 Welcome to lecture 38 on measure and integration. Today, we will start looking at the notion of absolute continuity of measures. We have started this concept in the previous lecture, but we will do it in detail in today's lecture. So, we will start with looking at what is called one measure being absolutely continuous with respect to another measure. And then we will prove an important theorem called Radon-Rickardim theorem for absolutely continuous measures. So, let us start recalling what is absolutely continuous measure. So, if two measures mu and nu given on a measurable space XS, we say nu is absolutely continuous with respect to the measure mu. If for any set E in the sigma algebra S, mu of E equal to 0 implies nu of E equal to 0. That means, if a set E has got mu measure 0, then it should imply that mu measure of the set E is also equal to 0. So, in that case, we write this relation by the symbol that nu with this special symbol, which is less than twice printed. So, it is called absolutely continuous with respect to nu and this is denoted by this symbol. And we looked at an example of absolutely continuous measures. So, we said let us take a function f, which is non-negative measurable on a measure space XS mu and let us integrate this function over a set E in the sigma algebra. So, this integral f is fixed, mu is fixed, E is varying and this is a non-negative number, which we denoted by nu of E. And we had shown when we defined the integral for non-negative functions that nu of E is a measure and it has a special property that if mu of E is equal to 0, then nu of E is also equal to 0. So, this measure nu of E, which is defined via the integral of a non-negative function f over a set E implies that this measure nu is absolutely continuous with respect to mu. Let us give a characterization of absolutely continuous measures in terms of what is called epsilon delta definitions, which looks similar to absolute continuity of functions. So, we want to prove the following, namely if mu and nu are two measures such that nu is absolutely continuous with respect to mu, then the following holds that if nu is finite, then for every epsilon bigger than 0, one can find a delta bigger than 0 such that whenever nu of E is less than epsilon, nu of E is less than epsilon whenever mu of E is less than delta. That means, given an epsilon you can find a delta, so that whenever for a set the measure mu of E is less than delta that should imply nu of E is less than epsilon. So, let us prove this result. So, we are given that nu and nu are measures on the measure measurable space X s and nu is absolutely continuous with respect to mu and nu is finite. To show for every epsilon bigger than 0, there is a delta bigger than 0 such that mu of E less than delta should imply nu of E is less than epsilon. So, the proof of this is by contradiction. So, suppose this claim is not true, that will mean what? So, claim is for every epsilon there is a delta, so that means that there is an epsilon such that then for every epsilon bigger than, no that means then there exists epsilon bigger than 0 such that for every delta there exists a set E belonging to the sigma algebra with a property whenever mu of E is less than delta, but mu of E is bigger than or equal to epsilon. So, if we assume that the required claim is not true that will imply that there is a number epsilon bigger than 0, so that for every delta you can choose a set E, of course this E will depend on delta such that mu of E is less than delta, but mu of E is bigger than or equal to epsilon. So, let us apply this result for, so apply this for delta equal to 1 over 2 to the power n. So, that will give us the following, so when I do that, so for every n for every n there exists a set E n belonging to S such that mu of E is E n is less than 1 over 2 to the power n, but mu of E n is bigger than or equal to epsilon. So, this is what we get if we assume our result is not true. Now, let us define a set E n to be equal to union of E n E k from k equal to 1 to k equal to n to infinity and a to be equal to intersection of A n's n equal to 1 to infinity. Let us note the following. The set A n, so mu of A n is less than or equal to sigma mu of E k k equal to n to infinity and mu of E k is less than 1 over 2 to the power k and summation n equal to 1 over 2 to the power n. So, this is going to be equal to 1 over 2 to the power n plus 1. So, mu of A n because mu of E k will be less than 1 over 2 to the power k. So, this says mu of A n is less than or equal to 1 over 2 to the power n and this A n is union of n to infinity. So, as n increases, these A n's are going to be smaller and smaller. So, A n is a