quality age three is usually just age two roughly speaking So they're sort of yeah Better quality. Well, it makes sense and it's sort of invariant under coordinate changes accepted Right, so if I wanted to make sense of age four over an age in an age three pullback tangent bundle and that is Probably going to depend on some choices of the trivialization Right, but when the age three, it's it's going to be independent. So That's Let's see good Okay, so this was the baby example and this was already sort of halfway to the real example So, ah, it's going to be very sloppy So I want to think about the bundle near a k-fold broken Fleur trajectory Because I'm lazy and I don't want to say SFT you could think, you know k-fold F a building right this a k-fold broken Fleur trajectory is the same thing as a k plus one floor building Because if you want to break k times you need k plus one trajectories and The first place where I'm going to be totally sloppy is that of course Fleur trajectories I have to mod out by r at least and I'm not going to do that. So Yeah The gluing parameters right yes, so for every Floor then when I glue together I need one gluing parameter exactly. Yes Yes, so K is in there in this. Yeah Well, no actually hard. No, it's the right K Because these come from nodes interior So every actual breaking every Fleur breaking that's what gives me the boundary and Yes Yeah, and then internal nodes are just going to give me right internal I mean I could have written C here actually fight Let me not do that right so Good so what is my bundle splicing here, right so so I'm going to have K gluing parameters I need some base space And some fiber space and I'm just going to Tell you what the projections are But first I should tell you what these Scare spaces are right. So what do I need to parameterize? Fleur trajectories right I need to vary these so really For every trajectory there I need to have a some section of a pullback tangent bundle and then Once I've varied them and I apply the Fleur Operator the Pater Kushi Riemann operator to it. I should end up in the fiber So these fibers then are just going to be same kind of product except age two and the fiber because I'm thinking about an operator of order one And then right I'm going to be lazy for now and just write epsilon E F goes to pi epsilon of E Hi epsilon of F and now I need to tell you what the projections actually are Right so really and each of these is a tuple right this K tuple and these are K plus one tuples so pi epsilon is the projection in the sort of Retract that Joel Defined at length so it should be the projection to The kernel of an anti-gluing along The kernel of the plus gluing right, so whenever I see Variations of the flirt trajectories right really I'm thinking okay a very and then I glue it together Right, that's sort of my chart map But that has ambiguity so what I want to do is I want to sort of reduce the ambiguity That's why I go to the kernel of the anti-gluing But by reducing the ambiguity I should like somehow not change what The point actually is in X that I have in mind. That's why I go along the kernel of the Gluing so I don't actually want to change the point that I have in mind Right and now you're going to say okay, there's Evidently Gluing parameters missing and I think it's just worthwhile to say this again. So So what are these so these are Auntie or pre-gluing off the Right of really the sections. I'm gonna I think probably Well, there's a lot of fuzziness here, right? So you have to ask yourself. Can I actually Glue these and then I should do this with some gluing parameters and those I like to be somehow large So there's always going to be some gluing profile in the picture That takes small parameters to large parameters and the choice of this gluing Profile is sort of that is that is one of the foundational choices in this whole subject There's pretty much just well, there's one for which you get smoothness and that's unfortunately not the sort of Obvious one you choose from Dylan Mumford space. So Eventually somebody needs to prove that the invariance that we get out of this are independent of the choice of gluing profile So yeah, yes I think it's pretty clear that something should work But then there's a question whether there's an actually in total order on gluing profiles Yes But it's basically like it if somebody gives you are the real line And you don't know what is and then you have to choose a chart So you take the identity or you take X goes to X So the difficulty is only one point Yeah Possibly yes, we don't know that Yeah Why right let's let's keep that for next week Right so Let's see What happens right so strictly speaking I told you how to glue in the base, but you can do exactly the same thing in the fiber You so for the fiber and this again goes back to sort of the vex question of what do I actually have a section off, right? I don't have a section that's just lives over sort of the broken flirt trajectories I wanted also sort of encode my section that lives over the unbroken flirt trajectories, right? And that takes a glued flirt trajectory and applies the flirt operator to it So you get just one section of one You know something that lives over a glued Curve and so in the fiber I kind of need to have the same domains that I have or the sort of glued domains Right I glue the cylinders so I need the same domain in the fiber that I have in the base So that's why somehow the Dylan Mumford parameter, maybe you could also think of the Epsilon's here on the splicing right the parameters are sort of the Dylan Mumford parameters and they affect The maps and the sort of zero one forms in the same way Right so why is that not defined on the whole space before I could define it on the whole space except There's nothing to do with Fleur theory So I should write I should maybe write this down right so Let's see. What do I have here? So my Fleur idea Right, so what is my D bar operator? really in terms of this pre-gluing and things really what it it sort of takes a big pre-glued Curve to D bar Off that pre-glued curve Right on most of it, right? It's just one D bar. It's not broken Right, so they open or the the main stratum of X is the unbroken one So this is the main stratum of my operator, right? Except when I'm near a broken one this sort of I want to pull all of this back somehow to Something here that should be in the fiber and something here that should be in the base right, so somehow this comes from a choice of gluing parameters and All these Fleur trajectories right so that I glue together to parameterize this point Then I apply the D bar operator and then I need to go back and so I kind of take the What what what am I doing here, right? I'm going to the fiber so to get to the fiber I need to take plus and minus inverse of Exactly this thing and I force minus gluing to be zero. So this then sits in that fiber Yes, right So that way I can this way I'm pulling back the D bar operator that just acts on this on a single Cylinder I pull it back to this sort of tuple of cylinders here by sort of first gluing Applying D bar and then sort of undoing the gluing so Right. However, as you can see right if this is a Fredholm operator, right? I'm Losing a lot of dimension here, right? So I have a massive kernel Then this is Fredholm and then I go back and I have a massive Well, this is never going to be surject if there's a lot of things I don't map to so I have Infinite kernel here and infinite co-kernel here and so the idea is going to be to Cancel the kernel here by the co-kernel there I'm not sure I'm going to get there So but this is so this is my real-life example and again, right? So here Y1 now comes from the local things that are V e and then The f's that are actually already in H3 and not just in H2 So that's really what I should be thinking of as that bundle Y1 So yes, right? Yeah, I fear that would come right so Yes, someone could say that But I haven't said what filling is so ask me that again once I've said filling Let me just quickly say what what can I say now, right? So let me go down my laundry list actually, right? I needed to tell you what an auxiliary norm is so What this is is a continuous function on this better quality bundle That's a complete norm in each fiber. So classically continuous norm in each fiber and in the example well You take The norm at a point. What is this V e f to be? F H3. Oh boy, and now I'm That's probably an exponential weight on here Because if I want this to be a scale space, right? If this is f0, then this is supposed to be f1 and so that needs to be Compact in f0 there was an extra as an extra condition on the norms Because the others can get some digital if If you have a sequence of vectors The base converges to x and the limit super is going to zero