 Welcome to lecture 11 on measure and integration. In the previous lecture, we had defined what is called an outer measurable subset. And we had started looking at the properties of the outer measurable sets. We will continue that study of properties of the outer measurable sets today. And if time permits in the end, we will specialize the case when this space is the real line. So, let us recall what we have been doing. So, properties of measurable sets we were looking at. Let us just recall what is an outer measurable set. A subset E of X was said to be outer measurable or mu star measurable. If mu star of any set Y is written as, can be written as mu star of Y intersection E plus mu star of Y intersection E complement. So, this condition must be satisfied for every subset Y of X. And then we said, let us denote by S star, the class of all mu star measurable sets. And we gave an equivalent way of verifying when a set is outer measurable. So, the condition is that a set E is measurable if and only if for every subset Y in X with mu star of Y finite, we have the condition 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. So, instead of just saying that for every subset Y, this equality must be true. We have to only verify for those subsets Y of X for which mu star of Y is finite. And instead of equality, we have to verify only bigger than or equal to one way inequality, because the other way round is always true for mu star being countably sub-additive. So, this condition we will use whenever we require. And so, the first observation that we proved last time was that A is the given algebra on which the measure is defined. So, the first claim we proved that every element in the algebra is also a measurable set. So, A is a subset of S star. The second property that we were looking at was that if S star is that S star is an algebra of subsets of X and mu star restricted to S star is finitely additive. We had already observed that a set E is measurable if and only if its complement is measurable. So, S star is closed under complements. Only we have to verify that it is closed under unions. And that proof we were working out in the last time and we had done it. Let us just revise it again, because we are going to need those inequalities. .. So, let us take let E 1 and E 2 be measurable sets to show that E 1 union E 2 is measurable. So, E 1 measurable implies that for every subset Y contained in X and let us also have that special condition less than finite, we know that mu star of Y is equal to mu star of Y intersection E 1 plus mu star of Y intersection E 2 sorry E 1 complement. This is true for every subset Y with that property. Let us replace Y by Y intersection E 1 intersection E 1 union E 2. So, replace this Y. So, then what do we have? We have mu star of Y intersection E 1 union E 2 is equal to… So, Y is replaced by Y intersection E 1 union E 2, but E 1 is a subset of it. So, the first term is just mu star of Y intersection E 1 plus the second term becomes mu star of E 1 union E 2 intersection E 1 complement. The first term will give you only empty set union Y intersection E 2 intersection E 1 complement. So, E 1 complement intersection E 2. So, that is what we get by using the effect that E 1 is measurable. Also, E 2 is measurable. So, thus for every set Y, a corresponding equation holds for E 2 complement, but we will replace Y by Y intersection E 1. So, mu star of Y intersection E 1 complement is equal to mu star of Y intersection E 1 complement intersection E 2 plus mu star of Y intersection E 1 complement intersection E 2 complement. So, using the fact that E 2 is measurable, we have written mu star of Y intersection E 1 as the set intersection E 2, the set intersection E 2 complement. Now, so in these two equations, look at this set, this term Y intersection E 1 complement intersection E 2 that is also sitting here. So, we will compute the value of this and put it in that equation. So, let us do that. So, from this second equation, we put the value there. So, we have got. So, from these two equations, thus mu star of Y intersection E 1 union E 2 that is the left hand side of this equation. That is the left hand side of this equation. So, that is equal to the first term. So, that is mu star of Y intersection E 1 plus the Y intersection E 1 complement intersection E 2 is equal to mu star of Y intersection E 1 minus that thing. So, mu star of Y intersection E 1 complement minus mu star of Y intersection E 1 complement intersection E 2 complement. Now, one should note down, note here that we have taken one term on the other side. So, this is possible because all the sets involved have finite outer measure. So, this is the equation of real numbers. So, we can take one term on the other side and so on. In general, that will not be possible if one of the terms is equal to plus infinity. So, the condition that mu star of Y is finite is being used here. So, we get using the fact that E 1 and E 2 are measurable, we get this equation. So, from here, let us take this negative term on the other side. So, implies mu star of Y intersection E 1 union E 2 plus mu star of Y intersection E 1 complement intersection E 2 complement is equal to mu star of this term Y intersection E 1 and the second term plus mu star of Y intersection E 1 complement. Now, using the fact that E 1 is measurable, this is same as mu star of Y. So, we have shown that for every subset Y with