Welcome to lecture 5 on measure and integration. If you recall in the previous lectures, we have been looking at the various classes of subsets of a set x with various properties. We looked at what is an algebra, what is a sigma algebra and a monotone class. Today we will start looking at functions defined on classes of subsets of a set x. We will look at first what is called set functions and then we will look at a very important example of a step function namely the length function. So, let us start defining with what are called set functions. Let us start with c, a class of subsets of a set x. Any function mu, so this is a Greek symbol called mu, so a function mu defined on the class of subsets c of a set x and taking non-negative extended real value functions. So, this interval 0 plus infinity both included denotes the set of all non-negative extended real numbers. So, a function mu defined on this collection c of subsets of a set x taking values in non-negative extended real numbers is going to be called a set function. So, it is a function whose domain is a collection of sets, that is why it is called a set function. Next we will be looking at some special properties of, we will be analyzing such functions. So, let us define a set function mu of course, where c is a collection of subsets of a set x and 0 to plus infinity is the non-negative extended real numbers. So, a set function mu is said to be monotone, if it has the following property namely for any two sets a and b in c mu of a is less than or equal to mu of b whenever a is a subset of b. So, it is some kind of a and monotone property that whenever a is a subset of b and both are in the collection c, we want that mu of a should be less than or equal to mu of b. So, this is called the monotone property. . Next, we will look at what is called finite additive t property of a set function mu. So, a set function mu is said to be finitely additive. So, I am emphasizing the point finitely and additive, if it has the following properties that mu of union of sets a i i equal to 1 to n. So, given any finite collection of sets a 1 a 2 a n in c, we want mu of the union of these sets is equal to mu of a i's. Of course, this will be whenever a 1 a 2 a n, this is the finite collection of sets in c such that their union also belongs to c for otherwise this number on the left hand side of this equation will not be defined and we want further these sets are pair wise disjoint. So, a i intersection a j is empty for i not equal to j. So, once again let us see what is finite additivity. Finite additivity means for any finite collection of sets in c a 1 a 2 a n in c such that their union is also an element in c and these sets are pair wise disjoint. For any such finite collection of sets, we want that mu of the union is equal to summation of mu of the individual a i's. Intuitively keep in mind mu in some sense is denoting the size of a set a. So, we are saying mu of the union is equal to sum of the individual sizes whenever the sets a i's are disjoint and this is we are requiring it for any finite collection i 1 to n. So, in a 1 a 2 a n is any finite collection of sets in c which are pair wise disjoint and such that their union is element in c. We want mu of the union is equal to summation of mu of the individual a i's. So, such a property is called finite additivity property of mu or one says mu is finitely additive. We can extend a generalization of this definition. We will say mu is countably additive. So, from finite we are going to countably additive if mu of union a n's 1 to infinity is equal to summation of mu n a n's. Of course, whenever a 1 a 2 a n is a sequence of sets in c such that the union is also an element of c and they are pair wise disjoint. So, countable additivity is a property about a sequence of sets a 1 a 2 a n so on in c which are pair wise disjoint and their union is an element in c. Then we want for any such sequence of pair wise disjoint sets mu of the union must be equal to summation of mu of a n's n equal to 1 to infinity. There is another notion of countable is sub additive if mu of a is less than or equal to summation 1 to infinity mu of a n's whenever a is a set in c and a is contained in union of a n's where a n's also in c for every n. So, in some sense if a is covered by a union of sets a n's then we want the size. So, that is mu of a to be less than or equal to summation mu of a n's n equal to 1 to infinity. So, this is called countable sub additivity because here we are just saying that mu of a is less than or equal to and we are not requiring that a n's are pair wise disjoint. So, this is called countable sub additivity property of the set function mu. A set function mu is called a measure on c and c is a collection of subsets on c. If mu has a property that it is countably additive. So, it should be countably additive and we want that empty set belongs to c and with the property that mu of empty set is equal to 0. So, mu is defined on a collection c of subsets and we want the properties that empty set should belong to c, mu of empty set should be 0 and mu on this collection should be countably additive. Such a set function is going to be called a measure on c. .. So, some examples of set functions. So, let us start with a very simple one. Let us look at a set x which is a countable set. So, its elements are x 1, x 2, x 3 and so on. So, x is equal to x n, n 1, 2, 3, so on and let us fix p n a sequence of non-negative real numbers. So, x is a set which is a countable set with elements x 1, x 2, x 3 and so on and we are fixing arbitrarily some sequence of non-negative real numbers. .. So, let us define for any set a contained in x, for any subset a contained in x, let us define mu of the empty set to be equal to 0 and for the set a if it is non-empty, let us define mu of a to be equal to summation over those p i's such that x i belongs to a. So, a is a subset of x, so some of the x i's will belong to a. So, look at those indices i such that x i belongs to a, pick up those p i's from the given sequence p n and add them up and that is called mu of a. So, mu of a is defined as summation over those p i's such that x i belongs to a. We want to check that this is a measure on the collection of all subsets of the set x. Well, that is quite obvious because mu of empty set is defined to be equal to 0 and let us observe that mu of the singleton, if a is a singleton set, then mu of the singleton set is going to be the number p i. So, if a set a is a countable disjoint union of sets, so let us check that this mu is a measure. So, we are defining mu of a to be equal to summation p i, where i is such that p i sorry i is such that x i belongs to a. So, mu is countably additive. So, let us take a set a. So, let us take a set a, which is union of a i's i equal to 1 to n, a i's belonging a i, a subset of a, a subset of where a i is any subset of x. To check that mu of a is equal to a i's belong to and we want that a i intersection a j to be empty. So, we want this to be equal to mu of a i's i equal to 1 to infinity. Let us observe enough to check when each a i is a singleton x i. So, let us check that case first. So, what is a? a is let us when a i's are some not x i, let us because x itself is x 1, x 2, x n. So, this will be the whole space. So, let us look at the special case when say a i is equal to some x k i, i bigger than x k i, i bigger than or equal to 1. Then the set mu of a is going to be equal to summation p of k i, i equal to 1 to infinity which can be written as limit n going to infinity of which can be written as n going to infinity i equal to 1 to infinity p i up to n. So, sum of a series these are non negative numbers. So, this sum is nothing but the limit of the partial sums because they are non negative. So, there is no problem in writing it that way. So, that means mu of a is equal to limit n going to infinity of sigma i equal to 1 to n of p i. But that means what we want to check that it is summation of mu of each a i. So, this is limit n going to infinity of summation i equal to 1 to n mu of a i because each one is p of k i. This is summation p of k i. So, this is k i and this is mu of a i. So, that is equal to i equal to 1 to infinity mu of a i. So, mu of a is equal to summation mu of a i whenever they are whenever a i is a singleton set. If not it is a finite set then each finite is a union of finite sets and then because for a non negative series you can add it in any way you like it is easy to check that mu of a is also equal to. So, it is easy to check that mu of a is equal to