Welcome to lecture 21 on measure and integration. In the previous lecture, we had started looking at the properties of sequences of integrable functions and we started proving an important theorem called Lebesgue's dominated convergence theorem. Let us continue looking at that and after that we will start looking at the special case of integration on the real line and that will give us the notion of Lebesgue integral. Let us recall what we had started proving namely dominated convergence theorem, which says that if f n is a sequence of measurable functions such that there exists a function g belonging to l 1, so that all the f n's are dominated by this integrable function g almost everywhere x for all n, then and if f n converges to f, then the function the limit function is integrable and integral of f is nothing but the limit of integrals of f n's.   So, to extend this case to almost everywhere we have to do only a minor modification. Let us define the set n to be the set of all x belonging to x where mod f n x is not dominated by g, so g of x or union the set of all those points x belonging to x such that f n x does not converge to the function f of x. So, on n complement we have f n x is less than g x and f n x converges to f of x. So, for every x belonging to n complement and mu of the set n is equal to 0 because we are saying that this mod f n x is less than g x almost everywhere and f n x converges to f of x almost everywhere. So, the set where it does not hold there is a set n and that set as n has got measure 0. Now, let us consider the sequence indicator function of n complement times f n. So, this is the sequence of functions which are dominated by. So, this is the sequence of functions for n bigger than or equal to 1 and they satisfy the property. So, namely this indicator function of n complement f n mod of that is less than g for all x everywhere and the functions converge to the indicator function of n complement f. So, by the our earlier case what we get is the following namely that indicator function of n complement times f is l 1 is integrable and integral of f n limit n going to infinity indicator function of n complement times f n the integral of that converges to the integral of indicator function of n complement times f. So, that is by the earlier case when everything is true for all points. So, that means, so this is same, but note that mu of n is equal to 0. So, that implies that the integral of f over n d mu is equal to 0 and we already know that on n complement f is integrable. So, that together with this fact implies integral of mod integral of mod f d mu is finite implying that f belongs to l 1 and this equation which said that integral of f n over n complement converges to integral of f over n complement and mu of n being 0 together gives us the condition that integral of f n d mu converges to integral f d mu because integral f n d mu is same as integral of f n over n plus integral over n complement and integral over n complement converges to integral over n complement of f and on n both are 0. So, this gives as the required result. So, that is how from almost everywhere conditions are deduced from the fact that something holds everywhere.  So, as I said this is one of the important theorems which helps us to interchange the limit and the integral sign. Let us look at some more minor modifications of this theorem. One thing more we can even deduce that integral of mod f n minus f d mu also converges to 0. So, to deduce that part we just have to observe that mod f n minus f is less than or equal to twice of g and mod f n minus f goes to 0.  So, that is another modification, another consequence of the dominated convergence theorem. Let us prove what I call as the series version of this theorem namely that if f n is a sequence of functions which are integrable and integrals of f n summation 1 to infinity. So, some of all the integrals of f n mod f n are finite. Then the conclusion is that the series f n x converges almost everywhere and if we denote the limit the sum being f of x then that function is integrable and integral f is equal to summation of integral f n. So, essentially this theorem says that if the summations of mod f n are finite then this the series f n x is this almost everywhere is convergent almost everywhere and integral of f is equal to. So, integral of f is equal to integral of summation of integral f n. So, that is again interchange of limit essentially. So, let us see how from dominated convergence theorem we can get this. So, let us define, see to show that this series is convergent almost everywhere we likely show that so we show is absolutely convergent. So, for that let us define say g n of x to be equal to summation mod f k of x k going from 1 to n, the partial sums of the absolute values 1 to n. So, let us observe that this sequence g n. So, note g n is a sequence of non-negative measurable functions and g n are increasing to some function. So, that is they are going to increase to k equal to 1 to infinity mod of f k of x and let us call that as g x. They are increasing to the function g of x. So, that implies by monotone convergence theorem we have integral of g n d mu must converge to integral of g d mu. But integral of g n d mu, so that is the same as saying integral g d mu is equal to limit n going to infinity of integral g n d mu. But what is integral of g n? g n is sum of absolute values of f k 1 to n. So, by linearity property this is nothing but limit of n going to infinity of summation 1 to n k equal to 1 to n of integral mod f k d mu. And this limit is nothing but 1 to infinity and that is given to be finite. So, let us write that, so this is equal to summation k equal to 1 to infinity of integral mod f k d mu, which is given to be finite. So, hence what we get is thus g is integrable, g belongs to g is a integrable function. So, saying that g is integrable implies recall that if a function is integrable and g is a non-negative function, so g of x is finite almost everywhere. That is non-negative function, which is an integrable function must be finite almost everywhere. So, we get that g is finite almost everywhere and what is the function g? g is nothing but the limit of the absolute values of f k x. So, that means proves that the series, so hence sigma k equal to 1 to infinity mod f k of x is finite almost everywhere x. And once a series is absolutely convergent, it is also convergent, so that implies that sigma k equal to 1 to infinity f k x is finite for almost everywhere x. So, let us denote this limit by f of x. So, this is f of x and so as observed earlier, so note. So, f of x, we can write also f of x as the limit n going to infinity of summation k equal to 1 to n f k x. And if these functions are called as something say f n, sorry phi n, then note that mod phi n, if we call this as phi n, then what is mod phi n?  