mu star of Y finite, its measure mu star of Y can be written as mu star of E 1 Y intersection E 1 union E 2 plus mu star of Y intersection E 1 complement intersection E 2 complement. But note, this set is nothing but E 1 union E 2 complement. So, this implies that E 1 union E 2 is measurable. So, that says E 1 union E 2 is measurable. Now, for the special case, if E 1 and E 2 are disjoint, that means E 1 intersection E 2 is empty set, that is same also implies that E 1 is contained in E 2 complement or E 2 is contained in E 1 complement. Either one is true. So, note this is true. So, in that case, let us go back and look at the first equation that we had. We had because E 1 and E 2 were measurable. So, we had this condition. So, in this equation, this is true for every Y. So, let us replace this Y by E 1 union E 2. That will give us the measure mu star of E 1 union E 2. So, in this, we are going to replace Y by E 1 union E 2. So, let us just put that equation here and look at what we are doing. So, in this equation, we are putting Y equal to… So, in this star, so in star put Y equal to E 1 union E 2. Keep in mind they are disjoint. So, the left hand side will be mu star of E 1 union E 2 equal to right hand side. The first term is mu star of E 1 plus in the second term because E 1 E 2 are disjoint, E 2 is a subset of E 1 complement. So, this implies E 2 is a subset of E 1 complement. So, that means this is nothing but plus mu star of E 2, E star of E 2. So, when E 1 and E 2 are disjoint, mu star of E 1 union E 2 is mu star of E 1 plus mu star of E 2. So, that means, thus mu star is finitely additive. So, we have proved that whenever our… So, we have proved this property namely, S star is an algebra of subsets of X and mu star restricted to S star is finitely additive. Next step is to go a bit further and we want to prove that whenever you got a sequence of sets in S star which are pair wise disjoint, then their union is also in S star and mu star of the union is equal to summation of mu stars of E n. That means, we are going to show that S star is close under pair wise disjoint union of sets, even countably infinite and mu star is countably additive. So, let us prove this property. .. So, let us take… So, a n belong to S star, n equal to 1 to n and so on, pair wise disjoint. That is, a n intersection a m is empty for n not equal to m. So, we start a 1 belonging to S star, a 1 measurable implies that mu star of any set Y can be for every Y contained in X. I can write this to be equal to mu star of Y intersection a 1 plus mu star of Y intersection a 1 complement. Now, use the fact that a 2 is measurable. So, leave the first term as it is Y intersection a 1 plus a 2 is measurable. So, measure of mu star of this set can be written as mu star of a 1 complemented intersection a 2 plus mu star of Y intersection a 1 complement intersection a 2 complement. So, this term mu star of Y intersection a 1 complement is written as mu star of Y intersection a 1 complement plus intersection a 2 plus mu star of Y intersection a 1 complement intersection a 2 complement. So, here we have used the fact that a 2 is measurable. Now, observe that a 1 and a 2 are disjoint. So, a 2 will be a subset of a 1 complement. So, this set is nothing but Y intersection a 2. So, I get this is same as mu star of first term a 1. The second term is mu star of mu star of Y intersection a 2 and the third term as it is mu star of Y intersection a 1 complement intersection a 2 complement. So, in the first we use a 1 is measurable, in the second we use a 2 is measurable and use a 1 and a 2 are disjoint. We continue this process. If we continue this process after n steps we will have this is equal to the second step gives you mu star of Y intersection a 1 plus mu star of Y intersection a 2. So, after n steps this will have mu star of Y intersection a i i equal to 1 to n plus one term will be there which is mu star of Y intersection a 1 complement intersection up to a n complement. So, let us write this last term in terms of union. So, this is equal to summation of i equal to 1 to n mu star of Y intersection a i plus mu star of Y intersection union a i i equal to 1 to n complements. So, this term is represented in terms of complements of the unions. So, this is after n steps. So, for every n we have got mu star of Y can be written as this and now this is true for every n. So, here I would like to write this union as 1 to infinity. If I do that I will be making this set bigger and hence the complements will be a smaller set. So, replacing this set if I replace this by Y intersection union i equal to 1 to infinity of a i complement this set is bigger than this is smaller than this set. So, mu star of this will be bigger than mu star of this. So, if I write mu star of this, so this term is bigger than this term. So, this will be bigger than or equal to summation i equal to 1 to n this term as it is Y intersection a i plus this. So, what we have done in the second term where it is union 1 to n I have taken union 1 to infinity and because of complements this term will be smaller. So, instead of equality I have got the inequality and this happens for every n. So, I can let n go to infinity. So, this will be bigger than or equal to summation i equal to 1 to infinity mu star of Y intersection a i plus mu star of Y intersection union i equal to 1 to infinity a i complements. And now mu star is countable a sub