summation mu of a i i equal to 1 to infinity whenever a i is are contained in x and a i intersection a j is empty. So, that says that this set function mu that we have defined is countably additive. So, this is what is called a discrete measure because it is given by a sequence and p i is called the mass at the point x i. So, as we observed mu of the singleton x i is equal to p i for every i and mu of the whole space which is equal to summation mu of the singletons mu of the singletons that is p i. So, mu of x is equal to summation of p i. So, obvious consequence of this is that mu of x is finite whenever this series is convergent. So, the mu is one says this discrete measure mu is finite that is mu of x is less than infinity mu of the whole space is finite if and only if summation mu of p i is less than infinity. If this summation of p i is if the series p i is convergent and the sum is equal to 1 then this measure mu is called a discrete probability distribution on the set x which is x 1, x 2, x n. A very special case which plays important role in the theory of probability and so on x is the set of the numbers 0, 1, 2 and so on. Let us fix any number p which is between 0 and 1 and define p k to be equal to n choose k. So, this is the binomial coefficient n choose k p to the power k 1 minus p raise to power n minus k k between 0 and n. This is called the binomial distribution because of this binomial coefficient appearing in the definition of p k. It is quite easy to check the summation of this p k is equal to 1 that is the summation of p k is summation k equal to 0 to n and this side is nothing but the sum is equal to of p plus 1 minus p raise to power n and that is equal to 1. So, this is a distribution which plays a very important role in probability. This is the probability distribution supposing you have got a coin and you are tossing a coin with probability p for head appearing then in n tosses this p k represents the probability that in n tosses you will get k heads. Another special case of this discrete distribution is called the binomial distribution is called the Poisson distribution which is characterized by the definition that p k is equal to lambda to the power k e raise to power minus lambda divided by k factorial. This is called Poisson distribution. This is another important distribution in the theory of probability. Finally, when we take only the finite number of points 0 1 up to n and p k is 1 over k and each point is given the same mass 1 over k then this is called the uniform distribution. They are special cases of discrete probability distributions. .. Next we give an important example of a measure which is defined on the collection of all intervals in the real line. So, to do that let us fix our notations. So, we will denote by i the collection of all intervals on the real line. For a interval i with n points a and b the left end point being a the right end point being b we will write it as i a comma b. So, a will denote the left end point and b will denote the right end point. We are not saying that this is a open interval a comma b. We are just saying that it is a interval with left end point a right end point b where the left or the right may or may not be or both way or may not be included in that interval. So, it is just an interval with n points a comma a and b left end point is a the right end point is b. So, on this collection of all intervals we are going to define a function. For example, recall that the open interval a comma a is the empty set and the interval 0 to plus infinity. So, the square brackets indicate that we are including 0 and we are including plus infinity. So, this is the closed interval in r star x belonging to r star the extended real numbers x bigger than or equal to 0 which is same as the open interval closed on the left 0 open on the right infinity in the real line union the special symbol plus infinity that we had added in the extended real numbers. So, with these notations we define what is called the length function on the class of intervals. So, it is a set function lambda defined on i taking values in 0 to infinity and is defined by take any interval i with left end point a right end point b. So, we define it as the absolute value of b minus a if a and b are both real numbers that means if a if the interval i is a finite interval with n points a and b then it is length is defined as b minus a and we define it equal to plus infinity in case either the left end point a is minus infinity or the right end point b is equal to plus infinity or both. So, length of i for a unbounded interval is defined as plus infinity. So, this function is called a length function on the class of all intervals and this length function going is going to play an important role for in our subject. So, let us study its properties. So, next we will be studying properties of this length function. The first property is the length function has the property that lambda of the empty set is 0 because empty set is an interval is an open interval with left end point say a right end point a. So, both is open interval a comma a which is empty set. So, by very definition that is equal to a minus a which is equal to 0. Next let us check that this is a monotone set function. So, namely length of i is less than or equal to length of j if i is a subset of j. So, to prove that so we want to check whenever we have got intervals i comma j and i is a subset of j this should imply that length of i is less than or equal to length of j. Since the intervals are defined characterized by the end points. So, let us case 1, let us say i is say infinite say i is equal to minus infinity to say a. Let j be equal to now since i is a subset of j obviously j has to start with minus infinity and can go up to some point say c where c is bigger than or equal to a. So, essentially what we are saying is if this is a and this side is the interval and this side is the interval all of it is the interval i and if j is to contain i then j must be somewhere ending somewhere here. So, that is c. So, clearly both are infinite. So, length of i is equal to plus infinity is length of j. So, that case is obvious. Let us look at the next case i is a subset of j. j is infinite whether i is infinite or not does not matter because the length of i is always less than or equal to plus infinity which is equal to length of j. j being infinite its length is always going to be plus infinity. So, this is obvious if this is the case. So, finally let us look at the case when both i and j are finite. So, here is let us say i has got the end point a and b. Now, i is subset of j that means the end points of j has to be somewhere here and it has to be somewhere here. So, if i is with end points a, b and j is with end point c, d then we should have c is less than or equal to a less than or equal to b is less than or equal to d. So, this implies this and that is same as saying that d minus c is bigger than or equal to b minus a and that is saying that the length of j is bigger than or equal to length of i. So, the monotone property is checked. So, the length function lambda is monotone property namely if i is a subset of j whenever interval i is contained in another interval j, length of i is less than or equal to length of j. Next, let us look at another property is called finite additivity property. So, what should be finite additivity property? We want whenever a interval i is a disjoint union of some other intervals then the length of i should be summation of length of j's. So, what we are saying if a interval i is written as a finite union of intervals j i, j i 1 to n where this j i's are pairwise disjoint then we want length of i to be equal to summation length of j i's. So, that is going to be called finite additivity property. .. So, let us check the finite additivity property. We want to check if i is equal to union of j i's all are intervals where j i intersection j k is empty then that should imply length of i is equal to summation length of j i's i equal to 1 to n. So, let us assume if i is infinite and i is equal to union of j i's j i's 1 to n then that implies at least one of j i's is infinite because if all of them are finite intervals the union will be again a finite interval. So, i infinite implies i equal to union of j i's implies at least one of this j i's has to be infinite. So, that implies lambda of i is equal to plus infinity is equal to summation lambda of j i's i equal to 1 to n because one of them is plus infinity. So, the case when i is infinite is okay. Let us look at the