decreasing sequence and it decreases to the intersection namely A. So, thus mu being finite, A n decreasing to A will imply, so this will imply mu being finite. So, this will imply that mu of A is equal to limit n going to infinity mu of the set A n. But mu of A n, A n is the intersection of A n. So, A n is union of E k and so let us observe that is bigger than or equal to mu and mu of A n is summation. So, it is bigger than or equal to mu of E n because A n is union of E n's because A n is union of E n's. So, A n includes E n. So, mu of A n is going to be bigger than or equal to mu of E n which is bigger than or equal to epsilon. So, that implies, so this along with this implies that mu of A is bigger than or equal to epsilon. But we just now observed that mu of A n was less than or equal to 1 over 2 to the power n plus 1. So, what does that imply? So, that implies that mu of the set A and because A is equal to intersection of this, so mu of A is less than or equal to mu of A n for every n. So, less than or equal to 1 over 2 to the power n plus 1 for every n. So, that implies that mu of A is equal to 0. So, what we are done is assuming that the required condition does not hold, we have shown that there is a set A, so that mu of A is 0, but mu of A is not equal to 0, it is bigger than or equal to epsilon. So, which is a, so this is the contradiction. So, that implies that what we assumed is not true and hence we have the required claim namely that if mu is absolutely continuous with respect to a measure mu and mu is finite, then for every epsilon bigger than 0, there is a delta bigger than 0 such that mu of A less than delta implies mu of the set A has to be less than or equal to epsilon. So, this is what we have proved. So, we have shown that if mu is finite, then the condition that mu is absolutely continuous with respect to mu implies this required claim. Now, let us prove actually the converse of this statement is also true namely for every epsilon bigger than 0, if there exists a delta such that mu of E is less than epsilon whenever mu of E is less than delta, then mu is absolutely continuous with respect to mu. So, let us prove the converse. So, converse says the following that we have the property that for every epsilon bigger than 0, there is a delta bigger than 0 such that mu of asset E less than delta implies mu of E is less than epsilon and we want to claim that this property means that mu absolutely continuous with respect to mu. Let us suppose, so let mu of E be equal to 0. So, if mu of E is equal to 0, then for every epsilon, for every epsilon whatever delta we take for every delta mu of E is equal to 0 is less than delta. So, that will imply by the given property that mu of E is less than epsilon. So, mu of E is less than epsilon for every epsilon. So, that implies mu of E is equal to 0. So, saying that for every epsilon there is a delta such that mu of E is less than delta implies the mu of E is less than epsilon implies that mu of the measure mu is absolutely continuous with respect to mu. So, the converse is easier and it is true even if mu is not finite. So, we do not use anywhere the fact that mu is finite. So, anyway we require mu to be finite. That means absolute continuity and mu finite implies this condition that for every epsilon there is a delta such that mu of E less than delta implies mu of E less than epsilon. For the converse we do not need this property. So, this is one characterization of absolutely continuous measures. Let us recall that we have defined we have characterized all measures on the real line on the Borel sigma algebra. We had shown that if a measure mu is absolute is given a measure mu on the collection of all Borel measurable sets in the real line there exists a monotonically increasing right continuous function such that for any interval a comma b mu of a b is given by f of b minus f of a. So, that is the that gave us the characterization of all tautably additive set functions on the sigma algebra Borel subsets of real line. Using that we would like to characterize what are all absolutely continuous measures on the real line with respect to the Lebesgue measure. So, the theorem states the following namely let f be a monotonically increasing right continuous function and mu f be the measure induced by this monotonically increasing right continuous function f on B R. Then the claim is that this measure mu f is absolutely continuous with respect to the Lebesgue measure lambda if and only if the function f is absolutely continuous on every bounded interval. So, we would like to prove this result. So, this will give us that what are all measures which are absolutely continuous with