So, this partial sums, which we have called as phi n are all dominated by g and phi n is converged to f, so by dominated convergence theorem, so what we have got is, so we have got, so all the phi n's are less than or equal to g and phi n's converge to f almost everywhere, so that implies by dominated convergence theorem that integral of phi n's d mu must converge to integral of f d mu, but this is nothing but this phi n, this is what we called as phi n that is summation 1 to n, so this is nothing but summation of 1 to n of integral k equal to 1 to n of integral f k d mu must converge to integral f d mu and that is same as saying that integral f d mu is equal to summation k equal to 1 to n of integral f k d mu. So, that proves the theorem namely, if f k is a sequence of functions which are integrable and the sum of the integrals is finite, then the series f n x n 1 to infinity itself is convergent almost everywhere and the limit function is integrable and integral of the limit function is equal to summation of integrals of f n's. So, this will refer to as the series version of dominated convergence theorem. There is another interpretation of the dominated convergence theorem, when the underlying measure space is a finite measure space, then one has that if x s mu is a finite measure space and f n is a sequence of measurable functions, so that all of them are dominated by a single constant m almost everywhere and f n x converges to f of x, then integral f n's converge to integral f. So, this is a particular case of dominated convergence theorem, when the underlying measure space is a finite measure space and the only thing to observe here is that because, let us see how does this follow from our is a dominated convergence theorem. So, we are given that mod f n x is less than or equal to m for almost everywhere x. Now, look at the constant function m, so look at the function g of x, which is equal to m for every x belonging to x. So, the constant function is measurable, so note that g is a non-negative measurable function, because it is constant function, note that so its integral g d mu is equal to the constant function m d mu and that is equal to m times the measure of the whole space x, which is finite. So, what we are saying is on finite measure space is a constant function is always integrable, so implies g is l 1, so f n x bounded by m, so that constant function and that is a integrable function. Once we have that and f n x converges to f of x almost everywhere, so now by dominated convergence theorem is applicable and that implies integral f d mu is equal to integral f n d mu limit n going to infinity. So, the main thing is on finite measure space is a constant function becomes integrable because of this reason. So, this is what is called bounded convergence theorem and it is quite useful when underlying measure space is a finite measure space. So, let us look at what we have proved till now, we have looked at the space of integrable functions and proved linearity property and an important theorem called dominated convergence theorem. So, if you recall for non-negative measureable functions we had two theorems one was monotone convergence theorem namely that was a theorem when f n is a sequence of non-negative measureable functions increasing to a function f then integral of f is equal to limit of integral. So, that means interchange of limit and integration is possible by monotone convergence theorem whenever the sequence f n is monotonically increasing and a sequence of non-negative measureable functions. The second theorem which involved sequences of measureable functions was again for non-negative measureable functions and that was called Fatou's lemma. So, there we do not emphasize we do not require that the sequence f n be non-negative and measurable. We only want the sequence f n to be a sequence of measureable non-negative measureable functions they need not be increasing. So, for the such a sequence we had that integral of the limit inferior of the sequence f n is less than or equal to limit inferior of the integrals of f n. So, that was Fatou's lemma and now we have the third theorem dominated convergence theorem which again helps you to interchange the notion of the notion of integral and the limiting operation under the condition that all the f n's are dominated by a single integrable function. So, these are the three important theorems which help us to interchange limit and the integral signs. Let us at this stage emphasize one more point about this technique of integration and so basically for integral we started with simple functions and then we go to non-negative functions and then we defined it for integrable functions. So, this process of step by step defining the integral can be is useful in proving many results and I call it as the simple function technique. So, this is a technique which is used very often to prove some results about integrable functions and non-negative measurable functions. So, what is the technique? Let me outline that and then I will give a illustration of this. Suppose you want to show that a certain property say let us all that property a star holds for all integrable functions. So, to prove that the property holds for all integrable function the technique is as follows basically that show that this property star holds for all non-negative simple measurable functions. So, if you want to show a property holds for all integrable functions first show that it holds for the class of non-negative simple measurable functions. And next show that star holds for non-negative measurable integrable functions by using the fact that non-negative measurable functions are limits of increasing limit of simple measurable functions. So, and use there one uses normally the monotone convergence theorem. So, using monotone convergence theorem one extends the property star from simple measurable functions to non-negative measurable functions or non-negative integrable functions. And then keeping in mind that for a function f it can be split into positive part and negative part. So, f can be written as f plus minus f minus and if a property holds for non-negative functions so about integrable. So, f for f plus that will hold for f minus that will hold and then conclude from there that it holds for f also. So, this is what I call as the simple function technique to prove results about integrable functions. To give an illustration of this let us look at the following result. Let us take a measurable space x s mu which is sigma finite measure space and let us look at a function f which is integrable on this measure space and is non-negative. So, we have got a sigma finite measure space and f is a non-negative integrable function on this measure space. Let us define nu of E for every set in the sigma algebra. Let us define nu of E to be integral of f d mu over the set E integral of f over the set E is denoted by nu of E for