 Welcome to lecture number 10 on measure and integration. Let us just recall that in the previous lecture, we had started looking at the notion of outer measure. So, let us recall how the outer measure was defined and what are the properties the outer measure has. Given a measure mu, on an algebra A of subsets of a set x, the outer measure induced by this measure is a set function defined on the class of all subsets of the set x. So, it is a function defined on the power set of x, of course taking non-negative values and it is defined as for a set E in subset of x, look at a countable covering of the set E by elements of the algebra A. Now, and look at the size, the measure of the set A i. So, that is mu of A i, add up the measures of all the sets A i, so that this union covers. So, that gives you a number which we can think of as approximate measure of the set E. So, look at the infremium over all such possible coverings of E. So, mu star of E is the infremium of all summation mu of A i such that union of A i is cover E and we proved properties of this set function in the outer measure namely mu star is monotone, it is countable is sub additive and on the sets in the algebra mu star is same as A. So, mu star extends the measure mu, but it is only monotone and countably sub additive. So, let us look at an example of this outer measure. So, in this example, we will start with the collection A of all subsets of the real line which are either finite or their complements are finite. So, recall in the beginning of the lectures, we have shown that this collection A is an algebra of subsets of A. So, this collection A forms an algebra of subsets of the set R. Let us define a set function mu on this, mu of the set A is equal to 0 if the set is finite and if the set in the algebra is not finite, then we know it is a complement is finite. So, in that case, we define mu of A to be equal to 1 if A complement is finite and we have also checked this example for this example that mu is a measure on this algebra A. So, I would strongly says that you try to prove it yourself once again that this mu is a measure that is mu on the algebra A is countably additive. So, let us denote by mu star the outer measure induced by mu on all subsets of the real line. So, mu star is the outer measure given by this particular measure and we want to find some properties of this mu R of this outer measure mu and if you recall just now we said outer measure always is monotone and it is countably sub additive. So, these properties are true for any outer measure. So, in particular this outer measure also we want to do something more and obvious let us note that mu star of x the whole space is same as mu of x and that is equal to 1 because mu star extends. So, and x belongs to the algebra actually x complement is empty set which is finite. So, by definition mu star of x should be equal to mu of x which is equal to 1 and if A is any subset of x and mu star being monotone. So, we know that mu star of A is less than or equal to mu star of x and that is less than or that is equal to 1. So, mu star of every set is going to be between 0 and 1. So, this is a property we have deduced from the general facts that mu is monotone and mu star of the whole space is equal to 1. We want to show that if A is a countable set in real line then mu star of A is equal to 0. So, let us see how do we show that. So, let us take a set A. So, let A contained in R A countable that means I can write A as a sequence. So, x 1, x 2 and so on. So, I can write A is actually equal to union of singletons x i i equal to 1 to n and by definition mu of the singleton x i it is a finite set. So, that is equal to 0 for every i thus mu star of A which is less than or equal to summation mu of x i. This is one covering and measure of each one of them being equal to 0. So, this is equal to 0. So, mu star of A is less than or equal to 0. We know it is always bigger than or equal to 0. So, that implies that mu star of A equal to 0 if A is countable. So, we have shown that for a countable set subset in the real line the outer measure which we defined is going to be 0 whenever A is countable. Let us go further and look at some other properties of this outer measure. We want to show that mu star of A is equal to 1 if and only if A is uncountable. So, let us prove that. Let A contained in R A uncountable claim mu of mu star of A is equal to 1. Note mu star of A is less than or equal to 1 that is obvious. That is by the definition we said mu star is monotone which is equal to mu of x anyway mu star of x. So, mu star of A is always less than or equal to 1 to show that mu star of A is also bigger than or equal to 1. Now, let us note A uncountable. So, look at A is uncountable. So, what can you say about A complement of the set? It cannot, if A is uncountable then we want to, if A is uncountable Let us