respect to the Lebesgue measure it ties up with the property of the function f being absolutely continuous. So, this relates the two concepts absolutely continuous measures and absolutely continuous functions on the real line. So, let us look at a proof of this. So, let us assume that f is monotonically increasing and right continuous and mu f is the measure on the Borel sets which is given by for the left open right close interval A comma B it is defined as if you recall we defined it as f B minus f of A and then we showed that this mu f is the measure is a measure which we called as the measure induced by the function f. So, this is given. So, let mu f be absolutely continuous with respect to the Lebesgue measure to show that the function f is absolutely continuous on every interval say A to B to show that f is absolutely continuous on every interval A to B what we have to do we have to show that for every epsilon bigger than 0 there is a delta. So, to show for every epsilon bigger than 0 there is a delta bigger than 0 such that if you take a sequence of intervals say A i B i are finite disjoint intervals these are finite disjoint intervals in the interval A B with the property with the property that sigma of B i minus A i is less than delta then that should imply that sigma i equal to 1 to n f of B i minus f of A i is less than epsilon. So, this is what we have to show. So, let us fix. So, let epsilon greater than 0 be given. So, let us start with an epsilon bigger than 0 be given. Now, since mu f is absolutely continuous with respect to lambda there exists delta bigger than 0 such that lambda of set E less than delta will imply that mu f of E is less than epsilon. So, this is by the property we just now proved that absolute continuity is equivalent to this property and mu of f on the interval A B is finite such that for every E inside belonging to the Borel sigma algebra of the interval A B this is true. So, in particular if E is a finite disjoint union of intervals A i B i, if A is a finite disjoint union of intervals A i B i inside such that lambda of E such that lambda of E such that lambda of E which is equal to sigma B i minus A i is less than delta. So, that will imply that mu of f of this set E, but mu of f of this set E is nothing but f of B i minus f of A i summation i equal to 1 to n is less than epsilon. That is because we are just now given epsilon we have chosen delta with that property. So, in particular we are applying this for a set E which is a finite disjoint union of intervals A i B i inside the interval A B. So, this will imply mu f and by definition mu f of E which is a finite disjoint union of intervals is nothing but sigma f of B i minus. So, that implies so hence this proves that f on A B to R is absolutely continuous. So, one way we approved that if so we approved that if f of mu of f the measure induced is absolutely continuous with respect to the Lebesgue measure then the corresponding function f the monotonically increasing right continuous function f which is inducing that measure is also absolutely continuous. Let us look at the converse of this statement that we are given that the function f capital F which is monotonically increasing and right continuous is also absolutely continuous. Then we want to show that the corresponding measure mu f is absolutely continuous with respect to the Lebesgue measure. So, let us prove that fact. So, let us prove that. So, assume that mu f is absolutely continuous with respect to lambda to show. So, we just now showed no we have to show the other way round. So, this is not this is just now we have shown. So, assume that f is absolutely continuous to show. So, we have to show that mu f is absolutely continuous with respect to lambda. So, this is what we have to show for the other way round proof. Let us assume. So, let E belong to the R and lambda of E equal to 0. So, to show mu f of E is equal to 0. So, this is what we have to show mathematically. Now, let us observe first of all. So, showing that mu f of E is equal to 0. So, enough to show that mu f of E intersection every interval A to B is equal to 0 for every interval A, B. So, then by we can split E into every interval. So, we can countable additivity will give us that this is 0 for every E. So, let us assume that we have got. So, enough to show that this is equal to 0. Now, since lambda of E is 0. So, lambda intersection A, B is also equal to 0. That means, this set E intersection A, B is a set of Lebesgue measure 0. So, by the properties of sets on Lebesgue measure 0, we can find. So, let us for a given epsilon, for a given epsilon, we can find intervals say A n, B n. We can choose them to be left