every set E in the sigma algebra S. Then we have already shown that this nu the set function nu is in fact a finite measure on S. So, this we have already proved but what we want to prove now is that further if g is any integrable function on the measure space x as nu this nu is the nu measure. So, if g is integrable on x with respect to nu then the product function f into g is integrable with respect to mu and this relation holds integral f d mu. So, integral of f with respect to nu is equal to integral of f into g with respect to mu. So, what we have done is by using a fixing a function f which is non-negative we have defined a nu measure on the measurable space by nu of E to be integral of E f d mu and we are saying if we want to integrate a function with respect to a function g with respect to this nu measure then it is same as integrating the function the product function f g with respect to the old measure mu. So, let us see how the simple function technique is used to prove this result. So, let us start. So, we want to show that for every g belonging to l 1 of x s nu integral of g d nu can be represented as integral f g d mu. So, where recall we defined nu of E to be equal to integral of f d mu over E. So, this is the property we want to prove. So, this is the property star we want to prove for every function g. So, as we said let us first check this property. So, step one let us say g is a l 1 function let us say g is a function which is indicator function of E. Let us take g as the indicator function of E, E belonging to s. In that case the integral g d mu the left hand side is nothing but nu of E because g is the indicator function. So, this integral of nu of E which by definition is equal to integral chi E of f d mu and so chi E is g. So, this is equal to integral g f d mu. So, what does it says? It says that the required property star holds when g is the indicator function. Now, let us take a non-negative simple function. So, that is step two let us take g is sigma A i chi E of E i i equal to some 1 to n where E i belong to s and our claim is that this property holds for this g also. So, we are saying that next step is to verify that the required property. So, integral of g d nu by definition is equal to integral of sigma A i indicator function of E i d nu and what is that? By generative property of the integral it is sigma i equal to 1 to n of A i that is the scalar times nu of E i because integral of the indicator function is the measure. So, that is equal to sigma i equal to 1 to n A i and nu of A i by definition is integral chi E i of f d mu that is the definition of nu of E i is this. So, which I can write it again as sigma i equal to 1 to n of A i we can take it out. So, let us so this is and again by the linearity property.  So, this is integral of g f d mu. So, what we are saying that if g is summation A i chi E i then using linearity property this is same as integral goes in. So, that is A i integral of the indicator function of E i that is nu of E i and nu of E i by definition is integral over E i of f d mu and again using linearity property of the integral I can shift it outside.  So, it says so the required property holds. So, star holds for non-negative simple functions g. So, that is what I said the simple function technique and now let us try to prove it that this property also holds when g is a non-negative measurable function. So, let us look at now g. So, let g on x to be measurable. Then we know by the property of measurable functions implies there exists a sequence S n of non-negative there is sequence S n of non-negative simple measurable functions such that S n increases to the function g. So, then by Lebesgue, by Morton Convergence theorem integral of g d nu with respect to nu must be equal to limit n going to infinity integral S n d nu. But for non-negative simple functions just now we proved this the star holds that means this can be written as limit. So, by step 2 I can write this as integral of S n times f d mu. So, with the integration of a non-negative simple measurable function with respect to nu can be converted into the non-negative simple measurable function multiplied by f d mu.  So, once again Morton Convergence theorem is applicable and this limit is nothing but integral of g f d mu that is so once again we have used. So, this step this step was by our step 2 that the property holds for non-negative simple measurable functions integral with respect to nu is integral with respect to mu of the product function. And now once again we are applying Morton Convergence theorem. So, first integral of g is equal to limit of integral S n d nu by Morton Convergence theorem. And now by our earlier step this is equal to integral of S n f d mu again by Morton Convergence theorem it goes back. So, that means so this implies that star holds for non-negative measurable functions. And now let us come to the last part namely so let us so final step 3 is let g belong to L 1 X S nu g be a integrable function.  So, by step 2 we know that integral g plus d mu is equal to integral of g plus sorry d nu.  So, that is by step 2 and now because g is L 1 so that implies g is equal to g plus g minus. So, g plus is in L 1 of nu and g minus also belongs to L 1 of nu a function g is integrable if and only if it is positive part and negative parts are integrable. So, that means these quantities they are all finite these are all finite quantities.   So, that means implies that g f is L 1 and by these two this is same as integral of g d nu. So, for a so this the step 3 for a g which is integrable we have deduced that this property is true. So, this is step 3 so this is what I call the simple function technique. So, let me go back and show you once again what we have done. So, we wanted to show that so this is property star we wanted to show for every function g which is L 1. So, first step is so this is my step 1 that look it at the functions g which are indicator functions. So, I want to verify this for the indicator function g to be the indicator function. So, when g is the indicator function so this left hand side is integral of over e of the constant function 1. So, this is equal to integral d nu is nu of e which by definition is integral f over e so which I can write as integral f e so that is true. So, step 1 is to verify the required thing holds for characteristic function and step 2 by using the property that the integral is linear we show it is true for every nonnegative simple function. So, take g a nonnegative simple measurable function and apply so g is equal to integral of nonnegative A i indicator function A i and interchange and show it required property holds. So, the step 2 was that the required property holds for nonnegative simple for nonnegative simple measurable functions and then using an application of monotone convergence theorem. So, that is step 3 that if g is a nonnegative measurable