take a covering of A i i equal to 1 to infinity. Let us take a covering where A i's belong to the algebra A. Now, note in this covering A uncountable implies at least one A i is uncountable. So, this is the observation which is going to be crucial. A is a subset of unions A i's which are in the algebra and each one of them, if A is uncountable then each one of them cannot be countable because if each one of them is countable then countable union of countable sets will be countable. So, A will be countable. So, that means there is at least one A i which is uncountable. Say A i naught is uncountable, but that will mean what? A i naught is uncountable means, so thus we have got that mu of mu star is A i naught is uncountable. So, mu of A i naught, what can we say about that? So, that cannot be equal to 0. Can we say that? It cannot be equal to 0. Let us see. One of the A i's naught is uncountable. So, that means this is not finite. This is because A i naught belongs to the algebra A. That is a crucial thing. So, that means either A i naught is finite or its complement is finite, but we know that A i naught is uncountable. So, that means A i naught is uncountable. So, that implies that A i naught complement is finite because it belongs to the algebra. So, either A i naught has to be finite or its complement has to be finite. So, this is finite. So, that implies that mu of A i naught by definition is equal to 1. So, that is what we have shown. So, thus if A is contained in union A i 1 to infinity, then for some i naught mu of A i naught is equal to 1. So, that automatically implies the fact that mu star of A, so any covering will have at least one of the elements. So, implies this is bigger than or equal to 1 because mu star of A is the infremum of because mu star of A, the summation i equal to 1 to infinity mu of A i is going to be bigger than or equal to 1 because A i naught is 1 at least, one of the terms is 1. So, for every covering A i of A, sigma mu of A i is bigger than or equal to 1. So, the infremum has to be bigger than or equal to 1. So, hence mu star of A is equal to 1. So, we have proved, therefore thus A uncountable implies mu star of A is equal to 1 and the converse is obvious because conversely if mu star of A is equal to 1, then A is then A is uncountable because mu star of A is equal to 1, then A is then A is uncountable because for countable sets we have already shown mu star of A is equal to 0. So, A is uncountable if and only if. So, we have shown, we have characterized all sets for which outer measure is going to be equal to 1. So, we have shown the fact that mu star the outer measure induced by the measure that we are looking at namely mu, if mu of A is equal to 0 whenever A is finite and mu of A is equal to 1 when A complement is finite and if you look at the outer measure induced by this measure, then that has the property that mu star of every countable set is equal to 0 and mu star of a set is equal to 1 if and only if the set is uncountable. So, that is the property that we have. So, mu star of A is equal to 1 if and only if A is uncountable. Now, let us we already know that mu star is going to be is countable is sub additive. We want to know is mu star additive. So, we will show that is not the case that mu star is not even finitely additive. So, to show that let us observe that the real line I can decompose into two disjoint sets minus infinity to 0, 0 close union 0 to infinity. So, the real line is written as a union of two sub sets and now the outer measure of the set minus infinity to 0 that is the uncountable set. So, that is equal to 1 and the outer measure of 0 to infinity is also equal to 1. So, outer measure of both of these sets is equal to 1 their union is r and so we get mu star and mu star of r is equal to 1. So, which is strictly less than 2 equal to sum of the outer measures of each one of them. So, that says mu star of r is strictly less than mu star of minus infinity to 0 plus mu star of 0 to infinity. So, this should be mu star and this also should be mu star. So, that says that the sets that mu star the set function mu star is not even finitely additive. But, let us observe some more facts about this outer measure. This outer measure on all subsets is not finitely additive, but it has some nice property it is countable additive on a subclass. So, let us look at the subclass to be S which is a subsets of r all those subsets of r where either a or a complement is countable. So, keep in mind a or a complement is countable we had shown that this collection forms a sigma algebra and our given algebra a which was a collection of all sets for which a or a complement is finite obviously is a subset of this class S of subsets which are a or a complement countable and that is of course a subset of all subsets. So, algebra a is contained in S this sigma algebra S and which is