open, right closed, n bigger than or equal to 1, 2 and so on, such that this set E, which is a null set intersection A, B is covered by these intervals A n, B n and the total length of this and sigma B n minus A n, n equal to 1 to infinity is less than epsilon. So, total length of this is small. This is by the property that the set by the property that E is a null set. So, E intersection A, B is a null set. So, by the definition of after measure, if you like, you can find a sequence of left open, right closed intervals, which cover this set and the total length of these recovering intervals is small, is less than epsilon. So, now let what we want to show? Our aim is to show that the function f is absolutely continuous. So, let us take any number say alpha, let alpha bigger than 0 be given, then we have to, it is enough to show that mu of that thing is equal to. So, what is to be shown? We are given that f is absolutely continuous, then by absolute continuity, by absolute continuity of f, there exist by the absolute continuity of f, there exist some delta bigger than 0 such that whenever say that whenever intervals say intervals A n, B n are disjoint in A B with sigma B n minus A n sum n equal to sum 1 to k, finite number of disjoint intervals A B less than delta will imply that f of B n minus f of A n n equal to 1 to k is less than alpha. This is by the absolute continuity of the function f, which is given to us. Now, we start with given any alpha bigger than 0, we find a delta with this property and now for this delta, we apply our earlier thing that lambda of this was a null set. So, for any given epsilon, we will use this when epsilon is equal to delta, because when epsilon is equal to delta, we will have the summation of these intervals less than delta and that will imply, so that will imply the corresponding, so by, so let us call this thing as star, so this thing as star, so by star with epsilon is equal to delta, so by star with epsilon equal to delta we have, so for, so we will have that sigma of f B n minus f of A n n equal to 1 to k is less than alpha for every k. So, this happens for every k, so that implies that sigma of n equal to 1 to infinity f of B n minus f of A n is also less than alpha, because this is happening for every k, but that implies that mu of A intersection AB, which was mu f of this, which was less than or equal to, because A intersection AB was contained in the union of intervals A n B n, so it is less than or equal to sigma mu f of A n B n, n equal to 1 to infinity and this is equal to sigma n equal to 1 to infinity of f of B n minus f of A n, which is less than alpha. So, what we have shown is that for every alpha mu f of E intersection AB is less than or equal to alpha, so that implies, so this happens, because this is happening for every alpha, so this implies that mu f of E intersection AB is equal to 0, because this is happening for every alpha, so we can let alpha go to 0. So, hence we have shown that lambda of E equal to 0 implies this mu f of A intersection AB is equal to 0 for every AB and so that implies, because this is happening for every AB, so this implies that mu f is of E is equal to 0, so hence mu f is absolutely continuous with respect to lambda. So, this proves the other way round theorem, so this completes the proof of the theorem that mu f is absolutely continuous with respect to lambda, if and only if the function f is absolutely continuous on every bounded interval. So, this completely characterizes and ties up the notion of absolutely continuous measures on the real line with respect to Lebesgue measure with absolutely continuous functions on the real line. Next, we want to prove a theorem called, a theorem called von Neumann theorem, which is a very nice theorem and gives a, it uses very nicely and very intelligently the fact that the dual of the L2 of a measure space is itself. That we called as the Ries representation theorem, namely that if T is a continuous linear functional on L2 of a measure space, then it is essentially given by the inner product, by the inner product. So, this is used very effectively, will not be giving a proof of this theorem, those who are interested can read the text book, but we will see how this theorem is used to prove some results about measures. So, let us first state this theorem called von Neumann theorem. So, the theorem says, let us take two measures mu and nu, which are sigma finite measures on measurable space x s. Then, it says, there exists mutually disjoint sets x i, a measurable sets such that the following properties hold. First of all, this x 1, x 2, x 3 give a partition of the space x. So, x is partitioned into three