function then we know it is a limit of nonnegative simple measurable functions increasing limit. So, an application of monotone convergence theorem together with the earlier step gives us that integral of g d nu is equal to integral of g f d nu. So, that is the next step to show that it holds for nonnegative measurable functions and once that is done the final step that it holds for all integrable functions is via splitting the function g into the positive part minus the negative part and g integrable means both are integrable and for each one of them the required claim star holds. So, by putting them together we get that the required claim holds property star holds for all functions g so which are l 1. So, this is what I normally call as the simple function technique. So, while proving results about integrable functions one uses quite often the simple function technique and while proving some properties about subsets of sets recall we had the sigma algebra monotone class theorem technique. So, for proving properties about sets one uses monotone class sigma algebra monotone class technique and for proving the results about integrals one normally uses what is called the simple function technique. So, with this we have defined and proved general properties about integral of functions on sigma finite measure spaces. Now, we will try we will specialize this property this construction when x is real line. So, we want to specialize this thing for the real line. So, let us see what we get. So, while you will be looking at the special case when x is real line the sigma algebra is l that of Lebesgue measurable sets and the measure mu will be the lambda the Lebesgue measure. So, we will be working with the measure space x s mu which is same as real line Lebesgue measurable sets and Lebesgue measure. So, the space of all integrable functions on this measure space R, L and lambda is called the space of all Lebesgue integrable functions and is also denoted by L 1 of R or L 1 of lambda. So, this is the space of all Lebesgue integrable functions. So, we want to study this space of Lebesgue integrable functions in some more detail. Let us first agree to call integral f d lambda to be the Lebesgue integral of the function f. So, whenever f is integrable or non-negative integral f d lambda will be called the Lebesgue integral of f. Sometimes we have to look at functions which are defined on subsets of E. So, for any subset E which is Lebesgue measurable L 1 of E will denote the measure space of all integrable functions on the measure space E. So, the underlying set is E, L intersection E is the collection of all Lebesgue measurable sets inside E and lambda is the Lebesgue measure restricted to subsets of L intersection, the sigma algebra L intersection E. Of particular interest for the time being is going to be the set when E is a close bounded interval A B. So, we will start looking at the space L 1 of A B that is the space of all Lebesgue integrable functions defined on the interval close bounded interval A B. We also have the space R A B namely the space of all Riemann integrable functions on A B. So, we want to compare these two spaces on one hand we have got the space of Lebesgue integrable functions on A B, on the other hand we have got the space of Riemann integrable functions on A B and we want to see the relation or establish a relationship between the two. That was one of the starting points for our discussion of the subject namely the space of Riemann integrable functions had some difficulties, some problems, some drawbacks and for which we want you to extend the notion to a larger class and this is the larger class L 1 of A B. So, what we are going to show is R A B the space of all Riemann integrable functions is a subset of L 1 of A B and the notion of Riemann integral is same as the notion of Lebesgue integrable for Riemann integrable functions. So, that is called the relation between the Riemann integral and the Lebesgue integral. So, to be more specific we want to prove the following theorem namely if f is defined on a close bounded interval A B is Riemann integrable function then f is also Lebesgue integrable and the Lebesgue integral is same as the Riemann integral of the function f. So, this is what we wanted to want to prove. So, let us start looking at how do we prove this. So, the proof of the theorem so we are given that the function f belongs to R A B it is a Riemann integrable function. That means so let us recall how is the Riemann integral of a function defined it is defined via limits of upper sums and lower sums of partitions. So, implies there exists a sequence P n of refinement partitions with norm of P n going to 0 as n goes to infinity partitions n going to infinity with the upper sums of P n with respect to f limit of that is same as integral the Riemann integral of f is same as the limit of the lower sums L P n of f. So, that is the meaning of saying that a function f is Riemann integrable. So, we can find Riemann integrable implies there exists a sequence of partitions P n which are refinement partitions refinement means P n plus 1 is obtained from P n by adding one more point and norm of this partitions the maximum length of the sub intervals goes to 0 and integrable means that the upper sums and the lower sums both converge to the same value and that is the Riemann integral of the function f. So, this is the property of saying that f is Riemann integrable.   So, let us say P n looks like A. So, interval is A to B. So, A the point x naught less than x 1 less than x n which is equal to B. So, let us say that is the partition P n. So, in the picture it will look like here is A here is B. So, this is x 0, this is x n and here is x 1, x 2 and so on. So, to construct the upper sums what one does? To construct the upper sums one looks at the maximum value of the function in this interval and the minimum values in this interval. So, let us write m k to be the maximum value of the function in the interval x, say x i minus k. So, let us say x k minus 1 to x k. I am just trying to make the intervals disjoint maximum in this interval. Of maximum in this interval, maximum of f of x, maximum in of f of x. Similarly, m k let us write it is the minimum in the interval x k minus 1 to x k of f of x. Only at the end points you have to make it close, but that is not going to matter much. So, that then we define what is U P n f that is essentially looks like summation of the maximum value into the indicator function of that sub interval and the lower sum with respect to P n f looks like summation small m k the minimum value of the function in that sub interval x k minus 1 and x k. So, let us do one thing. This is upper sums and these are not. So, let us let me so I am sorry this is not the upper sum. Let us call this when in the interval x k minus 1 to x k the value is capital m k. Let us call that as the function