case when i is finite. So, i finite i equal to union of j i's and j i's are pair wise disjoint. Now, let us say the interval i has got n points a and b and let us say the left end point is a the right end point is b. So, here is a and here is b. We want to compute the length of i. So, note length of i is same as the length of the closed interval a comma b. I can include the end points in the interval i because the length depends only on the values of the end points. It does not matter whether the end points are inside or not. So, what we are saying is without loss of generality let i be equal to a comma b. So, this is the interval a comma b. Now, i is equal to union of j i's. So, the point a belongs to this union. So, it should belong to one of the intervals j i's. So, it belongs to one of the interval j i's and actually it has to be end point of one of the intervals of j i's because the interval cannot start with somewhere else. So, let us say so without we can say we can assume i 1 sorry the interval j 1 is starting at a and ending somewhere let us call it as b 1. The point end point may or may not be included. So, the first interval i 1 we can assume it starts here and ends somewhere here that is b 1. Now, the point b 1 is again in that union. So, either it is already included in the interval i 1 or it should be an end point of another interval in the union j 1 j 2 j n's. So, the second one must start with here and end somewhere here that is b 2. So, what we are saying is i 2 some other interval we can rename it as i 2. So, it should start somewhere again at b 1 and end somewhere here and so on. So, here will be the last one a n and that should be b n. So, what we are saying is we can assume this is. So, going this way we can arrange the end points of j 1 j 2 and j n such that such that a is same as a 1 less than or equal to b 1 is equal to a 2 less than or equal to b 2 and so on. So, a n less than or equal to b n which is equal to b. So, we can rearrange this end points of these intervals because this is a union and that is a disjoint union. So, this is what is possible for us to arrange. So, that means and that clearly says that b minus a is equal to b n minus a 1 that is equal to summation b i minus a i i equal to 1 to n adding and subtracting these terms in between that is same as i equal to 1 to n lambda of j i. So, whenever i is a finite interval i is equal to union of j i is they are pairwise disjoint we have gotten that the length of this b minus a is the length of the interval i is equal to summation length of j i's. So, that means that the length function lambda is finitely additive. So, this is the property of lambda the length function being finitely additive namely if an interval i is a finite union of pairwise disjoint intervals then length of the interval i is equal to summation length of j i's. Next let us look at another property supposing i is a finite interval say that i is contained in union 1 to n i i's where a finite union of the intervals but we are not longer saying that they are disjoint then the claim is that length of i must be less than or equal to summation length of these intervals a i's. So, if you drop the condition that these are pairwise disjoint i is subset of. So, we are saying if an interval i is covered by a finite union of intervals then the length of i must be equal less than or equal to summation of length of this intervals i i's. Let us look at a proof of this the proof of this is once again similar to the earlier property. So, let us say i we are saying i is contained in union of i i's i equal to 1 to n. Obviously, if one of i i's is infinite then clearly this implies length of i is less than or equal to summation length of i i's that is obvious because one of these terms on the right hand side in the summation is plus infinity which is always greater than or equal to length of i whatever be i. So, let us suppose that each i i is finite and this is a finite union. So, that implies i is finite. So, without as before without loss of generality assume without any loss of generality that i is equal to a comma b. So, here is once again the same picture here is a and here is b. Now, the point a belongs to the interval i. So, this is my interval i. So, a belongs to i that means it belongs to this union. So, it will belong to at least one of the intervals i i's. Let us name any one of them which contains the point a to be i 1 and let us say the end points of that is a 1 b 1. So, the point a belongs to one of the intervals i i's because it is in the union. So, it will belong to one of them say i 1 and let us say the end points of i 1 or a 1 b 1. So, here is the end point a 1 here is the end point b 1. Now, the possibility is this b 1 is on the right side of b. So, one possibility is it is on the right side of b. So, either b 1 is bigger than or equal to b that means my picture looks like here is a 1, here is a, here is b and here is b 1. Then length of i which is equal to b minus a is less than or equal to b 1 minus a 1 which is equal to length of i 1 and which is obviously less than or equal to summation length of i i's equal to 1 to n. So, in the case b 1 is on the right side we are obviously through by this case. So, what is the other possibility case 2? So, suppose this is the picture. So, suppose this is the picture namely we have got a we have got b and here is a 1 and b 1 is on the not on the right side, but on the left side of b. So, let us take that as the picture. So, in that case the point b 1 belongs to that union. So, b 1 is in the interval a b. So, it will belong to that union. So, it belongs to b 1 belongs to i. So, it belongs to union. So, it will belong to one of the intervals in the i i's. So, let us call that as some interval i 2. So, b 1 belongs to i 2. So, that means a 2 must start here and b 2 either it will be somewhere here or it will be on the right side. So, and if it is on the right side of it that means what? So, let us say it is on the right side. So, here is b 2 instead of here. Let us say b 2 is here. Then the length of the interval i. So, length of the interval i which is equal to b minus a which is b minus a is less than or equal to b 2 minus a 1 b 2 minus a 1 which is less than or equal to b 2 minus a 2 plus b 1 minus a 1. So, b 2 minus a 1 is less than or equal to b 2 minus a 2 plus b 1. We are adding something bigger and then a 1. So, that is length of i. So, in that case length of i will be less than or equal to length of i 1 plus length of i 2. So, that is anyway less than or equal to summation length of i i's i equal to 1 to n. So, if you go on repeating this process. So, what does it mean? So, in the next stage what is the other possibility that b 1 is inside? So, that means here is here is a, here is b. If it is not outside, that must be inside. That means here is a 1, here is here was our b 1, here is a 2 and somewhere here is b 2. It is not on the right side, it is on the left side. So, once again b 2 belongs and then we can proceed in the same way. So, either at some stage will be through or eventually if not then we will have a 1 is less than equal to a, is less than or equal to a 2, less than or equal to b 1, less than or equal to b 2, less than or equal to so on, less than or equal to a n, less than or equal to b, less than or equal to b n. So, what we are saying is either will be through at some final stage or we can rearrange eventually after n stages the end points in that way. In that case again, lambda of i which is equal to b minus a, so here is a, here is b is less than or equal to same idea b n minus, less than or equal to b n minus a 1. So, go on adding and subtracting less than or equal to b n minus a n plus b n 1 minus 1 minus a n minus 1 and so on and plus b 1 minus a 1 and that is equal to sigma lambda of i j, j equal to 1 to n. So, it is just whenever we are on a finite stage the end points can be arranged nicely and we get the property namely the length function is having the property whenever a interval i is covered by a finite union of intervals and the length of i is less than or equal to summation length of i i's. Let us look at an extension of this property namely supposing i is a finite interval such that i is covered by a union of intervals i i's 1 to infinity. That means the interval i is covered by union of a countable union of intervals i i's then again the claim is length of i is less than or equal to summation length of i i's. So, let us prove this property and keep in mind here we are assuming our interval i is a finite interval. So, interval i is contained