inside of P r and it is actually we have also shown that mu star. What is mu star on S? If a set a is countable then mu star of a is 0. If it is not countable but then it is a complement is countable then mu if a complement is countable then the set a cannot be countable it has to be uncountable and for uncountable sets mu star of a is equal to 1. So, mu star restricted to S is the set function which is mu star of a set a is 0 if a is countable and mu star of a set a complement of a is equal to 1 if it is complement is countable. So, and that we have already shown is a measure on the class of on all subsets all sets in the sigma algebra S. So, given that measure in our example given the measure on the algebra a when we define the outer measure which is defined on all subsets of P r is not even finitely additive but if we restrict it to the collection of sets S which is a or a complement countable then on that class it is a measure and it extends. So, the given measure on the algebra does not extend to all subsets but at least it extends to a collection a sub collection of all subsets and that includes the original algebra. This is a situation which we are going to see is very common in our extension process. So, we are starting with a measure mu on an algebra and just now we said in the process of extension let us define outer measure on all subsets and here is an example which says to all on all subsets outer measure is not it may not be even finitely additive but at least this example says that we can probably find a subclass of all subsets of that set X which includes the given algebra and on that probably it is countably additive. So, the problem is to look for a collection of subsets on which it is going to be countably additive. Before we go over to that process let me give you an exercise which all of you should try namely if mu is a measure on an algebra then we define the outer measure as the inifrimum. Look at all coverings and look at summation mu of A i there was no condition on put on the sets A i and the exercise says if you put take only those coverings of E by elements of the algebra which are pair wise disjoint and then take the inifrimum over only such coverings that also will give you the outer measure. So, the exercise is that outer measure for a set E can be defined in terms of in terms of countable disjoint coverings of E namely take coverings of E by elements of the algebra and the elements of the covering are pair wise disjoint look at the sum mu of A i and take the inifrimum over such coverings. So, inifrimum is taken over all countable disjoint coverings in the original definition we did not put this condition and the exercise is both of these are same and the answer lies in the simple observation that when a measure mu is defined on an algebra and you look at the union of elements in the algebra any union can also be written as a union of pair wise disjoint sets. So, a union of elements of the algebra can be represented in terms of pair wise disjoint sets that fact we have used earlier also. So, using that you can try to prove this exercise here is another exercise which you should try to prove to get familiarize with the concept of outer measure. Let us let take x any non-empty set A is an algebra of subsets of a set x and let us fix any element x 0 in x any arbitrary element now given any subset A in the algebra either x naught will belong to A or x naught will not belong to A two possibilities. So, if x naught does not belong to A that particular element that you fix does not belong to A put mu of A define mu of A to be equal to 0 and define mu of A equal to 1 if x 0 belongs to A. So, whenever x 0 is in A the measure of that set is 1 otherwise it is equal to 0 show that this is a countable additive set function namely it is a measure because mu of empty set is automatically 0. So, we would like you to characterize that show that the outer measure look at the outer measure induced by this measure show that the outer measure has the property that outer measure of every set is either 0 or 1 again and outer measure of a set is equal to 1 if the element x naught belongs to A that is a property of mu also. So, we want you to show that mu star of A is equal to 1 if x 0 belongs to A can you say that the converse is true namely we would like you to also show that mu star of A is equal to 1 implies x 0 is belongs to A, but that will be you will need the condition that if x 0 belongs to A because x 0 the singleton x 0 may not be in the algebra. So, that makes a difference. So, check look at this exercise and try to prove the facts as for that will help you to understand what is an outer measure how does the outer measure of a set change with given conditions. Let us look at now the problem you are given a measure mu on an algebra A of subsets of a set x you have defined the outer measure which