Okay, good. Okay. Yes. There's something. Okay. Good. Okay Small print There's also right so I should say right so obviously you can see all the small print in the hoof of is that skits in the papers And if you're trying to find where that specific small print is I would recommend looking at the what we used to call the user's guide poly folds at a glance or two that We also posted on the on the web page that that is lacking this Small print. However, it always precisely sites the place in the publications where you can see all the wait First and second glance, right? Yes, because it became too long, right? so I Need to tell you what an sc plus section is so a section Well, first of all a section is something that gives you the identity when projected down and it's sc plus if It actually takes values in these better quality fibers and is sc infinity as a section of that better quality bundle Which is exactly you know a first order. So what you should think if here is a zeroes order operator compact operator Maybe compactly supported would be really nice The note here is that the classical perturbations if I change J this is something like J minus J prime DT That is evidently first order. So this is not sc plus So wiggling J is not what happens in this regularization theorem however a Homotopy of J's fits into this cubordism argument and so that makes sort of Changes of J built into the theory after all Well, it's first order so in particular this needs to take age three to age three And it yes, right. Yes Good, okay, so now I can Right, so now the I think the only thing left to say is what is Fred home am I missing anything else? But we know what compact means And I explained what controlling compactness means so So first things first so Now this is the this is the place where I think a grad student Hold up in a basement would have a little bit of trouble making up the whole theory So Fred home is not quite obvious. So one thing I Need as I need it to be regularizing which means that if I'm in the quality If my section takes values in the quality k fiber Then I actually want to have the base also quality k which is like saying if D bar u is in HK Then u is an HK plus one also known as elliptic regularity So that's something that's sort of true for the operators that we usually look at and then the key point is this filling that we've Alluded to and this at last count you needed a filling somehow strangely not just at Solutions but at everything that's sort of smooth. That's of quality infinity So near each Point I need a local trivialization and a filling and I'm going to say what the filling is So that the filled section is Fred home ha ha so this Fred home However, it's going to be easier to define than that Fred home Because this Fred home happens on splicings and retracts and this Fred home is going to be a map between Or map germ of maps right between banner spaces or SC banner spaces or SC who but spaces in this case So I'm going to leave that for the moment undefined and tell you what the filled section is so so in a local Chart Right. I'm going to this bundle splicing This sits in and now I'm going to forget that there are open sets. I don't want to write another one in here so that sits in a splicing that sits in here and this bundle splicing sits in the bundle splicing that sits in here and Right, so what I would really like to do is I would like to fill up my section here So that I have any chance of this being Fred home. So let's see. What could I do? Right. So first of all my problem is that this e is not this ve is not necessarily in O But I can map it down with my retraction and Then apply s which sits in here. So That's not so bad. Um, ooh, let me I Should say what this s actually does Yes, so I want to write so the section takes Ve obviously to ve and then something happens in the fiber. So I'm just going to write f for whatever s does in the fiber and So then I can apply f f is only defined on Oh, so I need to throw in a projection here. So I can do that But then I'm only ever going to hit the The splicing within this fiber never going to hit the complement. So I want to add something and I'm going to let whatever happened They're just depend on the complement of e so I can write each e as Pi ve plus 1 minus pi ve That's the splitting Here and I'm going to use the same splitting in F This is the splicing and the complementary splicing. I'm going to have right this guy here and What do I want right so? so F prime of v comma dot right maps the kernel of little pi v to the kernel of capital pi v and I'd like this in an appropriate sense to be an s c infinity family of Isomorphisms The idea being that I do need to soup up kernel and co-kernel to make s twiddle Fred Holm, but I don't actually want to change the zero set. So that's the main remark as twiddle inverse of zero is s inverse of zero and also kernel and the image of the linearizations of ds twiddle Isomorphic to the That's that is wrong kernel and co-kernel image perp are isomorphic to the co-kernel and co-kernel of ds So that allows me to dream of Fred Holm theory. So that is the theory Yes isomorphism, yeah, yes Yeah, yeah, I mean in my splice, I mean this is a simple if I mean I think so well So here's the example, right? So in the example actually listen one example is as good as all yeah Right so the filler is the linearized operator Where you may ask right? It's it's on the anti-gluing so So overall the filled section So if you look up there, I wrote down what actually the section is in Local coordinates so the section went from epsilon e To what was that? I just wrote down the fiber plus minus inverse of plus of e and Zero right so that was my that was the section and In order to get the filled section all that I'm doing is I'm writing down the minus gluing here By complete Breakdown, of course, there was a d bar operator here and so the minus gluing of e Always lives in So if h3 of r times s1 it comes from data that is very close to the breaking so It just lives over a fixed Hamiltonian orbit. Well, it's really a product right because they have various breakings so Point being so there's no r dependence here. And so when I take the linearized D bar operator on this There's a general theory that says that Our invariant operators here, you know if they are useful if they're fret home at all then they're actually isomorphisms so This is of the form DS plus a and Then you read my favorite paper by robin Salomon about I think spectral flow right and I Have successfully sidestep the question of what a fret home map between scale Hilbert spaces is Which is evidently in my notes so I'm just going to hand wave for like one minute and then you can ask me about details if you want them So the problem is really when you're proving Invisit function theorem that you need to do Newton iteration and the Newton iteration sort of comes from a contraction and When you write down the contraction that you get from a scale smooth family of from a scale smooth What what what do you actually mean by a fret home section, right? That's nonlinear usually we say well the linearizations better be fret home, right? so that's something that makes sense but the other thing we need is that Newton iteration works and that requires a certain continuity of the linearizations and Unfortunately, there's this scale shift and so you get a contraction property, but the Contraction goes sort of down in levels by one So if you start iterating you go further and further down in levels that does not bode well for any convergence So you kind of need to sort of force yourself back up So we need to add Something called a contraction germ property that sort of implements this Contraction So really what this is is a contraction a level preserving contraction in All but finitely many dimensions and so then you ask yourself well, how do you ever prove this? well in that sort of comes if Your fret home section is actually right. It's scale smooth That means you always have that loss of levels, but if you have classically see one in all But finitely many dimensions Then you can prove that that implies this contraction property well plus some small print and so that's this can be found in the Grim of Witten paper and Somehow I independently figured well but there should be a better definition of fret homelessness and I'm currently revising that paper in which I also attempt to prove the fret home property for flirt trajectories and I would end by saying that poly folds are awesome Because you actually get referee reports So if the referees are in the audience, thank you You know They did not complain of me failing to spread peace in harmony They actually wrote the read the paper and you know font mistakes and pointed out things and so I'm revising and that is Yes, beautiful thing. So I Wish all referee reports were like this and maybe I just need to keep doing poly folds So Yeah, well, I'm I'm going to scan these Yes, and then put them on. Yeah. I mean, however, I would say I mean all of this is sort of essentially just skimmed from the user's guide Yes, yeah, yeah, if you want to there's more than everything in there I Right he's asking me and I'm asking you because I don't know what the small print it is so I thought you just use cutter functions That works right it satisfies whatever small print you have awesome good Ah As the as the index of the linearized Fethom operator. Yeah, and then somewhere it's it's built in that that's actually constant Yes, and then you could ask how do you define orientations, right? So you need to construct the determinant line bundle and yeah