in union of intervals i i's i equal to 1 to infinity these are intervals i finite this implies length of i is less than or equal to summation length of i i's i equal to 1 to infinity this is what we want to prove. Obvious case note if any one of the terms on this side lambda of i i is infinite then we are through. So, that is note if i i is infinite for some i then what will happen length of i i will be equal to plus infinity which is bigger than or equal to length of i whatever it may be whether finite or infinite. So, that is so implies sigma length of i j j equal to 1 to infinity is also bigger than or equal to lambda because one of them is infinite so that case is obvious. So, let us assume not only i is finite all the intervals i i's are also finite and we want to check this property. So, what we want to check is the following i finite say n points say a b we can assume it is a closed interval because the length of i is not going to change each i j is finite with left end point a j right end point b j. We are not saying that we are assuming these i j's are open or closed or anything we are just naming the end points the left end point of i j. So, we are saying i looks like a and b and each i j is a j b j we are not saying that these end points are included and we are given that i which is a comma b is contained in union of i j j equal to 1 to infinity and if this was finite then we already know how to manipulate that that we have already done earlier in the previous case. So, the idea is from that infinite union bring it to a finite union and here is a closed bounded interval contained in a infinite union and we want to say this is going to be contained in a finite union. So, somewhere the compactness property of the interval a to b is going to be used. So, but for that we need the intervals to be open. So, let us make these intervals i j is open but of course the lengths will change. So, let us fix, let epsilon greater than 0 be fixed, select an open interval say call it as j j such that this j j includes our interval i j and does not change the length much. So, length of i length of this j j is equal to say length of i i plus epsilon. So, slightly increase. So, what we are saying in this picture take an interval from here to here the open interval from here to here call that as j of j. So, each i j i j which was from a j to here b j is enclosed in an open interval slightly bigger but the length. So, this is the length portion that you add that at the most is equal to epsilon. So, now what happens is the following. So, a a b is contained in union of i j's and each i j is contained in the union of j j's and each j j is an open interval. So, we have taken an open interval. So, we have got an open cover of the close bounded interval a b. So, implies by Heine Borel property of the real line which says whenever a close bounded interval is covered by an collection of open intervals then implies there exists some n such that finite number of them will cover it. So, a b will be contained in union of j equal to 1 to n j j's. Finite number of them will cover it and now implies by our earlier case that length of i. So, this was my interval i. So, length of i is less than or equal to sigma length of j j's 1 to n. Now, each one of them is less than or equal to sigma j equal to 1 to n length of i j plus epsilon. Now, we want to separate it out this summation and let it go to infinity and equal to 1 to infinity. But, the problem will come because of this summation epsilon added n times that summation will tend to become very very large. We do not want to happen that. So, what we go is we revise our construction. So, where we select it for a given epsilon fix, select an open interval j j such that this is so instead of epsilon for the interval i j, let us divide it by 2 to the power j. So, instead of having this extra length to be equal to same length as epsilon for every interval i j, for i j we want this extra length to be equal to epsilon by 2 to the power j. So, once we do that, we are in a better shape because now this estimate will be 2 to the power j. So, that means it is less than or equal to summation j equal to, I can put it 1 to infinity because it is less than or equal to lambda of i j plus summation epsilon by 2 to the power j j equal to 1 to infinity. Now, this series is convergent because it is a geometric series with common ratio 1 by 2 which is less than 1. So, this term is equal to epsilon. So, what we are saying is length of i is less than or equal to summation length of i j plus n number epsilon, but epsilon was arbitrary. So, let epsilon go to 0. So, we will get length of i is less than or equal to summation length of i j's. So, what we are saying is that the countable property that we looked at namely length of i is less than or equal to summation length of i i's whenever a interval i which is finite is covered by any countable union, then the length of i is less than or equal to length of i i's. So, this is we have extended that earlier property from whenever a finite covering is there, we have extended it to a countable infinite covering, but only for finite intervals. But we would like to extend this to even arbitrary intervals which are not necessarily finite. So, for that we will have to do a little bit of more work. So, let us look at the next property which says the following. Let i be a finite interval such that i is equal to union 1 to infinity i n where i n's are pairwise disjoint. Then at least we can conclude that the length of i is equal to summation length of i n's. So, whenever a finite interval is a countable union of pairwise disjoint intervals, then the length of i is equal to summation length of i n's. So, let us prove this property. So, what we have got i is equal to union of i j's j equal to 1 to infinity i j's are pairwise disjoint i finite implies length of i is equal to summation length of i j's. So, let us observe note, we have already proved just now that if interval is written as this a union of countable disjoint union, note length of i we have just now shown is less than or equal to length of i j's added up j equal to 1 to infinity, call it 1. So, length of i is less than or equal to this is just now we proved for finite intervals. So, only to show only to show that length of i is bigger than or equal to summation j equal to 1 to infinity length of i j. So, only this is to be shown and for that, so here is the interval a to b, i is finite. So, here is the interval finite interval i. Now, look at i 1 is a subset of i. So, it should be somewhere inside. So, somewhere is a 1, somewhere is b 1. Similarly, i 2 is also inside i. So, somewhere it has to be, either it has to be a 2 here, b 2 here or it could be here somewhere and so on. So, for every n, let us look at, let us consider the end points, the end points, a n, b n of i n's. We can arrange them, we can arrange, there are only finitely many of them such that here is a, here is a 1, here is b 1, here is a 2, here is b 2 and so on, here is a n and here is b n and here is b. That means, we can arrange them in such a way that a is less than or equal to a 1, less than or equal to b 1, which is less than or equal to a 2, less than or equal to b 2 and so on, less than or equal to a n, less than or equal to b n, which is less than or equal to b. Once that is done, so this implies by a simple algebra that means, length of i, which is equal to b minus a. Now, this is b, this is a, I am going to make it shorter b n and a 1. So, this is bigger than or equal to b n minus a 1, which is bigger than or equal to b n minus a n plus b n minus 1, the next one here minus a n minus 1 and so on plus b 1 minus a 1. So, this put together is nothing but, which is equal to sigma i equal to 1 to n length of i i. So, what we are saying is for every n, the end points of the intervals i 1 up to i n can be rearranged in this fashion. Hence, by looking at the ordering of this length of i is bigger than and this happens for every n. So, that implies length of i is bigger than or equal to sigma i equal to 1 to infinity, because it is happening for every n, I can let it go to infinity length of i. So, the other way round inequality is also proved. So, that means, we have proved that whenever i is a finite interval, which is written as a countable union of pairwise disjoint intervals, the length of i is equal to sigma length of i n. So, with that, we prove an important property of the length function for finite intervals. We will continue our study of the length function in the next lecture. Thank you.