additive. So, this term the first term is bigger than or equal to mu star of Y intersection union a i i equal to 1 to infinity second term as it is. So, mu star of Y intersection union 1 to infinity a i is complement. So, using the fact that for every n a n is a measurable set we are able to say that mu star of Y is bigger than or equal to mu star of Y intersection the union a i plus mu star of Y intersection the complement of the unions. So, that implies that union a i 1 to infinity belongs to a star is a measurable set. Not only that you can say something more actually. So, let us in this equation. So, this equation star this one let us put Y is equal to. So, in star let us put Y equal to union of a i's. So, what we will get? So, let us do that substitution and see what do we get. So, in this equation in star take Y equal to union of a i's then left hand side is mu star of union a i 1 to infinity is bigger than or equal to summation i equal to 1 to infinity mu star of Y is unions of a i's. So, this is just a i plus this is union and is a complement that is empty set mu star of that is equal to 0. So, that is equal to 0. So, what we get is mu star of the union of a i's is bigger than or equal to this also by sub additivity mu star of the union a i's is less than or equal to summation 1 to infinity mu star of a i's. So, these two implies that mu star is countably additive on s star. So, this is very nice. So, we have got the following property that if so what we have done till now is we have shown that s star as a consequence of all these properties now we can say that s star the class of all measurable sets is a sigma algebra of subsets of s of x and mu star on this is countably additive. So, we started with a measure mu on a algebra a of subsets of a set x we defined an outer measure via this on all subsets of set x then we picked up a subclass namely s star of sets which are mu star measurable and we have shown that mu star which in general is countably sub additive is actually countably additive on s star the sigma algebra of measurable sets. So, it is a sigma algebra why it is a sigma algebra because we have already shown it is an algebra and it is closed under countable disjoint unions. So, any collection any algebra which is closed under countable disjoint unions is automatically a sigma algebra that we have shown. So, this gives us a way of defining measures on extending measures. Before doing that let us observe one more thing let us look at sets e in x whose outer measure is 0 these are called sets of null outer measure measurable sets. So, the claim is every set whose outer measure is 0 is automatically measurable. So, let us check that. So, let e be a subset of x and mu star of e equal to 0 then for every y contained in x mu star of y intersection e is equal to 0 because y intersection e is contained in e and mu so because and mu is mu star is monotone. So, this is 0. So, thus mu star of y is given is bigger than or equal to mu star of y intersection e complement because again y intersection e complement is a subset of this and I can add 0 to it. So, that is equal to mu star of y intersection e complement plus mu star of y intersection e and that is precisely saying that the set e is measurable. So, that shows that the class of mu star null sets are also measurable. So, this class n is inside s star. So, let us summarize the process now what we have gotten. So, let us start we said that let us start with a measure mu on a algebra a mu is a measure a is a algebra of subsets of the set x then if mu is sigma finite then there exists a unique extension of mu to the sigma algebra generated by it and how do we conclude that. So, the conclusion for that is as follows. So, mu is on the algebra and it is sigma finite given. So, we define from it outer measure mu star which is defined on all subsets of x it is countably sub additive and we picked up the class of measurable sets s star. So, if we restrict mu to this. So, let us call it as mu bar there is a restriction of mu to the smaller class s star keep in mind s star is the class of measurable sets and so what is mu star mu bar. So, mu bar is equal to mu star restricted to s star then this is a measure s star is a sigma algebra and this is a measure and we know that this is an extension. So, from mu we come to mu bar an extension of mu from the algebra to s star and note a is all sets in a are measurable. So, the sigma algebra is also inside here. So, a is inside s of a which is inside s star which is inside all subsets of x. So, mu is defined here we get mu star here and we restrict we get mu bar and that is same as mu bar on s of a. So, we get a measure mu bar on s of a. So, that is same as mu bar on s of a. So, what is mu bar? mu bar is the restriction of the outer measure mu star to the sigma algebra generated by a and that is inside the class of measurable set. So, it is a well defined measure and because mu is sigma finite supposing there were another extension by some other method to the sigma algebra. Then by the uniqueness of measures on the sigma algebras we know that there is only one possible extension that we have already proved that in case an extension exists if two measures agree on the algebra they will also agree on the sigma algebra provided they are sigma finite. So, uniqueness follows from that theorem. So, we have got that if mu is a sigma finite measure on an algebra then we can extend it