One of the central assumptions governing these Nobel conferences is that the questions of science which they discuss cannot be separated from questions of values. The connection between science and values has already been especially clear this year in the discussions which have arisen from the previous presentations. This afternoon's final speaker, Dr. Holmes Rolston III, University Distinguished Professor of Philosophy at Colorado State University, is superbly qualified to help us understand something about what the battle within means, both for our understanding and for our actions. His earliest academic training was in physics and mathematics, which he followed with degrees in theology and the philosophy of science. Some of the fruit of this study can be seen in his extensive publications, some of which you can read about in your conference program. Deserving special mention are three books, Science and Religion, a Critical Survey, Philosophy Gone Wild, and Environmental Ethics. Still more of the fruit of Professor Rolston's labor can be seen in the positions of trust and authority which he has filled. He has held editorial positions with respected journals, Environmental Ethics, and Zygon. He has been a consultant to Congress and a presidential commission. He has been president of the Rocky Mountain Great Plains region of the American Academy of Religion and president of the International Society for Environmental Ethics. His success as an interpreter of the meaning of scientific endeavors for our social and moral life can be seen in the impressive list of places encompassing five continents to which he has been invited. To name just a few, the Royal Institute of Philosophy in Wales, the Chinese Academy of Social Sciences, the United Nations Conference on Environment and Development in Brazil, the South African Forum at the University of Cape Town, and the Center for Theology and the Natural Sciences University of California at Berkeley. Next year he will address the World Congress of Philosophy in Moscow. In the last few days we have discovered that this philosopher, hiker, expert on mosses, wears all of this learning, all of this achievement with enormous goodwill, grace, and modesty. Please join me in thanking Professor Ralston for his contributions to this conference so far and welcoming him to this podium for his talk, Immunity in Natural History. Dr. Ralston. I do deeply appreciate being here in a way as a philosopher amongst all these world renowned immunologists. I worry that I might be a fish out of water. But in a way it's the duty of philosophers I think, try to think a bit more comprehensively perhaps about the work of specialists. And that is what I would like to try to do today. Philosophers examine world views, and the world view of an immunologist it seems at first is that life is a kind of a microscopic war in a microscopic world. And if you read the folder which drew you to this conference, you would read the following. Even in the relative peace and calm of a normal day, the human body is constantly under attack. Viruses, bacteria, and other trespassers launch regular assaults against the body's immune system, which raises an intricate web of defenses to identify and repel these biological invaders. That's the picture of the body as a kind of sabbathole that's under constant attack. There are innumerable hostile bacteria, viruses, and parasites. They lurk everywhere in the environment. There is a battle going on within. These enemies float in the air and infest the water and pollute our food, and they cover every surface we touch. Even the body's own cells can turn traitors, such as cancer cells. Now, that's a vivid kind of imagery, but imagery sometimes needs philosophical analysis, especially imagery that begins to color our world views. When we speak, for instance, of ant wars, or selfish genes, or even queen bees, we borrow words from one domain of experience and transfer them to another. So what about this battle within? Does that picture need to be set in a larger world view? What is the place of immunity in natural history?  Perhaps the imagery of battle, war is too strong, but surely we are engaged in a constant struggle for health. There can be no doubt, can there, that there's a struggle for health? That's not metaphor, that's straight truth. And, of course, these diseases can reach epidemic proportions. After the Second World War, the flu virus killed 20 million people, more than died in the war. And we forget how feared were the Black Death, or smallpox, or diphtheria, or cholera, before modern medicine won the battle against them. And I still remember in my childhood friends who died of polio and others who were crippled before Dr. Salk, who is with us now and with whom I felt like I was sitting beside a childhood hero when I spoke to him a minute ago. So we do have a kind of battle that we fight against these viruses and microbes. In fact, your body is at work, even as I speak, this very moment, killing these killers. If we had lived in the medieval world, we should have been on the constant lookout against demons. But now in the place of superstition, we have science. The microscope has given us the truth, which is worse than demons, germs, killer microbes, microdemons, billions of them in every nook and cranny. These microbes were, in fact, the causes of those great medieval plagues, which they took to be the work of demons. The first noble truth of Buddhism is that life is dukkha, suffering. And though the Buddha had no microscope, Indianology confirms him. Or the psalmist was right. Life is green pastures, but surrounded by the valley of death. It is a table prepared in the midst of our enemies. Microscopic enemies, too, adds the immunologists. So you can praise God for your natural killer cells. Still, this idea of a battle within might not be the whole truth, even from within immunology. Another word that immunologists often use is the word self. Maybe we better take a closer look at that word. Immunology is the science of self, non-self discrimination. So let's think about self-identity. Life involves organization, information, reproduction. Now, it's impossible for an organism to do these things unless it has an outside and an inside. Really, the essence of life includes the definition of self from long self, setting of limits to the self. There's got to be some kind of defining envelope. And after that, an organism can take in nutrients from the outside and sequester them for its own uses. The boundary line demarcates an order contained and maintained against entropy. An order which, in the prospect of death, has to be protected and reproduced. Now, an immediate implication of the fact that life has to have its boundaries is that you can have invaders. Things inside that don't belong inside, non-self inside, other cells violating these limits. And so life has got to control passage across a defining membrane. That's part of the bigger truth that life is constantly self-defense. There are all kinds of things out there that we must protect ourselves against. Cold, wet, dry, wet especially, we've learned these last two days. Solar radiation, poisons, predators. Cells have to be repaired when damaged. Outsiders that get in have to be put out. Also, there can be insiders that no longer belong inside. The body has to recycle dead cells or to program the death of cells that are no longer needed. Sometimes, insider cells get out of control and, lest these become tumors, they must be stopped. Immunologists can think of this as killing these outside invaders or these inside cells gone wrong. But we can also think of it as a kind of ordering the self. In the multicellular organism, such as a mammal residing in a biologically complex environment, this ordering of the self can become quite complex. In a world with millions of species, making billions of kinds of organic molecules, the body has to produce a defense against almost any kind of organic molecule except those of...