Something that's extremely important when it comes to having football conversations and quality football conversations is context. And context doesn't just apply to football, but just in life in general. Say for instance, you going to work and you end up getting there 30 minutes late, your boss is heated. Why you here 30 minutes late? Your shift started a half hour ago, but you just showing up now? Oh no, no, no, no, I ain't taking that.  They just were in a lot of pain and you didn't know why, but you had to take care of that because that is of the essence. But all your boss sees is that you were late. They don't see the background behind it. But when they find out what the background was behind it, they're like, oh, okay, now I get it. This is exactly what's going on this week with Lamar Jackson. There have been several people that have been saying, oh man, Lamar Jackson had a bad game. Terrible game, they look at the numbers 22 for 38. Miss 16 passes, what? What is that? And you didn't even throw a touchdown.  What kind of garbage game is that, Lamar? That's what a lot of people are seeing. They're seeing it on surface value, but they're not digging. They've been not looking at the context. And Stephen A. Smith and Shannon Sharp, they have been two of the biggest that have done that this week with Lamar Jackson. And they have questioned whether he deserves the contract, whether the Baltimore Ravens are even getting their money's worth when it comes to Lamar Jackson. Something Stephen A. Smith brought up. He said, even though you got those drops, so this part is what makes it worse because he brought up the drops, but he pushed them to the side. He said, even though you got those drops at the end of the day, if you don't turn the ball over, you win anyway. This man has been turning the ball over, and that is not what they are paying him to do. Lamar Jackson has an NFL worth 11 turnovers in the fourth quarter or overtime of one score game since 2021. That is inexcusable. So with Lamar Jackson in this game this past, so now he obviously did bring up more than just this game. And the fumbles, they have been a problem, especially this year. Now with those fumbles, there have been some that were, hey, it's like Lamar, what's going on? Because there was the wine member that the handoff to Justice Hill, where Lamar dropped it too early. There was one where Lamar was in the pocket and he was throwing, but he ended up throwing a ball and somebody hit him while he was throwing. And so some of them have been on him. And then there was another one, I think it was in week one, where he was running with the ball and the Texan defender said, no, you ain't about to shake me and the Texan defender knocked it right out of there. So there have been some fumbles that have been on him. But if you look at a lot of the other fumbles too, they have not all been on him, especially the one in this game toward the end of the game, where Lamar Jackson, and especially when you look at it on an all 22, you don't see the different angles of it and stuff. Ooh, that makes it hurt that much more because Lamar Jackson, he had dropped back to pass. Ravens were down, he had dropped back to pass. He was looking, he saw O'Dell getting ready to break and he was getting ready to throw it. But Ronnie Stanley had got abused by Alex Highsmith and Alex Highsmith just came right around the corner, boop, knocked it out. And that's happened several times this year. So while some of the fumbles have been Lamar Jackson's fault for sure, there have been other ones where there's literally nothing that any quarterback could do about that. And again, I feel like a lot of people are missing the context. Now, a lot of people are talking about Lamar Jackson hasn't been worth the money.  And they're looking at this last game like, oh no Lamar, what a bad game for Lamar Jackson. But again, this is why context is so important. Context is everything. Because if you don't have context then you don't have a quality conversation. And the fact that Lamar Jackson's contract is even being brought up after a game, especially after a game like this. Like he continued to show you throughout this game why he was worth the money. But unfortunately in this game the receivers didn't cash in at all. There were drops after drops after drops. Now, one drop that I actually gotta take back and I gotta take it off as they flowers. And it wasn't even a drop is when they flowers tripped. When I was watching the game live, I was like, oh man, that was a perfect pass from Lamar. He put it right on the money. But they said that the wind actually made the ball sail on that play. I know seeing the all 22 a couple of people have shown and I know Ryan Mink, he highlighted it till he showed the goal post. And you know how on the goal post he got the little flag and that flag was blowing heavy. But the ball was going one way and then the wind kind of took it another way. And that's what threw off that pass with Zay flowers. But besides that, all those passes that were dropped, they were on the money. And if all those passes are caught, even if most of those passes, it ain't even gotta be all of them. But if most of those passes are caught, then it's a completely different ball game. Completely different ball game. Because as we highlighted before, there were 14 points. Well, since the touchdowns were six, but there were 12 points that were dropped, dropped. I mean, technically it was 18 points because Mark Andrews and Rashad Bateman, they both dropped touchdowns too. And Nelson Aguilore did as well. But with Rashad Bateman and Mark Andrews, those touchdowns were dropped back to back on that same drive. Ooh, and that makes it hurt that much more. But with people saying Lamar had a bad game, a lot of people, what they've been doing this week, they've been erasing the first two, three quarters, and they've been fast forwarding right to the fourth quarter. They've been like, all right, them first three quarters, it didn't happen. I ain't concerned about that. I'm only concerned about the fourth. Now Lamar Jackson did throw that interception to O'Dell Beckham Jr. And whoever you wanna place that on, it's all good either way. Whatever you wanna say, it was a bad pass from Lamar Jackson. I know some people feel that way. Some people feel like O'Dell Beckham Jr. He just never got position. Now you know what the Baltimore Ravens with that play, with that fade route that he threw, that it was destined for O'Dell Beckham Jr. They had that predetermined, I guess, that when the Baltimore Ravens got to the line, they were like, look, if we see this certain look from the Pittsburgh Steelers defense, then we going for it. We are going for it. We going for it all the way.  Let's get it. And he got that one-on-one looking. It was not pretty. But whoever you wanna put that on, hey, I get that. I get that. Cause I can see how people can think it was a bad throw from Lamar. And I can see how people think O'Dell Beckham Jr. just never got position. And this is why I'm still questioning it. And I'm really wondering like, what is the status of O'Dell Beckham Jr. How healthy or unhealthy is he? But I guess we'll continue to find out in the coming weeks. But to completely put this game on Lamar Jackson and say that the Baltimore Ravens are not getting their money's worth out of Lamar Jackson, I think that's crazy talk. Yeah, he did have his faults in this game for sure. But this game was majority, huge majority, not on him for losing. Because again, when you are on the money, you on the money, you hitting these people in stride, you giving them plenty of opportunities and everybody getting opportunities. Everybody getting opportunities. But those opportunities are continuously going to waste and you getting nothing from them. And it's tough. I don't wanna say there's nothing you can do, but what you can do is limited. Like, RG3, he talked about it, he said, man, Lamar here fighting for his life. He's saying he's fighting for his life. He said, I don't know what else he could do. It's tough. It's really, really tough. But my point is, I say all that to say, when you're having a conversation, especially the conversation that I've been saying, I see been seeing a lot of different conversations this week that have lost the context of this game. They've just looked at the final outcome and they look at the final couple of drives from the Baltimore Ravens, which one, they look at the interception in the red zone. And that wasn't good. That wasn't good at all because that interception, it takes away points. Then they look at the fumble, but they just look at Lamar Jackson fumble, but they don't look at, oh, Ronnie Stanley just got dogged from the jump. And that's happened a lot. Not just with Ronnie Stanley, but with our offensive line, a lot with some fumbles, it's just been a lot of them getting dogged from the jump. Some of them been Lamar too now. So, but some of them just been getting, they been getting dogged from jump. But then they look at the sack at the very end. So yeah, man, it's tricky, but really in my opinion, it ain't that tricky. It's not that tricky to just actually look at what happened. And again, with these, we've been talking about this for years when it comes to the big media, when it comes to them talking about different teams and whatnot, I know they have a tough job to do. So I gotta give them, they have an extremely tough job to do. They can't look at every single game and they don't look at every single game. So that's why when you have people that do actually look at the game, it's important to hear them. It's important in my opinion to listen to what they got to say. But I know with the big media guys like the Stephen A. Smith, the Shannon Sharps and whatnot, all of them, they can't sit, they don't sit there and watch every single play from every single game. Cause that'd be too much. They wouldn't have time to actually do what their job is. But when they do do their job, when they don't have context, I think it takes away a lot from it.


Hyponymy is a hierarchical sense relationship between lexemes based on the notion of inclusion. A less general term includes the meaning of a more general term. The less general term is called the hyponym. The more general term is known as the superordinate term or hyperonym. In complex hierarchies, hyponyms can also be hyperonyms. The lexemes and fly and bee are hyponyms of the hyperonym insect and fly in term is a hyperonym of fruit fly, blue fly and brown fly. The lexemes insect and fly are hyperonyms of the concepts that are lower in the hierarchy. Hyponyms of the same superordinate term are referred to as co-hyponyms, whereas their vertical relationship with a hyperonym is one of inclusion, and for example includes insect. Their horizontal relationship with each other is one of incompatibility. Hyponymy is a transitive relation. If A is a hyponym of B and B is a hyponym of C, then A is necessarily a hyponym of C. Oddly, there are sets of elements, for example the color words, for which no superordinate term can be found. And some linguists include the relationship of synonymy within hyponymy and define it as symmetric hyponymy, where A is a hyponym of B and B one of A.