to the sigma algebra generated by it. So, this is the extension process. So, one has to start with a measure mu on an algebra and recall we already have extended it from a semialgebra to the algebra generated. So, essentially it says that if we have a measure on a semialgebra of subsets of a set x and the measure mu is sigma finite then it can be uniquely extended to a sigma finite measure on the sigma algebra generated by that algebra. In fact, we have proved something more. So, we have actually shown that not only mu which is defined on the algebra extends to s of a the sigma algebra generated by it. Actually it extends to a class s star which not only includes s of a it includes also the class of null sets mu star null sets sets of outer measure mu 0. So, let us denote the class of the sigma algebra generated by s the sigma algebra generated by s of a and the null sets by a new name. So, what we are saying is one can show that this s star the class of all outer measurable sets which is a sigma algebra which includes s of a also includes n. So, it includes this union. So, this union we are writing it as sets of the type. So, s a union n it is not the union of these two classes it denotes sets of the type e union n where e belongs to s of a and n is a null set. So, take sets which are in the sigma algebra generated by a adjoin to it any null set mu star null set. So, look at this new collection then one can show that s star is same as e union n. So, it involves two things name one namely this collection is a sigma algebra and this sigma algebra is same as s star. We will not go into the details of this they are slightly technical we will assume this. So, but we will this gives us a new notion. So, let us define that. So, let x be a non-empty set s a sigma algebra of subsets of the set x the pair x comma s now onwards will be called a measurable space. So, a measurable space is a pair where x is a set and s is a sigma algebra of subsets of it. And suppose we are given. So, this elements of s normally are called measurable sets and so let us next suppose that we are given a measurable space x s and we are given a measure on the sigma algebra s then we get a triple x s and mu that is called a measure space. So, a measure space signifies a triple a ordered triple where the first element x is a set x the second one is a sigma algebra of subsets of the set x and mu is a function defined on the sigma algebra taking non-negative values and it is countable additive it is a measure. So, this triple is called a measure space. So, what we have done our extension process we can now summarize it as follows given a measure on an algebra a of subsets of a set x what we did we constructed two measure spaces one was x s of a the sigma algebra generated by it mu star which is a outer measure induced by mu and we know mu star on s of a is a measure and we also have the measure space x s star and mu star mu star on s star the class of all outer measurable sets. So, we get these two measure spaces keep in mind s of a is a subset of s star and we gave the relation between these two namely the measure space this is in some sense we can say it is a bigger measure space because the sigma algebra s star is bigger than s of a and this measure space has a special property namely that if we take any set e in x and mu star of e is 0 then e belongs to s star. So, not only so for example, this is a very special thing supposing you take any subset a of e then by monotone property mu star of a also will be 0. So, that also will be inside s star. So, s star includes all mu star null sets such a measure normally is called a complete measure space. So, our construction has given the measure space x s star mu star and it is a complete measure space namely all sets of outer measure 0 are elements of s star that is a nice condition to have we will see it a bit later on. So, this is called a complete measure space. So, a complete measure space is a space such that the sigma algebra s star or sigma algebra includes all null sets all sets whose measure is 0 in general a measure space need not be complete. So, for example, in particular for example, this measure space need not be complete in general. So, there is a theorem which says every measure space x s mu can be completed and this process of completion of a measure space is a slightly technical one. The basic idea is given a measure mu on a algebra s of subsets of a set x collect together all sets whose outer measure mu star is 0 and adjoin them add them to the sigma algebra s that means generate a new sigma algebra by looking taking together s and the sets which are null sets. So, that gives a bigger sigma algebra and on that bigger sigma algebra one can show we can extend that measure mu to the sigma algebra and the new measure space becomes complete. So, if the process is very much similar to looking at x s of a and mu star and visa v x s star and mu star. So, we will assume this theorem that every measure space x s mu can be completed. So, if you are interested in looking at the technical details for this look at the text book which we mentioned in the first lecture namely an introduction to measure and integration by me. So, we will leave these details for those who feel more interested in looking at the details. Next we will give there are some equivalent ways of describing the set mu star of e and that is that mu star of e can be also written