in general is countably sub additive. So, how to pick up sets how to pick up those subsets of x such that mu star restricted to them will become countably additive as happened in the previous example that we discuss. So, for that that is what is called the concept of a measurable set a subset E of x. So, mu is fixed mu on the algebra is fixed mu star is defined by that measure mu. So, mu is fixed. So, we are saying that a set E is mu star measurable mu star is outer measure induced by that measure is called measurable if for every y in x mu star of y can be written as mu star of y intersection E plus mu star of y intersection E complement. So, be careful this says we are saying E is measurable. So, E measurable means take the set E which you want to test whether it is measurable or not. So, divide any set y into two parts y intersection E and y intersection E complement then measure of the two pieces should add up to give you the measure the size of mu star of y. So, this is a picture here that you have got this is my set E. I want to check whether it is measurable or not. So, take this E. So, take any set y. So, this is my y. So, that gives me this piece which is y intersection E and this is the part that is y intersection E complement. So, the requirement is we are saying that E is measurable if using E y is the set cut it into two parts y intersection E and y intersection E complement. So, there are two disjoint pieces of y and we want you should have mu star of y is equal to mu star of y intersection E plus mu star of the other part. That should happen every for every y it should happen for every y that is important. So, that is called this we say that E is measurable if for any subset y divide y into two parts y intersection E and y intersection E complement we want measure of outer measure of y should be some of these two outer measures. So, that is when we say a set E is measurable. So, let us look at some. So, let us we will denote by S star the collection of all mu star measurable subsets. So, whenever a set E satisfies this condition we say that is measurable and let us put all the measurable sets in a collection and call that as S upper star. So, S upper star is the collection of all mu star measurable subsets of x. Here is one observation that a set is measurable if and only if mu star of y we want definition says it should be equal to the sum of the pieces, but it is enough to say that mu star of y is bigger than or equal to mu star of y intersection E plus mu star of y intersection E complement. That is because for every set y, y is equal to y intersection E plus a union y intersection E complement and mu star is always sub additive. So, that means mu star of y is always less than or equal to mu star of y intersection E plus mu star. So, the inequality less than or equal to is always true because mu is monotone. So, to verify whether a set is measurable or not one has to check only that mu star of y should be bigger than or equal to mu star of y intersection E plus mu star of y intersection E complement for every set y. So, only bigger than or equal to has to be checked and so we have to check only bigger than or equal to and in case mu star of a set is infinity, then this inequality is obvious. So, that means one has to check this inequality only for sets for which mu star of y is finite. So, that gives us a very simplification saying that a set E is measurable if and only if for every subset y of x with the property that mu of mu star of y is finite one has to verify that mu star of y is bigger than or equal to mu star of y intersection E plus mu star of y intersection E complement for every subset y with mu star of y finite. So, this is the condition we are going to use again and again to prove whether a set or check whether a set E is mu star measurable or not. So, we are going to now understand the properties of this class S star. What are the properties of this collection of measurable sets? So, the first observation is that every set in the given algebra is measurable. That means if A belongs to the algebra A, then this condition is always going to be true. So, let us verify that. So, let us show that if A belongs to the algebra, then A is measurable and that is for every y is subset of x with mu star of y finite. We should have mu star of y is bigger than or equal to mu star of y intersection E, y intersection A, sorry the set is A. So, y intersection A plus mu star of y intersection A complement. So, this is what we have to show. So, let us look at the proof of this. Now, we are going to use the fact that mu star of y is finite and mu star of y finite means that it is a infremium of some quantities. So, that crucial definition what is definition of infremium we are going to use. .. So, let epsilon greater than 0 be fixed. Then by definition of infremium there exists a covering. So, there