What do we do to ensure our empire goes to our children because of rules and regulations of insurance with our renewals and things like that? Do you have any insight on that? Man, I don't mean to hug a mic, but these are really, really good questions. I'm sorry. Okay, number one, every agent should be a LLC, okay? Because when you pass, a lot of carers will not pay the residuals anymore, but they will if it's an LLC. That's number one. So even if your tax person says, we only really need one, it's just going to be a pass anyway, still get it so that if something happens, you still have funds coming in through your family. As many of you know, last year my business partner, Vince Spombinato, spoke at 8% Nation. And he had just got done with his last round of chemo and had just gotten the clean bill of health that he had been cancer-free. Only to find out a month later, I'd come back and then ultimately took his life two months ago, okay?  And there's a huge difference there. A salesperson wakes up every morning unemployed. You have to get up, pound the pavement, pretend like I'm fired up and I'm really nigh, get the car, show up, convince them it's their signature, okay? Like that was my life, okay? I didn't forge this, Mary, okay? Like this is you, alright? That's a salesperson, but a business owner has infrastructure, processes in place so that when they're removed, their business still runs. I wake up every morning with cash flow, not figuring out where's my next lead going to be. That's the advice I give you. I'd answer the same thing. Obviously Secure is whenever I retire or pass away whichever comes first, we'll go to Cody. And so obviously I have an LLC and Cody has an LLC. And then we've also, in my trust that I've done, it's designated in there. And so just kind of done some brief planning on it.  They'll still have to have an insurance license, I believe. The insurance companies still require them to have an insurance license. So you wouldn't be able to pass your business to someone who's not a licensed individual. They would have to be a licensed individual. As far as I understand it, the insurance company would require them to be licensed. So they could be an agent for the business that runs the LLC and that's how they would keep their residuals going? Correct, because commissioners continue to come into the LLC. I tell people all the time that individuals, they're going to pass away. Corporations never die. As obviously a big example, Sam Walton, he passed away. Obviously he founded Walmart. But when Sam Walton passed away, Walmart didn't shut the doors. Walmart continued to the bigger and better than they've ever been. It's obviously a big example, but that's an example.


I don't know about you guys, but I am super pumped for Top Gun 2, but then it occurred to me, Adam, Top Gun 2 came out back in 1990. It was called Days of Thunder. I don't know what my problem is, but for some reason, young Adam skipped out on all the classic Tom Cruise movies until just recently I had not seen Top Gun, I had not seen Cocktail, I had not seen Days of Thunder. I've since rectified all those issues, Cocktail's fantastic, Top Gun, not great, Days of Thunder I'm somewhere in the middle on. I think it's better than Top Gun, yeah I do, even though planes I think are cooler than race cars, this movie works better for me. It's like director Tony Scott and TC, short for Tom Cruise, looked back on what worked in Top Gun and more importantly what didn't and tweaked the formula just enough to work better. We still have that tacked on love story, but this time it's implemented better within the storyline, it makes sense that he's talking to the doctor and how she's implemented into this picture. We still have our rivalry on the road, this time instead of Iceman, we have Rowdy Burns played by Michael Roker. Tom Cruise is the exact same character again, except his name's not Maverick, it's Cole Trickle. Who I read is based on an actual race car driver, but it doesn't matter, it's Tom Cruise playing a hot headed, cocksher racer who can do anything until it's proven he can't and has to take a reality check. We got a very young John C. Riley in this, we got Randy Quaid who's somewhat normal in this movie for once, and then there's the best part of this movie for me which is Cole's relationship with Robert Duvall's character Harry, the mentor-mentee thing really works. We get to see the frustration with Harry when Cole doesn't listen to him because this guy knows what he's talking about, he gets cars and what makes them run and how the driver needs to respond to it and treat it like an extension of themselves. And like Top Gun, we got Hans Zimmer back this time behind the wheel of a car making badass music. No days of thunder doesn't have Kenny Loggins, there's no danger zone to be heard, put through some freaking guns and roses here. The plot of this is very straightforward, Cole is a fantastic stock car racer who's trying his hand at the big leagues. He's now a NASCAR driver competing in the Daytona 500. One thing that's held up impeccably well were the visuals in Top Gun. That plain porn, it's top notch, especially with the orange motif Tony Scott goes for. He brings that here again to NASCAR, the oranges are back, the beautiful sweeping shots of the racetrack are in full effect. We got race car vision, we got crashes, and TC was right on the money when he hand picked Nicole Kidman for this role, that's right, he's the one that casted her in this thing, perfectly done, much better than Charlie. The Cole rowdy rivalry is far better too than it was in Top Gun, which never really made much sense to me. Here, these guys are measuring each other up, they're really at it to win, all they want to do is prove who the best driver is. Yes, that was kinda what it is in Top Gun, but we have more time with these guys, we even see them race down a hospital aisle in their wheelchairs. It does unfortunately lead to a terribly corny sequence later where they race each other to a restaurant for lunch, just so badly done, and it's mainly because the music they play is borderline 70s porn. I might get copyright strike because that was pretty much spot on. I totally get why Days of Thunder got a bad rap, it's not ranked very well, a lot of people dismissed it as a copycat of Top Gun, it absolutely is a copycat of Top Gun. I'm just telling you, it's a better version than Top Gun, it holds up better. If you sat these two next to each other, I'm going Days of Thunder every time, no it's not as iconic, no it doesn't have the cool shades throughout the film, no it doesn't have the grade one liners like you're writing checks your body can't cash, what it has though is a more competent screenplay, more likeable characters, an easier to follow storyline, amazing visuals still especially for a 1990, and just an all around better film. So if you're looking for a Tom Cruise to watch before you head out to Maverick in a few weeks, I highly suggest Days of Thunder. If you're familiar with the film, let me know in the comments if I'm spot on on this or if you have a different take, is it really just a poor man's Top Gun and I'm seeing things that aren't there, or do you agree it's the better version of the two? And please think about subscribing to the channel, Adam Does Movies, as I post movie content all the time here. Love to talk about it, love to have you, hopefully I'll see you next time. If there's other videos you can watch over here, I suggest you do that. Come on man, come on, just click on the videos over here.


Hey! I'm Anfa. Welcome to my Trap Drums mini-series. In this video I'm going to synthesize a kick bass or an 808, if you will, from scratch. You'll learn how that sound works, so hopefully you can put your own spin on it. By the end of this mini-series we'll end up with this. As you can see there's a kick bass, a snare and a hi-hat. I'll show you how to synthesize all of that with just one plugin. I'm going to use the Zenfusion synthesizer, my open-source weapon of choice for sound synthesis and Ardour, my favorite open-source DW. Still you should be able to more or less follow this series with any decently capable synthesizer plugin and at all. If you'd like to learn basics of Zenfusion this mini-series should also give you a head start. And lastly you can download the MIDI files, Zenfusion presets and the whole Ardour session to play around with on your own. The link is in the video description. Enough talking! Let's get synthesizing! I've prepared myself a short MIDI loop. For now it's just the in-it patch of Zenfusion FX or Zenfusion which is just a sine wave. So let's open up the interface. Let's go to add synth and here let's first pitch shift this two octaves down. That's more like it. Now in the amplitude I'm going to rise it to make the whole thing louder. Now the amplitude should slowly decay so let's use this envelope. I'm gonna move the sustain all the way down and extend the decay time. Maybe a bit shorter. That's about right. Now we need to add a pitch envelope to give it that kick character. So I'm going to use this pitch envelope. Now tuning the envelope is a very important step so I'm going to spend some time on this. Let's go to the free mode where I can freely move all the points, add new points and just tweak the envelope as I want it. I think that sounds nice. Okay now I want to add a little bit more punch to the sound. So let's go to the second voice. Let's enable it and it's just another sine wave but let's turn it down. You see I'm just in the voice tab turning down just the voice not the whole instrument. Now let's go to the modulation tab and change it from the oscillator or sound to white noise. Now I can go to voice, amplitude, envelope. Let's activate it and we can shorten the noise part itself. I'm gonna also reduce the release time so we can easier see the whole function here in this window. And that's the base of our kick bass. Now let's go to effects. I want to add some distortion. Let's go to part insertion and insert a distortion. It's a bit too much. Let's lower the drive a bit. Also I want to move up the Lopez filter so it is open. It doesn't cut off any high frequencies. I'm also going to lower the amount a bit. So it's parallel. This is basically the dry wet control because for example give it more distortion but mix it back with less amount. I like that. And here's our 808 like kick bass sound. In the next video I'll show you how to synthesize this sweet snare. See ya!