as infremium of mu star of a where a belongs to s of a and all e are inside a. So, look at all elements from the sigma algebra generated by a which include that set e and look at the mu star of a and take the infremium of them. So, in some sense mu star of a set can be approximated by sets from mu star from elements of s of a and a similar result is true for belonging to elements which are measurable sets. So, these are technical things. So, these are some facts which will not prove and most probably will not be using them in our course, but they are nice to know that relation between mu star of e and mu star of sets in the sigma algebra s of a and s star of a. Here is another fact which again will not be proving and most probably will not be using namely that for every subset e in x you can find a set in the sigma algebra s of a the sigma algebra generated by a such that the set e is a subset of f. So, f is which includes e and the outer measure of the two are same and that also in turn implies that outer measure of f minus e is 0. So, essentially it says for every set e contained in x there is a set in the sigma algebra s of a such that the difference has got outer measure 0 such a set is called a measurable cover of e. So, such a set because f covers e and a similar results for a set inside. So, if e is in x then you can find a set k inside e say that the difference mu star of e minus k is 0 and such a set is called a measurable kernel of e. So, given any set e there is a cover by a measurable set and there is a smaller set inside which is a kernel. So, a difference is of sets of measure 0. So, these things will not prove will prove a result which will be needing later on and that relates the outer measure with measure of the set inside the algebra that we have started with. So, mu start with a measure mu on algebra a of subsets of a set x let mu star be the induced outer measure. So, suppose we have got a set e such that mu star of e is finite. This set need not be in the sigma algebra. So, take any set e say that mu star of e is so we do not need this condition that e should be a measurable set. So, take any set whose outer measure is finite then given any epsilon you can find a set in the algebra a such that mu star of e symmetric difference that set f epsilon is less than epsilon. So, this is a very nice result which says any set of finite outer measure as I said this condition is not there it is not needed for any it is a typo for any set of finite outer measure. You can find a set in the algebra such that mu star of e symmetric difference the measure of the symmetric difference is small. So, let us look at a proof of this result. So, what we are saying is so let us take a set e contained in x with the condition that mu star of e is finite. So, it says given epsilon bigger than 0 there exists a set f epsilon belonging to algebra such that mu star of e symmetric difference with f epsilon is less than epsilon. Let us see what we are saying. So, we are saying that this is a set e it says given a set e with the condition that mu star of e is finite. I can find a set call this as f epsilon such that what is the symmetric difference? Symmetric difference is e minus and f minus. So, that is the portion. So, this portion is f epsilon symmetric difference e. So, it says that the sets so this is the common portion. So, it says that the measure outer measure of the sets where which are outside the common portion is small. So, essentially almost you can say that e and f epsilon are same. So, let us prove this property. So, to prove this let us observe that mu star of so mu star of e is finite and what is mu star? Mu star of e if you recall mu star of e is equal to infimum of sigma mu of a i i equal to 1 to infinity where this a i's union of a i's cover. So, union of a i's cover the set e and a i's in the algebra. So, this being finite. So, given epsilon a small quantity bigger than 0 there exists a covering. So, there exists sets a i belonging to the algebra such that e is contained in union of a i's and mu star of e which is infimum plus the small number is bigger than sigma mu of a i's. That is by the definition of the infimum. Infimum is finite so this. So, note this implies because this is finite. So, this implies that the series i equal to 1 to infinity mu of a i is finite. So, there exists as a consequence of this there exists some n naught such that tail of the series so n naught plus 1 to infinity mu of a i is less than say epsilon by 2 that is because the series is convergent. So, once that is done we define. So, let us define the set f epsilon to be equal to union of a i i equal to 1 to the stage n naught. So, note this set belongs to the algebra because it is a finite union of elements in the algebra. So, it belongs to the algebra. So, let us calculate look at the set e minus f epsilon. So, what is that? So, that is e minus union i equal to 1 to n naught a i. Now, the set e is contained in union i equal to 1 to infinity a i. So, this is contained in this minus union i equal to 1 to n naught a i. So, I can say this is contained in union i equal to n naught plus 1 to infinity of a i. So, that implies that mu star of e minus f epsilon is less than or equal to mu star of this set union i n naught plus 1 to infinity a i and which by sub additive. So, this was subset of this. So, mu star of this is less than or equal to by monotone property and by sub