exists sets A i belonging to A with the fact that the set y is covered by union of A i's i equal to 1 to infinity and mu star of y which is infremium plus a small number does not remain the infremium. So, it is bigger than or equal to mu of A i, i equal to 1 to infinity. So, here we are using the fact that mu star of y is infremium and that is a finite quantity. Now, A i's are in the algebra, A is in the algebra. So, I can write this as this is equal to sigma i equal to 1 to infinity mu of union of A i's intersection A, the set is A 1 to infinity union of the sets union i equal to 1 to infinity A i intersection A complement. So, what I am saying is sorry not the union this is wrong. So, let me let me just simply write it as. So, let us observe what we are saying we are saying because of this each set A i. So, let us note let us note A i I can write as A i intersection A union A i intersection A complement and these are two disjoint sets and mu is a measure all the A i's A everything is in the algebra. So, using the fact that mu is a measure I can write it as implies that mu of A i is equal to mu of A i intersection A plus mu of A i intersection A complement. So, now mu of A i intersection A complement. So, that means summation mu of A i i equal to 1 to infinity is equal to summation i equal to 1 to infinity mu of A i intersection A plus summation i equal to 1 to infinity mu of A i intersection A complement. Now, let us note that the set A intersection Y is covered by union of A i intersection A i equal to 1 to infinity, because Y is covered by union of A i's. So, A intersection Y is covered by this and A complement intersection Y is union of I equal to 1 to infinity A i intersection A complement. These are sets in the algebra, because A belongs to the algebra. That is a crucial thing. So, this will imply that mu star of A intersection Y is less than or equal to summation of mu A i intersection A complement I equal to 1 to infinity and mu star of the second one gives me A complement intersection Y is also less than or equal to using this summation I equal to 1 to infinity mu of A i intersection A complement. So, look at this equation, look at this equation, look at this equation. So, summation mu star of A i is bigger than this sum and that sum is bigger than or equal to mu star of A intersection Y and this sum is bigger than or equal to mu star of A intersection Y, A complement intersection Y and we had mu star of Y plus epsilon was bigger than this summation. So, that summation supporting these three equations together. So, if we call that earlier equation as 1, call this equation as 2, call this equation as 3 and call this equation as 4, then putting all these four equations together what we have is the following that mu star of Y plus epsilon which was bigger than or equal to summation mu star of A i, i equal to 1 to infinity that is that is equal to actually summation of mu star of A i intersection A i equal to 1 to infinity plus 1 to infinity mu star of A i intersection A complement and that is bigger than or equal to mu star of Y intersection A plus mu star of Y intersection A complement. And now epsilon is arbitrary, so let epsilon go to 0. So, this inequality will be still maintained will imply that mu star of Y is bigger than or equal to mu star of Y intersection A plus mu star of Y intersection A complement and that will imply that A belongs to S star that is A is a measurable set. So, hence we have proved that the algebra A is included in the collection S star that is what we wanted to prove. So, this is the proof of the fact that the algebra A is contained in S star every element of A is measurable. The next property that the class of measurable sets is closed under complementation namely if E is measurable then E complement is also measurable that is obvious because in this criteria if you want to check if E is measurable then this is what we require and to check E complement is measurable the same thing is required because this will become E complement and E complement of complement is E. So, it is the same criteria same equation to be verified. So, obviously because the definition has inbuilt E and E complement symmetric with respect to E and E complement that says the set E is a set E is measurable if and only if it is complement is measurable or the collection S star of measurable sets is closed under complements. Next, we want to check the property. So, the collection of all measurable sets one it includes the class of all subsets in the original algebra A and we want to check now that it is an algebra of subsets of X that means and mu star restricted to S star is finitely additive. So, two things we want to check one S star is an algebra and mu star restricted to S star is finitely additive. So, let us see what we have to check for this. First of all we want to check that S star is an