Hola, mi nombre es Belgica Paola Rodriguez Flores and today is day 17 of self-tape every day in July. It is the last day of the special skills week and I have had so much fun filming all of these different special skills and I do have a few other sports and a few other special skills that I have that I am a beginner or intermediate at but I'm just not filming this week but I will film them in the future so I can have them on my casting websites and that's exactly why I'm doing all of this to practice, to maybe give you some inspiration so you can practice as well but also to have clips of me doing those special skills. So I did save the best for last because I am the most proficient at Spanish compared to all of my different special skills. I am bilingual, I speak English and Spanish and Spanish was actually my first language until I was four and so for this clip I am going to start it in English and briefly say how proficient I am at it and then I am going to in Spanish say what part of Mexico I'm from as well as my favorite drink in Mexico so that I can give them an idea of my culture and my accent most of all because I do have an accent from the state that I'm from but also because I've been in California for so long my accent is a mixture of you know if you are from that state and you come to California you get this accent or at least that's what happened to me you know that's how I got the accent that I have in Spanish so I'm super excited to do this one because I know it's going to be easy and I'm not going to get sweaty for this one so yeah let me show you my clip and I decided on a blue backdrop even though I'm wearing a blue dress I just thought it would look cool and this is a dress I know there's shirts like that look like this so I'm hoping people just think I am wearing a shirt I guess it doesn't really matter they're just seeing this much but because they are just seeing this much of me I thought I could get away with it I'm exploring this month with wearing different things using different different backdrops and if you haven't heard already I am doing a giveaway of a savage paper backdrop it's not sponsored I know if I'm mentioning it it's so much but like I just want you to submit to the giveaway I'm giving away a 53 inch paper backdrop like I said not sponsored but Savage is giving these for free to me so that I can give them to you so thank you so much Savage paper backdrops or Savage Universal for giving us these backdrops I think they're so much fun to film in front of and I've had a lot of fun so far trying different colors here is the clip of me speaking Hi my name is Belgica Rodriguez and I am fluent in Spanish it was my first language y yo soy del estado de Jalisco en México en un pueblo chiquitito que está cerca de tequila si conoces a tequila y la bebida que me encanta tomar diario cuando soy en México es te juino si no lo conoces es una bebida de lote y en muchas partes le ponen mucho limón mucha sal de la gruesesita del mar y también el raspañelito y queda riquísimo cada vez que soy en México tomo un litro al día mínimo what did you guys think if you speak Spanish what do you think of my accent what country are you from do you guys speak another language let me know in the comments and thank you so much for watching at the end of every video I feature another channel this is today's feature if you would like to be featured on my next video make sure you're subscribed like this video and leave me a comment about what your channel is about


What's up internet? Nalala mo nang nasa school ka, tapos may test, tapos hindi ka talaga nag-aaral, so kailang mong fallback dun sa mga general rules na maalala mo. For example, maraming ganun sa law school. Sa negotiable instruments, usually yung holder in good faith and for value, siya yung tama. Sa land titles, kung may parcel of land na binenta to multiple people, first-in-time, first-in-right, usually. At kung wala ka talaga maalala at all, tulat mo lang sa test paper mo, Duralex-Sednex. May mga general rules fit sa computer, and yung number one dito, higher number, better. Let's say between two computer parts, PC100 or PC50, all things being equal, ipiliin mo yung PC100. Pero tama nga ba yung general rule na to? Pag-usapan natin, pero bago noon ang video na to ay handog me. Pusnay di kalao na windows. Pero sand ko hukuan ang legit windows? Daming options, daming prices. Aka dito. Gudi na lang, may cdkeyoffers.co. Madali lang ang order. Search for the software you need. Add to cart. Dahan ka sa payment options nila. Wala apag five minutes. Finished. May legit, working cdk na para sa windows mo. Gamitan lang aming code para makakuhapan ng discounts. Kaya kung nagahalap ka ng legit, mura, at original software. Check out cdkeyoffers.co. Let's go part by part, starting with the CPU. Yung usual metric for speed sa CPU is gigahertz. And hindi ba 2 gigahertz is better than 1 gigahertz? Yes, some of the time. Sa gigahertz kasi, comparable lang siya between CPUs of the same family. So let's say, Ryzen 5000 series, yung R5 5600X, 4.6 gigahertz yun. Ryzen 9 5950X is 4.9 gigahertz. So yes, masabilis nga siya kaisa sa Ryzen 5 5600X. Pero hindi mo sila mo compare dun sa Intel 12 Gen. Na magkaibang CPU family. Hindi nga, gigahertz only works within the same types of chips. So useless na yung compare yung gigahertz ni Ryzen versus gigahertz ni Intel versus gigahertz ni Apple. Kahit, let's say, mabilis yung megahertz ni Intel, hindi hindi kabi yun nun na automatically masabilis siya kaisa sa kaya AMD compared to an AMD CPU with a lower gigahertz. Core count, yes, usually higher number better because more cores let's you do more multitasking. So better nga siya in the sense na masabilis nga nang kalagin performance. Pero hindi lahat na mga program or applications nagagamit niya lahat nung cores. For example, for gaming, a lot of games don't use a lot of cores. So yes, may speed boost sa games yung R950-950X. Pero price to performance, lamang parin si Ryzen 5 5600X which has a lot less cores. Because cores are better for multitasking. Individual programs like games aren't designed to take advantage of a lot of cores. So to to nga naman, higher number better, more cores better pero kailangan mo rin tingnan yung price to performance. Value by ba siya para sa use case mo? From other boards, yes, usually higher number better. Speed of RAM supported, higher number better. Transfer speeds from the PCIe bus, higher number better. So dito, to to yung rule. Higher number better except of course for temperatures. You want the VRM temps to be as low as possible. Number of faces of the motherboard. Mas marami faces, mas fine yung control niya dun sa power na pumapasok dun sa CPU and other components. So yes, a motherboard would generally pass the higher number better rule. Yung GPU din, usually higher number better. Speed of the GPU, higher number better. Size of its memory, higher number better. Number of fans, higher number better. Number of temps for cores or the AMD equivalent, higher number better. So yes, a GPU follows the general rule, higher number better except dun sa pressure. Siyempre gusto natin, lower number better when it comes to the price of a GPU. PSU, yes, generally higher number better especially dun sa color rating ni 80 plus. An 80 plus gold is more efficient compared to an 80 plus white or 80 plus bronze. Titanium yung pinakakayas na 80 plus rating. Generally, the higher the color, the more reliable a PSU is kasi kailangan mas magandang components niya para umabot dun sa higher efficiency rating. Generally, also higher watt capacity better. So 850 watts can deliver more power compared to a 650 watt PSU. Hindi naman makakasama sa computer kung mataas yung watt capacity yung PSU. Di nasasaktan yung computer, let's say, 850 watts yung PSU mo. Di siya na overpower. Walang nadadamage na component kasi ang daming watts na pinisteed ng PSU. Don't need to worry about that. Although again, isipin mo rin yung price to performance. Syempre mas mahal yung mga higher watted PSUs. 80 watts, 1000 watts, 1200 watts. 