phi k and when you are taking the minimum value in that interval and summing up let us call that as psi k. So, these are functions because they are linear combinations of indicator functions and the upper sums and lower sums are nothing but the upper sum P n f is nothing but the Riemann integral a to b of phi k x dx and the lower sum P n f is equal to the integral of Riemann integral of this function psi k of x dx that these functions phi k and psi k which are linear combinations of indicator functions are in fact, non-negative measurable functions on the measure space AB the interval AB. So, note so that is the observation that we should note and then let us note down. So, note phi k and psi k are measurable functions are measurable non-negative sorry are simple measurable functions say that phi k is less than or equal to at every point x is less than or equal to f of x is less than or equal to psi k of x and as far as the integral is concerned the integral a to b of f x dx is between the upper sum and the lower sum. So, that is the phi k was maximum sorry this should be bigger than or equal to like this because phi k is taken as the supremum. So, this is so this is upper sum P k of f and that is bigger than or equal to the upper sum sorry bigger than or equal to the lower sum with respect to P k of f and in the limit both of them are converging. So, here is the second observation is that the upper sum with respect to the partition of f is same as so what was it? So, that was equal to sigma m k into the length of the interval x k minus x k minus 1. So, that is the upper sum that is also the Riemann integral. In fact, this is also equal to so length. So, this is the length of so you can write this as the length so m k times the length of x k minus 1 and x k which is same as the Lebesgue integral of the function phi k d lambda. So, this is this is the important observation that we should keep in mind that the building blocks for Riemann integral which are these step functions are also Lebesgue integrable and the Riemann integral of the step functions phi k and psi k are same as the Lebesgue integrals of phi k and psi k. So, similarly the lower sum p k f is equal to integral of psi k d lambda and now essentially the idea is to put them together. So, because phi k and psi k they are between these two. So, let us look at integral of so look at the sequence so consider the sequence psi k minus phi k minus psi k. Recall phi k is bigger than f x is less than psi k. So, phi k minus psi k is non-negative for every k and saying that the upper sums and lower sums converge to the same value is saying that the integral of phi k minus psi k phi k minus psi k d lambda that goes to 0. So, that goes to 0. So, because the phi k d lambda is the upper sum this is the lower sum and that goes to 0. So, that implies that limit so that means this implies that the limiting function f is trapped in between so that means limit phi k x is equal to limit psi k x almost everywhere. Why is that? So, that we can reduce from the fact that applying Fatou's lemma. So, to reduce this look at the limit inferior of phi k minus psi k integral d lambda will less than or equal to limit inferior of integral phi k minus psi k and that is 0. So, this is 0. So, that means and this is and so this says that integral of a non-negative function is 0. So, the function must be 0 almost everywhere and that is same as saying this must be 0 almost everywhere and f is trapped in between. So, that implies that limit phi k x limit of phi k x is equal to f of x is equal to limit psi k x for almost everywhere x. So, that proves is equal to so we are falling short of time. So, that means that the function f is measurable. So, we will continue the proof of this tomorrow in the next lecture. So, our aim is to prove that the space of Riemann integrable functions is inside the space of Lebesgue integrable functions and the Riemann integral is same as the Lebesgue integral. So, we will continue the proof in the next lecture. Thank you.


one Shaber 1000 here. Today we're gonna open up this package. I think it's my fish scale and I'm sorry about the noise behind me but I got the air conditioner on because it's very hot out. So let's get this opened up. Okay guys this came in really early it wasn't supposed to be here until like Saturday so I'm gonna go ahead and get this opened up here and we're gonna try it out and see how well it works yep that's what it is pretty cool it's supposed to do 110 pounds I do believe so it does let me see okay it's got a blue or yeah a blue back like function tear function auto power off data lock function overload indication low battery indication it's powered by three three volts it's a two triple a batteries I got some if it didn't come with any high precision strain gauge sensor system capacity 10 kilograms five grams 20 kilograms five grams 40 kilograms 10 grams okay whatever that means but I do know it will do cool does come with the batteries this was cheap this was like eight bucks seven something eight bucks it will do get these instructions open here Wow it's got a temperature test I'll have to read all this stuff up here and but it does have it does what is it kilograms ounces and pounds so well it feels feels pretty nice feels really nice there we'll get a screenshot of this okay have false out and let's let's put the batteries in it now the real test of course will be when we go out fishing with it which will be soon now unfortunately this will not help us catch fish but it will however help us measure the fish that we do catch so all right batteries are in that's on right unit there's kilograms there's pounds the fuck was that something just made a real weird noise over here in the bush that was weird there's ounces I do like the back light so I can see at night night fishing so and I'm not sure what that is like I said I have to go over so there's pounds we'll put it on pounds that see I don't know if it remembers it or not but okay but what we're going to do is I've got a gallon of water over here and we're gonna weigh it I've got some I look this stuff up so we can get close of course you know cold like frozen water is going to weigh a little more than room temperature water and like in the US a gallon I got this on my phone here a gallon is 128 fluid ounces or 3.785 liters in the UK a gallon is 160 fluid ounces or 4.546 liters okay so I mean everything there's going to be some variables but US gallon of water weighs around 8.33 pounds or 3.78 kilograms at room temperature so now also like this you know I you know I don't know where a gallon where they're measuring the gallon from this is one gallon this is one gallon or 3.785 liters now where do they fill it to I got this filled up with water okay where do they fill it to so we're going to call it around 8 pounds okay and then we'll check it and see what what the gauge says like in