Welcome to lecture 9 on measure and integration. As you recall, we have been looking at classes of subsets of a set X called semi-algebra, algebra, sigma algebra and so on. And then we also looked at set functions defined on these classes with properties. So, in particular we define the concept of a measure. A measure is a set function defined on a collection of subsets such that the measure of mu of the empty set is equal to 0 and mu is countably additive. Today, we are going to start the process which is called extension process. So, the topic for today's discussion is going to be extensions of measures. Basically, the question arises from some properties on the real line. Let us look at mathematically the question. We know that notion of length is defined for all intervals. So, the question is, can the notion of length be extended to arbitrary subsets of the real line? That means, can we define the notion of the length for an arbitrary subset of the real line? Of course, it should be compatible with the definition of the length for the interval. So, the need for such an extension, one of course, is purely a mathematical curiosity that we have the notion of length for an interval, can we define it for a arbitrary subset? Another reason which is more important is that it arises from some problems in Riemann integration. The concept of Riemann integral which is defined for a class of functions fails to satisfy some nice properties, like if a function is the fundamental theorem of calculus does not hold for Riemann integrable functions. So, in order to remove those difficulties, one started for looking for an extension of Riemann integral and that led to the problem of extending the notion of length from a class of subsets that is intervals to all subsets if possible. If you are interested in looking at more details about that, why Riemann integral should be extended to a wider class of functions and how that leads to the concept of extending the notion of length to arbitrary subsets, read chapter 1 and 2 of the text book that we have mentioned earlier, namely an introduction to measure and integration by myself Inderke Rana. . So, let us start with the question of what is an extension. So, let C 1 and C 2 be two classes of subsets of a set X and let us assume C 1 is a subset of C 2. We have two measures or two set functions mu 1 and mu 2, mu 1 is defined on C 1 and mu 2 is defined on the collection C 2. So, mu 1 and mu 2 are set functions mu 1 defined on C 1 and mu 2 defined on C 2 with the property that mu 1 on C 1 is same as mu 2 for subsets of C 1. So, mu 1 and mu 2 agree on subsets of C 1 because C 1 is a sub collection of C 2. So, in such a case we call mu 2 is an extension of mu 1. So, on C 1 which is a smaller class mu 1 and mu 2 are same. So, and mu 2 is defined on a bigger class that is C 2. So, we say C 2 and mu 2 is an extension of the measure of the class set function C 1. So, the problem is given a measure mu, we start with a measure mu on a semi-algebra C of subsets of a set X. We want to show that there exists a unique extension to a measure mu tilde on A of C the algebra generated by it. So, this is going to be our first step of extension theory namely given a measure on a semi-algebra. We are going to extend it to a measure on the algebra generated by that semi-algebra. So, let us see how this process is carried over. So, recall that a set E in the algebra generated by a semi-algebra where characterize such sets can be given by a representation E is equal to union i 1 to n E i. So, every set in the algebra generated by a semi-algebra is a finite union of sets form the semi-algebra and in addition they are pair wise disjoint. So, this was the result we had proved that the algebra generated by a semi-algebra is nothing but all finite disjoint union of sets in the semi-algebra. So, let us take any set E in the semi-algebra. So, we define mu tilde of E to be sigma i 1 to n mu of E i and the claim is that this is the unique extension which we are looking for. So, let us see how do we do it. So, we have got mu on C and this is a semi-algebra. So, we define mu tilde on the algebra generated by C. So, this is algebra generated by C. So, we want to define a function here a set function which should look like an extension. So, if E belongs to F of C then we know this set E looks like a disjoint union of element C i i equal to 1 to n for some n C i belonging to C. Now, why we defined it the way we have defined mu tilde? See, if mu tilde of E is going to be defined and it is going to be a measure on the algebra F of C then we know that every measure is also finitely additive. So, by the finite additive property of mu tilde which we have not yet defined, but the finite additivity property will say that this should be equal to mu tilde of the union C i i equal to 1 to n and this being finitely additive we should have i equal to 1 to n mu tilde of C i. But mu tilde is going to be an extension. So, that means mu tilde on C i is same as mu on C i's. So, this is same as 1 to n of mu of C i. So, that actually fixes what is going to be the definition of mu tilde of E. So, if E is a finite disjoint union of elements which is C i's then mu tilde of E must be given by this and that also shows the uniqueness of the definition of mu tilde. So, mu tilde should be defined by this it is necessary and we will show that actually this definition works also. So, let us prove this property that mu tilde. So, first we want to show that mu tilde is well defined. So, what does that mean? Suppose E is a set which is in F of C then we know that E can be written as a finite union of set C i's finite disjoint union of set C i in C. But it is possible it can have some other representation. So, it is possible that is also representable as j equal to 1 to m of some sets D i, where C i's belong to C and D j's also belong to C. So, to show then because our definition dependent on the representation. So, we should show that mu of C i summation i equal to 1 to n is same as summation mu of D j, j equal to 1 to m. So, this is we should show then only we can claim if we can able to show this then only we can claim that our function mu tilde i is well defined. So, let us show this. Now, note because E is given by these two different representations. So, I can write union C i i equal to 1 to n also as union C i intersection union D j's j equal to 1 to m. So, that is equal to union i equal to 1 to n union j equal to 1 to m C i intersection D j. Similarly, union D j's j equal to 1 to m is also representable by the same way because the two sets are same. So, it is same representation. Now, let us compute sigma mu of C i i equal to 1 to n. I can write it as sigma i equal to 1 to n. Now, this mu of C i is this is what union of C i intersection D j. That is where we are using this representation that we just now wrote j equal to 1 to m. This is a disjoint union C i's belong to the semi-algebra D j's belong to the semi-algebra. So, this intersection belongs to the semi-algebra and their unions is C i which is also in the semi-algebra and mu is a measure on the semi-algebra. So, it so this is also finally additive. So, it is i equal to 1 to n. So, I can write this sigma equal to j equal to 1 to m mu of C i intersection D j. Similarly, we can also write j equal to 1 to m mu of D j to be equal to this is summation j equal to 1 to m and mu of D j. So, that I can write as union of D j intersection C i i equal to 1 to n. Now, again by finite additive property this is j equal to 1 to m sigma i equal to 1 to n mu of D j intersection C i. Look at this equation 1. Look at this equation 2. 1 and 2 imply that sigma i equal to 1 to n mu of C i is equal to sigma j equal to 1 to m mu of D j. So, that says implies mu is well defined. So, what we have shown is the following that we take any set in the algebra generated by the semi-algebra. So, that has got a representation in terms of the elements of the semi-algebra. So, any element E in the algebra generated by the semi-algebra can be represented as a finite this is not union of elements in the semi-algebra say C i. So, pick up any such representation and define mu tilde of E to be equal to sum of mu's of this P C i sigma i equal to 1 to n. It does not matter which representation you choose you will always get the same sum. So, that means mu tilde of E is well defined. Now, let us look at the next property namely that mu tilde which is defined on the algebra generated by the semi-algebra is finitely additive. So, let us prove that property. So, we want to prove that mu tilde. So, mu tilde is finitely additive. So, to prove that what we have to show? So, let E be written as a union of E j's j equal to 1 to n where each E j belongs to the algebra generated by C. And of course, E also belongs to the algebra generated by C. So, we want to show to show that E mu tilde of E is equal to summation j equal to 1 to n mu tilde of E j. So, this is what is to be shown. Now, to show any such