algebra. We have already shown A is inside S star. So, that implies implying the empty set and the whole space that belong to A and hence A is in S star. So, empty set and the whole space belong to it. We just now observed that E belonging to S star implies E complement belongs to S star. So, if E is measurable E complement is measurable that also we have checked. So, let us check the third property namely if E 1 and E 2 belong to S star we want to check this implies E 1 union E 2 also belongs to S star that means union of measurable sets is again measurable. So, this is what we want to check. .. So, let us look at a proof of this. So, to check that E 1 E 2 is measurable we have to check to check for every Y contained in X mu star of Y finite. We have to check that mu star of Y can be written as mu star of Y intersection the set that is E 1 union E 2 plus mu star of Y intersection E 1 union E 2 complement. So, this is the property that we have to check. So, what we will do is we will compute each one of the term and show it is equal to mu star of Y. So, for that we start. So, note E 1 is measurable. So, that implies that mu star of Y we can write it as mu star implies for every Y mu star of Y is mu star of Y intersection E 1 plus mu star of Y intersection E 2. Now, this is important that this happens for every Y. So, I can change Y according to my requirements. So, what I want to do is I will change this Y to Y intersection E 1. See, I want to compute Y intersection E 1 union E 2. So, let us change this Y to that. So, that implies that mu star of Y intersection E 1 union E 2 is equal to here I should replace Y by Y intersection. So, mu star of Y intersection E 1 intersection E 1 union E 2, but E is a subset of E 1 union E 2. So, that is just Y intersection E 1. Is that clear? Because if I replace Y by Y intersection E 1 union E 2, then this intersection with Y 1 with E 1 is just Y intersection E 1 plus what is the second thing? Let us write. So, mu star of sorry this one is E 1 complement. I am sorry we made a mistake saying it is measurable mu star of Y is mu star of Y intersection E 1 plus mu star of Y intersection E 1 complement. Now, when we replace Y by Y intersection E 1 union E 2, so this is same as this plus the second term is Y intersection E 1 union E 2 intersection E 1 complement. So, let us simplify that. So, what we have gotten is the following that mu star of Y intersection E 1 union E 2 that was the left hand side is equal to mu star of Y intersection E 1 plus what is this? Now, E 1 union E 2 intersection E 1 complement. So, when I take E 1, E 1 complement that is going to be empty set. So, this set is nothing but mu star of Y intersection E 2 intersection E 1 complement. So, we have computed mu of Y intersection E 1 union E 2 to be equal to this. Now, I also want to compute what is mu star of Y intersection complement of this? What is the complement of this E 1 union E 2 complement? So, what is that going to be? That is going to be mu star of Y intersection by using our proper De Morgan laws for set theory. This is E 1 complement intersection E 2 complement. So, I want to compute mu star of E 1 complement intersection E 2 complement. How can we compute that? Recall, saying that E 1 was measurable, we had that. So, if I replace Y by Y intersection E 2 complement, then I will get the required set here. So, use this equation. So, since E 1 is measurable, we have mu star of Y. So, we will just keep it here to follow. So, mu star of Y intersection E 2 complement. So, let us look at Y intersection E 2 complement is equal to mu star of Y intersection E 1 intersection E 2 complement plus mu star of Y intersection E 2 complement plus mu star of what will be this set? Y intersection E 2 complement intersection E 1 complement. So, that is what we will have. So, this is what I wanted. Now, let us observe in this equation. All the numbers are real numbers because of the assumption that mu star of Y is finite. So, this is a subset. So, this is finite. This is finite. All are finite numbers. So, I can interchange them. I can take one term on the other side if required. So, let us do that. So, from here, we compute. So, implies mu star of Y intersection E 1 complement intersection E 2 complement. This set is equal to mu star of Y intersection E 2 complement minus, take it on the other side, it is mu star of Y intersection E 1 intersection E 2 complement. So, we have gotten the required quantities. So, we wanted mu star of, we wanted what is mu star of Y intersection E 1, E 2. So, that is lying here and we wanted that is lying here, the second term. So, let us add these two terms. So, adding, so add, call it as this equation as 1, call this equation as 2, add 1 and 2 and that will give you that mu star of Y intersection E 1 union E 2 plus mu star of Y intersection E 1 complement, intersection E 2 complement. So, this