80 plus titanium compared to 80 plus gold. Baka naman hindi mo kailangan yung sobrang high-powered na PSU. May recent video kami na for a high-end computer okay nang yung 650 watts for added headroom or added allowance. Usually, most enthusiasts get 750 watts. Anything beyond that is really mining territory. Sa RAM, yes, higher number better when it comes to the size. So definitely 8 gigabytes is better than 4 gigabytes. 32 gigabytes is better than 16 gigabytes. Although tandaan na mas maganda kung nakadual channel meaning you have two sticks of RAM. So 1x8 versus 2x4, piliin mo yung 2x4. Actually, even then, higher number better. The more sticks you have, generally better. Kasi magagamit mo yung dual channel ng MOBO. Hindi mo magamit yung kung single stick RAM ka lang. Pero sa speed ng RAM, higher number is not necessarily better. Sa Ryzen architecture ngayon, the sweet spot is around 3,600 to 3,800 megahertz dahil sa infinity fabric clock ng AMD, ng Ryzen series, anything beyond that actually makakasamapa sa computer mo. Might take the system instability or it might even crash the computer. Hindi na kasi instinct yung infinity fabric clock dun sa RAM. So may mga clear instances na higher speed is not necessarily better for RAM. CL timings, opposite naman to. Actually, mas gusto mo yung masbababa na CL timings. Lower timings usually mean that they're tighter, meaning that the communication of the RAM is faster to the motherboard. For example, mas hinahanap yung CL16 by enthusiasts kaisa CL18. Odo para sa akin, medyo debatable, gano ba lang yung advantage yung sobrang tight timing CL16 compared to CL18? Meron talaga, although if you're not an enthusiast na magkitingkar around manually with the timings, you'll probably be happy already with CL18. SSDs, yes, usually higher number better. Usually mga metrics yung SSDs, data transfer rate, data read rate. All of those things, usually higher number better. Number of cycles, kasi usually silasabi na mga SSD manufacturers, oh, itong SSD ko, rated up to ilang read write cycles. So the higher the number, the longer the SSD will last. Sa monitor, yes, usually higher number better. Unang-unang yung size niya, 27 inches is a big leap from 24 inches. Hindi mo mai-sip na, gano ba laki talaga yung quality of life increase? Tatlong inches lang. Pero sulit na sulit talaga yung 3 inches, mararam daman mo talaga yun. From 22 to 24 to 27, just those little increments of inches, mas lumalaki talaga yung screen space and new immersiveness na mga games, yung productivity mo kasi mas maramig kang windows na mo mabuksan. And then of course, the jump to 32 and even larger monitors. Kung kaya nung desktop mo, yung actual desktop, yung size nung desktop mo, and kung kaya nung budget, really, there's no disadvantage to getting the largest monitor you can. Other metrics for monitors, higher number better also, refresh rate, resolution. All of those things, usually higher number better. Of course, you could be overpaying again. Dito ulit lumalabas yung price to performance factor. Kung dik ka naman gamer, di mo nung traga feyangan nung 144Hz monitor. Just for office, 60Hz or 75Hz would do. So yes, while on paper, higher number might be better for refresh rate, depende talaga yun sa feyangan mo. Ano yung use case mo, and piliin mo yung tamang monitor para hindi masayang pera mo. Air cooler, usually higher number better, the more towers it has, the more fans it has, even yung physical size niya, yung height niya, and yung width niya. The larger the number, usually the better performing the air cooler. Of course, inyat ka lang kung bumili ka nung big boy air cooler like the NH-T15, the Qwais Dark Rock Pro 4, baka hindi mo na mapasok yan sa case mo. Better nga yung cooling performance, pero di naman pala siya tugma sa system mo. AIOs, higher number better, it's true, mo specifically sa rad size. Nagula talaga ako when we tried it out in our previous videos, laki ng performance increase from a 240mm rad to a 360mm rad. Para sa akin, di ko nang ma-recomend yung 120mm rad. Medya mahi na talaga yung performance niya. You'd be better off with an air cooler. That's why hardware sugar doesn't stock anymore, 120mm rad AIOs. It's 240 and up for us. Yung price to performance, medyo alanganin talaga yung 120mm. But 240 and above is okay, and there's a big jump in performance between 240 and 360mm. Yung case, di naman talaga uplikable yung rule na yun. Sizing of cases, in demand ngayon yung very small cases or just smaller cases than the usual ATX format. Outside of the computer, pero still very important to everybody, is internet speed. Yung talaga usually higher number better, 300Mbps is better than 100Mbps. Pero yung exception dun is ping. Dun lower number better. Ping is basically the amount of time it takes for you to communicate to the server. And it's possible to have high internet speed. So obram bilis yung internet speed mo, kailong mataas rin yung ping mo. Medyo mapapaisip ka ng kontin. Kasi mabilis na nga internet speed ko eh, pero bakit mataas pa rin yung ping ko? That could depend on the server that you're talking to. Mahalaga talaga yung ping sa games. That's why big games have multiple servers around the world, so that gamers can connect to the server that's closest to them to reduce ping time. So yes, you do want higher number for your internet speed, but for ping, you want it as low as possible. So yun yung round up natin. And the general rule seems a little shaky to be honest. Kung testo, medyo 50% of the time lang talaga siya na tama. Medyo mahirap kasi ikahon yung computer sa mga general rules or basta, palaging tama yung isang rule na to. And even when higher number is correct, depended din yung same case mo. For example, nga dun sa monitors, yung refresh rate, mas magadangan naman talaga yung 144 hertz kais sa 75 hertz. Pero kung di ka raman gamer, useless lang naman yun. Magbabayad ka for a feature that you don't really need. So kahit talaga sa mga situation na, sobrang inaw talaga na higher number better, be sure to understand the spec and what it's actually trying to say and then isipin mo kung kailangan mo ba talaga yung bilis na yun. What is the higher number representing and do you need it in your life? Pero for sure, general rule, size matters. Yung inches ng monitor, laking quality of life improvement yun. Thanks for watching. And thank you to our top fans na afford na namin to kupuwa na to and edyo clap. Thank you to Leah Magnaya, Ian Meru, Jan Ruben Ocha, Christian Espinosa and Raphael James. Thank you for supporting the channel.