property we have to go back to the definition of mu tilde of any set. So, since E belongs to the algebra generated by C that implies let us write each E j. E j belongs to the algebra. So, each E j can be written as a disjoint union of E j and say k, k equal to 1 to n j, where E j k belongs to C for every j and k. So, every element E j is in the algebra generated by C. So, it must be a finite disjoint union of elements of C. So, that implies that the union E j, j equal to 1 to n is equal to union j equal to 1 to n union k equal to 1 to n j of E j k. Then this is my, so this is my over set E. E is equal to union. So, we have represented E as a finite disjoint union of elements of C. So, that implies that mu of E, mu tilde of E, I can choose any representation. So, in particular this, so it is equal to summation j equal to 1 to n summation k equal to 1 to n j of mu of E k j. And now using the finite additive property of mu, we will write this. So, this is equal to, look at this sum. So, look at this sum. That is nothing but j equal to 1 to n mu tilde of E j. That is by definition, because E j is a union of E k j over k. So, by definition I can take that representation and write this is equal to this. So, that says mu tilde of E is equal to this. So, hence mu tilde is finitely additive. So, we have proved that mu tilde is finitely additive. The uniqueness we have already shown. So, thus we have shown that a measure which is defined on a semi-algebra can be in a unique way extended to the algebra generated by it. And basically the idea is, because every element intuitively keep in your mind that mu of asset is the size. So, any element in the semi-algebra, in the algebra generated by the semi-algebra is a union of disjoint pieces in the semi-algebra and size of each of them is known. So, the size of the union must be equal to some of the sizes of the individual pieces, because they are disjoint. So, that was the idea and that helped us to extend a measure from a semi-algebra to the algebra generated by it. So, that is the first step of the extension theory. So, as a consequence the length function can be extended. That we have already shown, length function can be extended from the collection of all intervals to the collection of finite disjoint union of intervals that is algebra generated by it. So, now we will go to the next step of the extension. So, we will start with a measure which is defined on an algebra and we want to try to extend it to the sigma algebra generated by it. So, the next step in the extension theory is we can the length function. For example, we would like to say can the length function be extended to all subsets of the real line. We have done it from intervals to the algebra generated by intervals. There is a theorem by a mathematician called S. M. Ulam and that theorem was proved in 1930. It says that under the assumption of continuum hypothesis it is not possible to extend the notion of length to all subsets of real line. And this is a very important theorem. So, it uses two things namely one is what is called continuum hypothesis. I will not go into the discussion of what is called continuum hypothesis at this stage. I would say that one should read about this theorem from the text that we have just now mentioned an introduction to measure and integration. So, this is very nice and important theorem which says as a consequence that it is not possible to extend the length function to all subsets of real line. So, the question comes if we cannot extend. So, in general a given a measure mu on an algebra of subsets of x, we would like to extend it to a bigger class than a. It cannot be done it for all subsets anyway. But let us try to intuitively follow our idea of measuring the size of an object. So, intuitively given a measure mu on an algebra a collection of subsets of a set a of a set x, mu of a is the size of the set a which you can measure. And given an arbitrary set e one may not be able to measure its size exactly using the mu, but we can at least try to approximate. So, let us define what is called the outer measure induced by a measure. So, let us take mu and algebra of subsets of a set x, a and algebra of subsets of a set x and mu a measure defined on it. For any subset e in x let us define what is called mu star of e. So, what is mu star of e what we do is given the set e. So, here is a set e you cover it by sets a i is in the algebra you cover it by the sets in the algebra. Take a covering of e by the sets a i's in the algebra and you know what is the size of the set a i. So, let us take the size of the set a i and add up all these sizes. So, what do you think this sum will represent? This sum will represent in some sense the approximate size of the set e of course it depends on the covering a i. Now, what we do is we take the infimum of all these approximate sizes. That means, we take the infimum of these numbers over all possible coverings of the set e and define that number as mu star of e and we will try to analyze what are the properties of this mu star of e. So, first of all let us give it a name this mu star of e is called the outer measure induced by mu. Why outer? Because we are covering e by sets. So, these things cover e we are going may be we are going outside e. So, this is outer and measure because we are trying to measure the size of this in terms of induced by mu because in terms of the known sizes mu. So, once again let us recall and look at carefully what this mu star is given is at e arbitrary subset in x cover it by elements a i whose sizes you know. So, take a covering of e by elements in the algebra look at the sizes of a i's add up all this that is the sum mu a i that is approximate size and take the infimum of all these approximate sizes. So, that we are going to call as the outer measure induced by. So, the first property we want to say is mu star is well defined. Well, what is the meaning of mu star is well defined? Let us go back to the definition this is mu star is infimum of some numbers and infimum of a subset of numbers exists in the real line if it is non-empty and it should be bounded below. Of course, all these numbers are going to be bounded below because all are non-negative numbers. So, it is bounded below by 0. Why is this non-empty? Why is this collection non-empty? Because a is an algebra. So, the whole space belong to it. So, keep in mind a is an algebra and in the definition of an algebra the whole space x is an element. So, e is covered by x itself. So, and x belongs to the algebra. So, at least there is one number in this collection over which you are taking infimums namely mu of x. So, it is a non-empty collection of extended real numbers. So, its infimum always exists and hence mu is a well defined number. Of course, it could be equal to plus infinity. Keep in mind the numbers here they are all extended real numbers. So, this is this set is a collection of non-negative extended real numbers and their infimum always exists and infimum could be equal to plus infinity. So, we have shown that mu tilde is mu star the induced outer measure is well defined. The next property. So, mu star is a well defined set function on the class of all subsets of the set x and we want to show some properties of it. So, the first property is mu star of empty set is equal to 0. That is true because empty set belongs to the collection A in the algebra and mu star of there is equal to mu of A and that is equal to 0. And for any set that is a non-infinimum of non-negative numbers. So, this infimum has to be bigger than or equal to 0. So, that first property is obvious. Second property we want to check that mu star is monotone. So, let us check that mu star is a monotone function. So, let us take let A and B be subsets of x and A a subset of B to show mu star of A is less than or equal to mu star of B. Now, what is mu star of A? This is in all these properties we are going to use the definition of infimum critically. So, what is mu star of A? mu star of A is defined as by our definition. It is the infimum over sigma mu of E i's say 1 to infinity, where this set A is contained in union of E i's disjoint union. Of course, E i's belong to the algebra. And what is mu star of B? That is the infimum i equal to 1 to infinity of mu of say f i's, where B is contained in union of f i's i equal to 1 to infinity disjoint union, where f i's also belong to the algebra A. Now, note if A is given to be a subset of B, if A is subset of B and B is covered by a union j equal to say 1 to infinity, then that implies A is also inside. So, this is also inside f of j. So, what we are saying is every covering of B is also a covering of A. So, and this is the infimum over all possible coverings of B and this is the infimum over all possible coverings of A and every covering of B is also a covering of A. So, here we are taking infimum over a larger set and here we are looking at the infimum over a smaller collection of numbers. And whenever you take