additive property this is less than or equal to sigma i equal to n naught plus 1 to infinity mu star of a i and that if you recall we have less than epsilon by 2. So, this is less than epsilon by 2. So, we get that mu star of e minus f epsilon is less than epsilon by 2. Let us also compute the measure of the other part namely we want to compute also mu star of f epsilon minus e. Let us compute that we want to compute what is this equal to. So, f epsilon minus e is union i equal to 1 to n naught a i minus e and note. So, this is a subset of union of i equal to 1 to infinity a i minus e and e is a subset of this. So, that implies that mu star of f epsilon minus e is less than or equal to mu star of this. That is sigma i equal to 1 to infinity mu of a i minus mu star of e and that if you recall is by the way we started we had summation mu star of a i this relation. So, this says sigma mu of a i minus mu of e is less than epsilon. So, we could have started with if required by epsilon by 2. So, then we have gotten this is less than epsilon by 2. So, we are getting that mu star of f epsilon minus e is less than epsilon by 2 and we have already shown that mu star of we have already shown that mu star of f minus e epsilon is less than epsilon by 2. So, putting these two together. So, call this as 1 call this as 2. So, by putting 1 and 2 together mu star of e delta of epsilon is less than or equal to mu star of e minus f epsilon plus mu star of f epsilon minus e and both of them are less than epsilon by 2 plus epsilon by 2 which is equal to epsilon. So, that proves the required property which we want you to prove namely that given epsilon bigger than 0 there is a set f epsilon which is in the algebra a such that mu star of e delta f epsilon is less than epsilon. So, this is an approximation property which we will be using later on to prove some facts. So, this is the process of extension theory. So, the process of extension theory gives us ways of constructing triples which are measure spaces and at this point it is worth mentioning there are measure spaces of importance in other subjects called probability theory. A measure space x s mu where mu of x is 1. So, that is a totally finite measure and mu of the whole space is equal to 1 is called a probability space and the measure mu is called a probability. So, a measure space where mu of x is 1 is called a probability space and mu is called a probability. The reason for this terminology is that such triples play a fundamental role in axiomatic theory of probability. Whenever you want to describe a phenomena a statistical phenomena which depends upon some randomness one has to construct a probability space to analyze it. So, this gives a mathematical model in the theory of probability to analyze statistical experiments. So, there so let me just give you a few things more. The set x denotes in the triple x s mu x represents a set of all possible outcomes of the experiment. For example, you are tossing a coin. So, all possible outcomes are head or tail or you are throwing a die and there are 6 possible outcomes the number 1, 2, 3, 4, 5 and 6 or you are observing the temperature of a particular place every day at a particular time. So, the observation will be a real number. So, in any particular experiment the all possible outcomes of that experiment are they constitute a set and that is a set x and all the sigma algebra s represents the collection of events of interest in that experiment. So, any subset of the set of outcomes in the experiment is called an event. So, for example, when you are tossing a coin there are 2 outcomes possible head and tail. So, if you look at the singleton h that is an event when you toss head can come or you toss tail can come or if you are throwing a die then the outcomes possible are 1, 2, 3, 4, 5 and 6. Look at the subset 1, 3 and 5 of x the set of all odd outcomes. So, when you throw is possible to find out whether that event has occurred or not that means whether the outcome was the odd number or not. So, that is a subset of the set of all possible outcomes. So, in general when you want to describe its statistical experiment one has to construct a class of subsets of that set x of interest and one requires because of mathematical considerations that that class should be a sigma algebra. So, the sigma algebra represents the collection of events of interest in that particular experiment and finally, for every event e for every event e of interest you want to assign some what is the possibility of that event happening a probability of that event taking place. So, a probability is a measure defined on the sigma algebra of all possible events of interest and taking non negative values and of course, probability of the whole space the chance of the whole space happening is 1 and probability of the empty set is 0. So, the probability is a set function defined on the collection of all events of interest and we want that to be a measure. So, that is the reason that the triple x s mu is called a probability space and gives a mathematical model for analyzing statistical experiments when mu of x is equal to 1. So, in till today's lecture we have constructed measure spaces and from the next lecture onwards we will specialize this measure space when x is real line and that gives a important example of a measure space and a measure called Lebesgue measure. So, we will do that in the next lecture. Thank you.