kilograms or whatnot so let me bring you up here a little bit okay so we're going to get our little tester here let's open this up we'll hook it around these tie straps and let's see what it says I don't know if you guys can see that this says 9.2 pounds ish so I might have like filled it up more than what they usually do so now we'll hit the tear button and we'll go to go to kilograms and see what it weighs four kilograms ish 4.1 cool so it should be pretty close what's the ounces 146 ounces so it's more than what I'm supposed to be lifting anyway and you know you got your on and off button now like I said if you really really want to get precise on this thing you know where where do you measure where you know I mean because it's going away different different temperatures it's going away different so I mean just hang this up and you fill this up until it's you know about what it should be or what but for way and fish it's going to be okay I mean I don't see an issue you know what I mean yeah I wish it would have a memory to where it would stop you know as soon as you took the pressure off I don't know if I can get it to do that let me read it and see okay guys so it does have a weight lock on it you hold the tear button for two seconds there it's off the L stands for lock now let's turn it on hold it for two seconds now it's on also this will give me the temperature but it only gives it to me in Celsius I'm in the United States so but let's see what the temperature is we hold the unit button for two seconds is 23 degrees Celsius okay so now that we got the lock on let's weigh this water again lock should appear up in the corner and it did and now it stayed whoops I should have stayed it did so it's 9.14 9 pounds would be 14 ounces so there it is now it's supposed to have a buzzer I was reading this says something about a buzzer and then the way something else you just hit your power button it clears it it hold the tear button for two seconds and it'll turn the weight lock on or off and then the unit button for two seconds and it will it will tell me the temperature there's really not much more about it but it does say something about the power on put the scale in a vertical position without weight in the hook press on off key the screen will display shoot eight eight eight eight and the buzzer will utter the sound of hint which it didn't do that shut this off so it says it's supposed to supposed to have a buzzer I don't hear buzzer so but as long as it measures okay I'm good with that I shouldn't be I mean I know it's only eight bucks but damn you know come on it was supposed to do it I wanted to do it but so yeah there's not a whole lot about it and it says wait two to three seconds until the you know zeros come up displaying the screen displaying on the screen scale will be ready for weighing and then it tells you about the tear zero put the container on the hook when weighing bulk goods press tear key and the screen will display 0.00 or 0.000 the scale is ready for weighing put the goods to the container the screen will display the weight of the goods only after finishing okay and then again it tells me about the weight lock I just showed you guys that hook in the goods after the scale is ready max weight shall not exceed full load plus 90 so the screen will display an O dash LD which is overload and buzzer will warn when it's overloading but I don't have a buzzer so I mean it's not buzzing I don't I don't know why but the scale will hold the weight of the goods automatically the buzzer will utter the sound of hint and display hold or the lock symbol which we got in the L on mode so that's that's it temperature test press unit key for two seconds the temperature will appear on the screen the tolerance of temperature is plus or minus 2 degrees so Celsius so it's it's only in Celsius that's pretty much it let me see what would we say it was 20 let's check that again 20 what was it 22 degrees Celsius okay let me see what that is for sure well alright 23 degrees okay I'm gonna check it okay guys I can tell you right now the temperature thing is way off because if you can see the temperature 88 degrees Fahrenheit but it's saying 23 degrees Celsius is seventy three point four degrees in Fahrenheit it's a lot hotter than that out here so I don't know you know I'm not gonna be using that using that at all because that's just way off I use my phone one I can but yeah so there you go let's do our final thoughts and like I said there's going to be more you know uses of this and we'll update it as we're using it out in the field okay guys so there you go this just one of the cheapest digital ones I could find it just says electronic portable scale I don't know what company it is let's see here we got another little booklet here thank you for your purchase thank you your purchase made our day we hope this package makes yours well the buzzer don't work so I gotta be looking at it I mean you're supposed to be able to hold it and then the buzzer will sound when it locks on but it doesn't so I guess I don't know there's a couple different there's like four there's four different they're from Pennsylvania there's four different websites on here but I like I said it doesn't really give me a manufacturer name or anything like that on it so I don't know I can't tell you that I just you know just looked it up on the internet and found the cheapest one so you know eight bucks I wish it worked the way it's supposed to I wish the temperature work you know it says you got to hold it vertical I'm gonna turn it on and there should be a buzzer sound and it didn't and then like I said you hold the unit button for two seconds and it gives you the temperature and that temperature is way off it's like 10 degrees off so this should be reading what at least 42 43 Celsius right now 70 no no I'm wrong about that it should be reading higher than that because it's 88 degrees out here this thing saying it's 75 something so it should be reading higher than that I don't know maybe at least 10 yeah it should be reading 10 degrees higher so it should be reading 33 degrees Celsius in there whatever but it's supposed to shut off on its own after a couple minutes if you don't use it so it'll be alright just to measure the fish or to weigh the fish we catch you know by law you don't go by weight you go by the length of the fish but we just like to know how much they weigh instead of just guessing you know that's about three pound two pound whatever monkey license no we should catch him big catfish how much they weigh so now I'll be able to tell her with some degree of accuracy it seemed to work all right it was pretty close on that jug of water like I said I don't know when this was full of thinner I think it was filled about right here so I got a little bit more water in it so that would that would you know cause a little discrepancy so we're talking what about eight pounds is what that weighs so it's pretty close that was reading nine pounds like I said I could have put them you know it doesn't take much make a pound of water so anyway guys thanks for watching stay tuned for more videos of this thing when we catch fish more fish to weigh it so that being said thanks for watching and she bears a myth man legend gone for now guys and we'll see you in the next one y'all take care be safe bye bye now