infimum over a smaller collection of numbers, that is always bigger than or equal to infimum over a larger collection of numbers. So, that is a simple property about infimums. If you are taking a infimum of a larger collection, then that tends to be smaller than the infimum over a smaller collection. So, that property implies that mu star of A has to be less than or equal to mu star of B. So, that is purely a property of the infimum over what collection you are taking. Every covering of B is also a covering of A. So, coverings of B form a subset of coverings of A and hence this property is true. So, that is the monotone property namely mu star is monotone. Let us look at the next property namely mu star is countably sub-additive. So, we want to prove mu star is countably sub-additive. So, that means what? To show, so whatever to show that if A is a subset of X and A is contained in union of Ais, Ais also a subset of X, then we want to show that mu of A is less than or equal to summation mu of Ais. So, this is what is to be shown. Now, let us observe. So, note we want to show one number mu of A is less than or equal to sum of these numbers. If one of these numbers is equal to plus infinity, then obviously this property is true. So, note if mu of A i is plus infinity for sum i, then clearly mu star of A is a number which is less than or equal to plus infinity which is at least one of the mu Ais. So, that is less than or equal to mu of mu. So, everything about star, so it is mu star we are looking. So, let us just write mu star. We are trying to prove that mu star is countable. So, mu star of A i is equal to 1 to infinity. So, what we are saying is this inequality is obvious if one of the terms in this sum is equal to plus infinity. So, let us take the case when all of them are finite. So, let us assume, suppose mu star of each A i is finite for every. Now, what is mu star of A i? mu star of A i is infremum of a certain collection. So, here we are going to use the property of something being infremum and that being finite. So, let epsilon greater than 0 be arbitrary, of course, fixed. You choose arbitrarily and fix it. Then mu star of A i is the infremum over all summations approximate sizes. So, then there exists at least one covering. So, there exists sets say A ij, j equal to 1 to so on in the algebra A such that this A i is contained in this disjoint union of A ij. So, and mu star of A i, which is an infremum, if to this I add the small number epsilon, this becomes bigger than summation mu of A ij, j equal to 1 to infinity. So, this is let me stress here. This is the kind of definition or this is the kind of analysis we will be coming across and we will be doing again and again. So, let us be very clear about this. We have got some number, which is the infremum over some collection and if this infremum is finite, then the infremum plus a small quantity epsilon cannot be the infremum, because that is on the right side of it. So, that cannot be the infremum of that collection. Otherwise, alpha plus if alpha is the infremum, then alpha plus epsilon will be the infremum, which contradicts the definition of the infremum. So, if alpha is the infremum, alpha plus the small number epsilon, any small number epsilon cannot be the infremum. That means what? That means there must be a member of the collection over which you are taking infremum, which so that alpha plus epsilon becomes bigger than that number in the collection over which you are taking infremum. So, that is what we are saying that if because mu star of A i is finite, so given epsilon, the infremum plus epsilon must be bigger than a member of the collection over which you are taking infremum. So, what is the collection that is obtained by taking a covering, a disjoint covering of a disjoint covering, not only disjoint, actually any covering we are taking. So, any covering and such that this is true. So, given epsilon, there exists a covering A ij, j equal to 1 to infinity of A i, say that mu star of A i plus epsilon is bigger than this, and this happens for every i. So, if we add up, so I add these equations over i, so summation over i equal to 1 to infinity, mu star of A i plus sigma alpha over i is bigger than sigma over i equal to 1 to infinity, sigma over j equal to 1 to infinity mu of A ij. That is what we wanted, mu star of A i is bigger than something, we have got that kind of inequality. Now, the problem is this, we are going to add epsilon infinite number of times, so this will tend to become infinity, and we do not want that. So, we go back and refine our estimates. So, given epsilon bigger than 0, this we can do it for any epsilon. So, in particular whenever you are looking at for A i, given epsilon, there should exist a covering, say that we will refine it, we will make it 2 to the power i. So, we will change our epsilon, that is true for every epsilon. So, in particular it should be true for this. So, what we are saying is, given epsilon, there is a covering, say that A i is covered by that collection, and mu star of A i plus epsilon divided by 2 to the power i is bigger than the approximate size, that is summation mu of A ij. This is for every i, now if I add here is epsilon to the power 2 i, so that means we have got, this is now convergent, so that means, so that implies that sigma i equal to 1 to infinity mu star of A i plus epsilon is bigger than this sum. And now note, if i and j both vary, so this is for every i. Now, if I take the union over i's, that will be union over this, so I will get a covering of union A j's, which will be covered by this, and A is inside this. So, what we are claiming is, this is bigger than mu star of A, because A is contained in union over i union over j A ij's and this A ij's belong to C. So, A is covered by this countable union and this is one approximate size for mu of A. So, that is always bigger than or equal to mu star of A, because that is an infimum. So, this quantity implies that this is always bigger than this. So, I can claim that mu star of summation is bigger than this quantity. Now, this epsilon is arbitrary, that was fixed arbitrarily. So, I can let that go to infinity. So, one writes, so now letting, so letting epsilon go to 0, we have sigma mu star of A i i equal to 1 to infinity. This epsilon becomes 0 eventually. Now, I will write bigger than or equal to, because in the limit it can become bigger than or equal to mu star of A. And that shows, so hence mu star is countably sub-additive. So, that we have proved is countably sub-additive. I just want to go through the proof of this once again, because this is an important kind of analysis, we will be doing again and again. Let us just revise the proof once again, that mu star is countably sub-additive. So, to show that mu star is countably sub-additive, we have to show that if A is a subset of x and A is contained in union A i's, A i's contained in x, then I have to show that mu star of A is less than or equal to summation mu star of A i's. Now, to show this, the first observation, which should, should keep in mind that whenever you are trying to show that one number is less than or equal to summation of a collection of numbers, then an obvious case may arise, namely one of the numbers may be equal to plus infinity. So, if mu of A i is equal to plus infinity for some i, then clearly this side is equal to plus infinity and mu star of A is always less than or equal to plus infinity. So, we get mu star of A less than or equal to plus infinity and which is always less than this sum. So, that means that property is true. So, the obvious case is mu star, then mu star of A i is finite for some i. So, what is the other possibility? Other is that mu star of A i is finite for every A i. Now, here is the main construction, part of the construction that we are going to use, namely it is an infremum, which is a real number. So, given epsilon bigger than 0 arbitrary, we can find a covering A ij of the set A i, such that mu star of A i plus this small number and that small number will make it dependent on i, the stage at which you are doing epsilon divided by 2 to the power i, bigger than the approximate sizes over which you are taking the infremum. So, once again the property of infremum being a real number is used here, nothing more than that. So, once that is done, you add both sides. This is for every i, take the summation on both sides. So, summation mu star of A i plus summation of this over i is bigger than summation of mu of A ij. Now, this is a convergent series, its sum is equal to epsilon. So, this is mu star of A i summation plus epsilon and this quantity on the right hand side is an approximate size of A, that is this is bigger than or equal to mu star of A. Mu star of A is the infremum over all such numbers, because A is covered by union over i union over j, A i is covered by A ij's. So, union over A i's will be covered by this union and A is inside it. So, this implies that summation mu star of mu of A ij's