is equal to, there we got Y star, sorry, mu star of Y intersection E 1 plus mu star of Y intersection E 2, intersection E 1 complement plus Y mu star of Y intersection E 2 complement minus mu star of Y intersection E 1 intersection E 2 complement. So, this is what we have gotten and we want to check that this should come out to be equal to mu star of Y. Now, let us again try to use, so this is mu of intersection E 2 complement here and that is E 1 intersection E 2 complement. Now, let us observe, till now we have not used anywhere the fact that E 2 is measurable. So, let us try to use that fact that E 2 is also measurable and so that we can simplify this quantity. Now, observe E 2 measurable implies the following fact, we want to simplify this. So, let us look at what is going to be E 2 Y intersection E 2 and Y intersection E 2 complement. So, let us try to E 2 measurable means, so for every Y we have got mu star of Y is equal to mu star of Y intersection E 2 plus mu star of Y intersection E 2 complement because of measurable. Now, I want to use this to compute one of the terms here. So, let us replace Y by Y intersection E 2. So, that implies I can replace this by mu star of Y intersection E 2 will be equal to, that will not give us anything. Let us replace this by Y intersection E 1. So, implies mu star of Y intersection E 1 is equal to mu star of Y intersection E 1 intersection E 2 plus mu star of Y intersection E 1 intersection E 2 complement. So, what is mu star of Y intersection E 1 intersection E 2 complement? That term is here. So, that we want to be the negative sign. So, if I take it on the other side, so that means minus mu star of Y intersection E 1 intersection E 2 complement is equal to, I bring it on the other side. So, that is minus mu star of Y intersection E 1 plus mu star of this term which is Y intersection E 1 intersection E 2. Now, this is what we have reached here. So, this is the value that I was looking for. So, let us put in this value. So, this required quantity I will just take it here is equal to. So, this required quantity is equal to mu star of Y and here is minus mu star of Y. So, those two terms will cancel out. So, let me just write that is mu star of Y intersection E 1 plus mu star of Y intersection E 2 intersection E 1 complement that we already had. So, plus mu star of Y intersection E 2 complement and minus. So, that is equal to minus mu star of from here Y intersection E plus mu star of Y intersection E 1 intersection E 2. Now, these two terms cancel out. So, what we are left with this? So, this is equal to mu star of Y intersection E 2 intersection E 1 complement and Y intersection E 2 intersection E 1. So, look at these two terms. So, these two terms with this Y intersection E 2 intersection E 1 complement plus Y mu star of Y intersection E 1 and E 2. That means, these two terms are nothing but mu star of Y intersection E 2. So, n 1 term is here. So, this is mu star of Y intersection E 2 plus what I am saying is that this plus this term is nothing but mu star of Y. So, this is complement mu star of Y intersection E 2. Is it clear? This term as it is. Now, look at the fact that E 1 is measurable. So, mu star of Y intersection E 2 is mu star of Y intersection E 2 intersection E 1 complement plus mu star of Y intersection E 1 intersection E 2. Now, once again using the fact that E 2 is measurable, that is equal to mu star of Y. So, we have proved the required condition that mu star of Y is equal to mu star of Y. So, we have proved that this is mu star of Y is equal to mu star of Y intersection E 1 union E 2 plus mu star of Y intersection E 1 complement intersection E 2 complement. So, that means we have proved the fact that S is an algebra of subsets of the set X. So, what we have shown is E 1 E 2 belonging to S star imply E 1 union E 2 also belong to S star. Here, let me just comment that this proof looks a bit technical, but it is not so difficult. E 1 measurable gives you one condition that mu star of Y is equal to something. E 2 measurable gives you mu star of Y is equal to something. Now, these sets Y are arbitrary. When we have given E 1 and E 2 are measurable means mu star of Y is equal to mu star of Y intersection E 1 plus mu star of Y intersection E 2 E 1 complement. So, you can change this Y to Y intersection E 1 Y intersection E 2 and so on. So, write down the three equations which are given, write down the equality that will be proved and just manipulate. This is only a simple algebra which is required. So, today what we have done is we have looked at, we have defined the concept of what is called a measurable set for a outer measure mu. We have shown that the original elements of the algebra are already measurable sets and the class of all measurable sets form an algebra. So, we will continue the analysis of this class S star in our next lecture. Thank you.