John Harbaugh started the press conference today being straight up, straightforward, no beating around the bush. He said Lamar Jackson and Gus Edwards both tested positive for COVID-19. And this comes after they said Lamar Jackson tested negative and negative and negative and negative and negative and negative and negative and negative and negative and negative and negative and negative. Then he got one positive. So now he's on the list and he will be out for the next 10 days of training camp. It is unfortunate because he's missing time away from the guys, he's missing the team camaraderie, he's missing the vibe, he's missing the chemistry, both him and Gus Edwards. But it is, even though it's a very unfortunate situation, it's extremely unfortunate, this is the best case scenario that it could have been for anybody on the team getting it. This is the best case scenario. The fact that it's happening right now at the very, very, very beginning of training camp, this is the best case for this scenario because if it happened later and it just, we don't, and then of course, if it happened in a regular season, we just don't want to even get into all that. So hopefully Lamar Jackson and Gus Edwards are both straight, hopefully they both good. But one thing too, hopefully both of them don't have any long-term effects because we have heard of that with enough people, with enough regular people and with enough athletes as well, the long-term effects because they have a very high motor, high energy job and COVID can really drain your energy. So hopefully those guys can really be full go healthy and no lingering long-term effects. So shout out to Lamar, shout out to Gus Edwards, y'all boys, rest up. Now, football is a very, very, it's a nasty business. And regardless, if you are ready or not, as a football player, the show must go on as they always say. And that's exactly what happened today at training camp. The show, it went on. They said that Trace McSawley, they said he had a pretty good day. They said the offense had a pretty good day overall and the offense would probably be the winners today. They did make some nice comments about Brandon Stevens. Said he made some nice plays, but they said the offense were pretty much the winners. Now, somebody from offense, and of course, I think they said it only happened two or three times. I mean, that is a lot for one practice, but again, it's better that it happens now in training camp and practicing all that than in the regular season. So please get it out your system, Bradley. Get it out your system, the bad snaps because they said it was three of them today. They said two of them were over Trace McSawley's head and then one of them was really high, but Trace McSawley reached up and caught it. So it's, again, better that that happened now rather than later. So again, get all the bad stuff out. Get it all out. And Trace is a little shorter than Lamar Jackson, so maybe that's why, but... It'll be fine though, man. I hope. All right, Hollywood Brown. Hollywood Brown was apparently very active today, catching a lot of deep passes. I think they said he caught two touchdowns. So that is a beautiful thing. They said James Proshay caught a deep pass. They said he got behind Jimmy Smith. I said, okay, cool. Nice, go ahead, Proshay. Do your thing, baby. They said Miles Boykin, he ran some nice routes, but he had a drop. They said that it wasn't the best pass though. So, hey, we'll see what happens with that. So J.K. Dobbins continued to do his thing. I mean, what else did we expect? They said Devin Duvenay, he looked good. I did not hear anything about Sammy Watkins or Rashard Bateman. I know Rashard Bateman was there for sure too, and I didn't hear anything about Sammy Watkins being absent, but I just, I didn't hear anything about those two. So again, not concerned. And remember, y'all, whether something good is something good for the offense or something good for the defense or something good for special teams. There ain't no overreacting with that because we know them, yeah, y'all got it. But if you hear something good for either offense or defense during training camp, it's great. It's wonderful. It's amazing, but you don't put too much stock into it. Now, it's great. We want these guys to be looking good at training camp because this is where it starts. But the real stuff happens on the field. It starts here, what the real stuff happens on the field. And that leads me to Mr. Hollywood Brown who spoke to the media today very briefly. All the, the press, the presser today was only like 22, 21, 22 minutes. And some of that had wait time in there as we waited for the next person to speak, but it was a pretty short presser. Not really much out of it, but Hollywood just say something that had some significance. He spoke about Greg Roman's offense because he was asked, are there many changes? Are there many new things that have been added to the offense this year? And Hollywood said, well, there's some stuff here and there, but as far as he said, we're doing a lot of the same stuff. But I really did appreciate how Hollywood, he took ownership because we of course know the comments from last year about the soldiers and whatnot. And we know who they were directed to and whatnot, but Hollywood took ownership and he talked about, hey, we need to get it done in practice first. And the more times we get it done in practice, that repetition, more times we get it done in practice, the more confidence that Greg Roman will have in us to do the same thing out there and call it a place when it's game time. And that's true. That's true. That's with anything. Once you do something over and over again, you have that repetition with it. So you get more and more comfortable with it. So when it's crunch time, when you have to do that thing for real, when it's not practice time anymore, things can flow that much better. And that's very, very important. That's super important. So shout out to Hollywood. Again, glad he had a great day. And like John Harbaugh has been saying for years, he's been saying for years. It's all about stacking, stacking great practices. That's what you gotta do. And then of course, it'll translate to the, well, you hope it'll translate to the field. You wanted to translate to the field for everybody. Marlon Humphrey. Marlon Humphrey and Chuck Clark, they also spoke to the media. They needed one more guy to fill in for Lamar since he wasn't speaking to the media today, which is all good though. But Marlon Humphrey, he was already scheduled to speak. It's supposed to be Hollywood, Humphrey and Lamar. But yeah, we know. But when Marlon Humphrey was speaking, he just talked about getting back to it. He talked about his family. He gave a shout out to his mom, his dad, his brothers and sisters, his entire family.   