summation over i and j is bigger than mu star of A and once that is done, that means that we have got and let go epsilon go to 0. So, you get this quantity. So, that says that mu star is countably sub additive. So, let us so we have proved this property that mu star is countably sub additive. So, mu star now the only thing left to be shown is that that mu star actually is an extension, otherwise all this process will be a waste. So, we want to claim that mu star is indeed an extension of mu. Mu star is not a countable additive, but at least we should check it is an extension and it is countably sub additive that we have already checked. So, we want to check that mu star of A is equal to mu of A if A is in A. So, to check that let us look at the proper definition. So, we had mu star of A is equal to infimum over summation mu of A i, i equal to 1 to infinity where A is contained in union of A i's of and A i's belong to C belong to the algebra A. Now, so if A belongs to the algebra then A is actually equal to A. So, A is contained inside A. So, this is one of the elements here in this covering A itself covers it. So, it will appear in one of it will be one of the elements over which you are going to take the infimum. So, that implies that mu star of A which is the infimum is less than or equal to mu of A. So, that property is by the sheer fact that A is covered by itself and A is in the algebra. So, that is we want to prove other way around inequality to show that mu of A is less than or equal to mu star of A. Now, once again we want to show that one number is less than the other number. So, there is an obvious possibility case 1 that mu star of A is equal to plus infinity. So, in that case this is plus infinity and mu star mu of A is always less than or equal to plus infinity which is equal to mu star of A. So, that is obvious. So, the obvious case is when mu star of A is equal to plus infinity. So, let us look at case 2, mu star of A is finite. So, in that case again we are going to use the definition of infimum. So, mu star of A is the infimum of all possible approximate sizes, summation so on. So, let epsilon greater than 0 be arbitrary. Then there exists a covering. So, there exists sets A j's belonging to the algebra such that A is contained in the disjoint union of A j's and the infimum says that mu star of A plus epsilon cannot be the infimum that has to be bigger than summation mu of A j's. So, there is at least one such covering possible so that this is infinity. This is not necessarily disjoint. Win can make it, we will see it later on. So, this is finite. Now, note A is contained in union of A j's and all of them are elements in the algebra. We assumed A is in the algebra. So, everything is in the algebra. So, and mu is a measure and we showed every measure implies mu is countably sub additive and that implies that mu of A is less than or equal to summation mu of A j's, j equal to 1 to infinity. So, look at this equation 1. Look at this equation 2.    So, let epsilon go to 0 and that implies that mu star of A is bigger than or equal to mu of A. So, that proves the other way round inequality also in the case when mu star of A is less than or equal to mu of A, because A is one of the members which is covering it. So, mu of A is element. So, mu of A is mu star of A is an infinitum. So, that is less than or equal to. That is obvious property and to show that the case when it is finite, mu star of A is finite. We look at once again the definition given epsilon is bigger than 0. There is a covering so that this holds. The infinitum plus epsilon is bigger than one of the elements over which you are taking the covering. Now, using the fact that mu is countable as a derivative, this is bigger than or equal to mu of A and hence that proves the required property. So, what we have shown is that the mu star is indeed an extension of mu of A. So, let us go back and have a look at what we have done is the following. We started with a measure mu on the algebra A. Measure means it is mu of empty set is equal to 0 and mu is countably additive. We are trying to extend it. So, we try to find out the size of any set by looking at sizes of sets in A. So, take any set E, cover it by elements in the algebra A and look at the sizes of mu, call it as mu of A i. So, take the summation. So, this gives an approximate size of the set E. Look at the smallest possible of these numbers, call it the infremum. So, mu star of E, the induced outer measure is defined as the infremum over all these summations and these summations arise from coverings of E. So, this is called the outer measure and we showed it is well defined. We showed it is, it has the obvious property namely mu of empty set is equal to 0, mu star of A is bigger than or equal to 0. It is monotone and so that means mu star of A is less than or equal to mu of B and mu star is countably sub additive. And finally, it is an extension. So, let me point it out that we have taken mu star of A as the infremum over those summations and we have taken the coverings which are countable in number. One can ask the question, cannot we take only finite coverings instead of countable coverings of E? So, let us give an example to show that that is not possible to do that. The finite coverings will not suffice. So, let us look at the set E, the case is the real line. We will look at the set E which is rational's intersection with 0 to 1. So, we are looking at all the rational's in the set in the interval 0, 1. Clearly lambda star of E we expect it to be equal to 0. Why we expect size of this? Because it is a countable set and the length of each singleton is equal to 0. So, we expect the length of each when added together this also should remain small or lambda star of E is equal to 0. It is quite natural. Now, suppose we define. So, this is when lambda star is defined by taking countable coverings. Now, let us take a finite covering of E by interval. So, E is covered by finite number of intervals union E i. We claim that in that case this number the approximate size of E will always be bigger than or equal to 1 because of the following reason. What is E? E is rational's inside 0, 1 and suppose E is covered by union i, j is j equal to 1 to n. So, if possible let sigma lambda of i, j, j equal to 1 to n be less than or equal to 1. Now, these are finite collection of. So, here is 0, here is 1 and i, j's are intervals of course intervals in 0, 1 which are covering. So, i, j's are in 0, 1 which are covering the side E which is rational's in 0, 1. So, now let us say that i, j say for the sake of just definition it is a, j, b, j. It does not matter whether open or closed. You can just assume to be open. It does not matter much actually. Then we have got these numbers between a, j's and b, j's. So, look at all the left end points and look at the smallest of them. Let us say the smallest is here that is a. So, what is a? a is the smallest of the numbers a 1, a 2, a n and b, j. Look at the largest of b, j's and call it as b. Then this a, b is equal to, may be closed does not matter, is equal to union of i, j's or at least it will cover the union of i, j's, j equal to 1 to n and they cover E. Now, so and that covers E. Now, if this is less than or equal to, this is the smallest and that is the largest ones which is covering. Now, this number a, so that means what? That means b minus a is less than or equal to summation length of i, j's, j equal to 1 to n and if that is less than or equal to 1, that means b minus a is strictly less than 1. That means it has to be like this, but then there is a rational here between 0 and 1 which belongs to E and E is inside a, b. So, that is not possible, so that will be a contradiction. So, this situation is not possible that means whenever these are covering, we have to have lambda of i, i is bigger than or equal to 1, but that means all approximate sizes of E is bigger than 1. So, that will imply that lambda star of E is bigger than or equal to 1. That means, but that is not possible because we just now said lambda star of E is equal to 0. So, in the definition of the outer measure, we cannot limit ourselves to only finite coverings. We have to allow all countable coverings also. So, today we have tried to go beyond an algebra. So, we started with a semi-algebra and a measure on it. We extended it to a measure on the algebra generated by it as a first step. As a next step, we started with a measure on an algebra and we showed that by an example on the real line, Ullam's theorem that you cannot extend it to all subsets of real line. So, let us try to go as much as far as possible. So, we define given a measure mu on an algebra, we defined the notion of an outer measure for any subset A of that set X and we showed this outer measure has some nice properties. It extends one given measure, it is monotone, it is countably sub additive. So, in the next lecture, we will see how to get from it an actual extension which is a measure. Thank you.


Hello and welcome back here for another episode of Random Drifting here on Aesthetic Corsa. We're doing a random drift track with a number generator and a random car. So make sure you guys follow me on all social media, all of which are found in the description box below. Let's do it. So we are randomly selecting a number here on this dock to find out the track to run 166. And we are looking at Rick's tandem park. I believe I have this track. I may not have this track. So we're going to go to Rick's tandem track first. We may get a secondary track set up or just run it in Rick's. So we're going to download Rick's tandem track. So now we need to get to random cars. So we have random car numbers nine through 109. So we're going to change this up nine through 109. And we're going to just randomly generate this and see what we come up with 74 ACW Nissan 180SX type four BN Sport type four. Why does that car? Okay, I would say why does that car seem familiar, but we're going to double check it. Make sure we don't already have that car. So we're going to pull up and get our track installed Rick's tandem park. And then we're going to get the ACW RPS S 13. That name sounds super familiar to me. So we're going to go RPS in here. I don't think we have we I don't believe we have this car. We do not. So we will be running the S 13 here in this episode at Rick's tandem park. Once we load in, we'll take a look at the car real quick, and then we will go from there. There it is. So it's just a standard 180SX. So I'm not sure why my Contamander decides to load very slow recently, but it is. So it's a basically a street car.  So we're going to go full send on it. We are looks like we have 469 horsepower. So let's get upset on Rick's tandem track. Rick's tandem park. And that is where we're going to go. All right, so we're going to get the wheel cam and everything turned on. And I believe this car seems like it might just be a world of drift towards car modified with maybe a little bit more power or possibly just some liveries and stuff like that. But we will divulge that once we get into the car. It does look like it has world of drift tour gear sets, world of drift tours tires. So this may just be a liveried out world of drift tours car. Correct me if I'm wrong, but we're going to go full send on it. Definitely going to have to increase that game volume a little bit more. This car is pretty quiet. I'm like maxed out in my game audio maxed out in all that audio. Well, I had the wrong button. Full send. Mike Pizowski chilling right there. I don't know why it went into first instead of third. So this does not feel like a world of drift tours car by any means, but the track is actually really flowing. Third gear hero. All the all the tire noise from this car throughout here. But I've been noticing recently that my shifter is actually missing shifts. So I'm going to have to be taking it apart and adjusting it to make it not miss shifts. I believe I've driven on this track before. That actually seems like a very fun track for street cars. We liked a little bit of elevation changes when it switches into first person here in a second. My fancy view button on my screen deck because trying to find the button on the wheel when driving is definitely not easy. So overall impressions of this car. It's actually, can't tell us quite well, it is very much a street car given the low power and the world of drift tires and everything like that. Actually, I do like this track a lot.  Did you hit that right? That's third though. So for a random car pulling a random track, I do actually quite like this track. And you know what we're going to do? You know what we're going to take the track with a pro style car and see if we can flow this track in a pro car. The pit section. There's another little road that I kind of want to take a look at. Looks like you can probably cut it short. A little wall tap. If you can kind of cut the track short. Okay. So we're going to go pull a pro car and we're going to see what the pro car does and see how the pro car does on this track because I feel like the pro car might be too grippy and too fast for this track. So we switch up to a pro car and then we're going to be right back. So we did bring out my pro party car, my Unito Eurofighter and we're going to see what this car can do here on this track. So let's go full send with this. We may have to go fourth. I said fourth. Not second. Gear box. Yep. Fourth gear seems to be good. Oh man. These, these corners are really tight with this grip of this car right now. I had this car so gripped up that it's, oh yeah. It's not happy with some of these transition changes and stuff like that. But we'll get it. We were just in the street car. So we got to get the flow of this car back. We guys remember this car. This car is definitely very aggressively loud monster. I think I can handle it once the tires get warmed up. So if you guys like this, you guys want to keep seeing these random drift drives. Let me know down in the comment section down below and make sure you follow me on social media. A lot of things coming here to the channel. So I hope you guys enjoyed this episode here on a set of courses and random drifting. So as always I think you guys have come back and watching. I'm Emil Rabbit. See you guys on the track.


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Hello, this video will demonstrate how to record your ID core amplifier via the USB audio and then monitor through your existing sound card using a Mac. Firstly, plug in the ID core via the USB connection. Then open up your favourite digital audio workstation or DAW for short. For this example, I will be using GarageBand. Create a new session and save it. We then need to add a new track. We will be using the real instrument track. In the settings window that will pop up, select the ID core as the instrument connected with. Then select Stereo 1 and 2 and then select to hear the sounds from the built-in output. Also, we need to then tick the box saying I want to hear my instrument as I play and record. Set up your sound that you want to record and then set the level so that the sound never clicks using the volume control on the ID core amp. Lastly, hit record and you're ready to rock.