They will let you know. And for me, I appreciate that. I loved that. I know there's some people who will be like, no, I don't want to, they wouldn't want to tell whether their kid had a bad game or something or their friend or something. Whether it's a bad game or they're doing something in not the best way. Some people don't like letting other people know, hey, you're doing a bad job. You did a bad job with that. That's not good. I love that. I appreciate that. I'm very straightforward with Carter. Not in a mean way, but I'll let him know. Carter, come on now, man. That wasn't good. It wasn't good because in my opinion, I think that builds character and that lets them know everything ain't gonna be peaches and cream in life.  Not that life is gonna be so bad, anything like that, but you got to be able to take that too. You gotta have a good balance. And of course, when he does good, I surely let him know that too. So he gets a good balance of both. And it sounded like that's what Marlon Humphrey's parents and his family did with him. Because he said that his mom was like, he said there was always somebody in track that was faster than him. It was always somebody that was faster than him that was better than him.  You are going to outwork them and look where he is now. I think it pretty much worked out for him. But anyway, so shout out to Marlon Humphrey. Now something that him and Chuck Clark talked about. Shout out to Chuck Clark with the new cut. That's how I used to cut my hair too before I really knew how to cut it that much. But anyway, but Chuck Clark, he went bald. He cut the billow. He said he keeping it clean. So I appreciate it, Chuck. Shout out to Chuck Clark. Anyway, something that both him and Marlon Humphrey said that really stuck out to me and something that we really got to appreciate and hope that it continues is Tayvon Young and him being healthy. Marlon Humphrey talked about what Tayvon Young and Chuck Clark said the same thing with Tay-Tay being back, that this allows the defense to do multiple things. This allows them to do a lot more. Marlon Humphrey said he gets to go back to the outside and Tay-Tay of course will be that slot corner and Jimmy's egg, Marcus Peters, and it's just a beautiful thing. Health is a beautiful thing. And like we talked about in a previous video, that health, health in so many different positions for the Ravens, that's, I think that's got to be our biggest concern as Ravens fans is health. That's it, the health with the offensive line. Ronnie Stanley, we hope he's back healthy. With the tight ends, Nick Boyer, we hope he's back healthy. With the wide receivers, we hope Sammy Watkins can stay healthy. With the cornerbacks, we hope Jimmy Smith and Tayvon Young can stay healthy. I'll be, that's the biggest concerns just health across the board. So hopefully these boys can just be able to stay healthy because it is a long season and the season just got long with them adding another game. Kaleas Campbell talked about and he was looking ahead which I appreciate and I like and that's planning ahead. Hopefully the Ravens can find themselves in a position to do that. We'll see. But Kaleas Campbell was thinking ahead when he talked about with the longer season, they asked him, oh, what are you doing to stay in shape and prepare for the longer season? He said, oh, all we can do is just embrace it. But hopefully we're in a, well, well, hopefully we're in a position because his voice is deeper than that. I can't even talk as deep as his voice is. But he said that hopefully the Ravens are in a position to where they can have the number one seed so they can have the bye week. So that can sort of give them another break as they head into the post season. So I like that thinking. Hopefully they get it if they don't. Okay, as long as they make it in, that's the biggest thing. So yeah, anyway, that's what we're at right now with the Ravens. Training camp continues. It was really nice to see fans out there. It was great to just hear that and then to see the videos of the fans and stuff. So shout out to all of y'all that'll be attending the training camp. Shout out to all of y'all that'll be attending the open practice and just getting to see the guys live. So unfortunately, won't get to see Lamar Jackson there, which it sucks. It sucks because that's a quarterback. That's the team starting quarterback. But still go out there, have a good time and make the most of it. I love y'all team, keep it clean. I appreciate y'all and we out.


My name is First Lieutenant Isabella Prad. I'm with the Trelly Company 315, I'm in 3rd platoon, I'm 3rd platoon split-tune leader. I'm almost 19 alpha. So we've been out here what I think is about day 17, so in July we went out to Gunnery, which was an individual tank with their individual crews go out and get qualified to fire as a crew. Then at the beginning of October we did platoon operations, that's all four of our tanks, maneuvering, working together, going against another platoon of tanks, getting qualified as tanks. And now we're finishing up our field exercise with a company live fire, so now each step is kind of like bringing up another notch, right? So now we're getting trained as a company to maneuver, engage targets together. Back in July when we shot that Gunnery, she ended up shooting the best officer in the company. So having a lieutenant that trusts me about my job, it just makes it so easy to work. She's super caring about her soldiers. I can't really say nothing bad about her because she is one of the best defense I have. I have excellent NCOs, I have excellent soldiers that teach me every single day, but then we are able to bounce ideas off each other. I think that's why my platoon is so successful because me and my platoon, we have an awesome line of communication with each other. I'm trying to improve them, they're trying to improve me, I think we just work really well together. So I think I got really lucky joining Force Comm when I did, because we were becoming modernized and we were getting the new generation of Abrams tanks. With these ones, there's a lot of new qualities that the other ones didn't. So for one, I know our new paint, it's dark green now, it's thermal reducing. I think that is so cool, I think that will give us such an edge on the battlefield because we have tanks, we have huge engines that are huge, produce a huge heat signature, so it's used to reduce that, so if we're ever doing night operations, it's harder for the enemy to engage us or identify us.