parts, x 1, union x 2, union x 3. One x 3 and x 1, so nu of x 3 is equal to 0, that means nu has no, for any subset in x 3, the measure of a set is equal to 0. So, that means nu at the most gives values to subsets on x 1 and x 2. On the other hand, mu of x 1 is equal to 0, that means mu gives values to subsets possibly in x 2 and x 3. So, nu of x 1 and nu of x 3 is equal to 0, mu of x 1 is equal to 0. On the set x 2, there is a function g, which is a non-negative measurable function such that for every subset E of measurable subset E of x 2, nu of E can be written as integral of the function g over the set E. So, it in some sense describes, so it says, let us take, this is my set x and we have got two measures mu and nu. It says we can decompose into three parts x 1, x 2 and x 3 into three parts x 1, x 2 and x 3 into three parts x 1, x 2 and x 3. On this part, nu of x 1 is 0, on this part mu of x 3 is equal to 0. So, in this part nu is 0, in this part mu is 0. On this part x 2, this nu of any set E, for any set E here, it can be written as integral over E of g d mu. So, this property holds on this set. So, this is a decomposition of the phononium 1. So, let us just recall. So, the phononium 1 decomposition theorem says that given two measures nu and mu and nu, which are sigma finite of course, we can decompose x into three parts x is equal to x 1, union x 2, union x 3, nu of x 3 is equal to 0, mu of x 1 is equal to 0 and on the middle part x 2, the measure nu can be represented in terms of the measure mu by the property that nu of E is equal to integral g d mu. So, this is called phononium 1's theorem and it is very useful theorem. We will see soon. So, using this theorem, we prove what is called the Lebesgue decomposition theorem. So, Lebesgue decomposition theorem says the following that suppose mu and nu are two sigma finite measures on a measurable space x s, then it implies that there exists sigma finite measures nu A and nu S with the following properties. Namely, this measure nu can be written as a sum of two measures nu lower A and plus nu lower S. So, the measure nu can be written as the sum of two measures nu of A plus nu of S and what are the properties of these two measures? The first property says that measure nu of A is representable in terms of the measure mu via integrals. So, it says nu of A for any subset E is integral of a non-negative measurable function f over the set E. So, it says there exists a unique non-negative measurable function f such that to compute nu of A of any set E, we just integrate f over the set E. The second part says it describes what is the measure nu S. It says there is a set A such that mu of A complement is 0 and nu S of A is equal to 0. That means mu and nu S are setting on disjoint sets. Mu of A complement is 0. So, that means mu sits on A and nu S of A is equal to 0. So, nu S is on A complement essentially. So, this is what is called Lebesgue decomposition theorem. So, we want to show that how it arises out of as an application of von Neumann's theorem. So, this is also part of the consequence that such a decomposition is unique. .. So, let us look at by von Neumann's theorem. Given the measures nu and mu, we have disjoint sets x 1, x 2 and x 3 such that the following operative holds. We recall that x is equal to x 1, union x 2, union x 3 and nu of x 3 is equal to 0 and mu of x 1 is equal to 0. So, mu does not give any mass, it does not give any measure to this set x 1 and nu does not give any measure to this set x 3. On x 2, we recall that for every set E measurable set, if we look at E intersection x 2, then we can compute this measure as an integral of a non-negative measurable function g over the set E intersection x 2. Moreover, recall this is a non-negative, g is a non-negative measurable function and g is equal to 0 on the complement of the set E 2. So, that is the because the measure nu on x 3 is equal to 0. So, this is the by von Neumann's theorem. Now, it is easy to define what should be our measures nu A and so on the set x 2, we should define nu A to be equal to nu on x 2. So, let us define, let us put A equal to x 2 union x 3 and on this set for every set E belonging to S nu of A of E is defined as nu of A intersection E and on the complement of this part. So, nu S of E that is going to be E intersection x 1 because x 1 is the complement of A that is A complement. So, define nu A nu the measure nu as nu of A intersection E restrict nu to E and restrict nu to x 1 to get the measure nu S and now obviously these two are nu S and nu A are measures that is obvious from the definition and also is clear from the definition that the measure nu is nothing but nu A because nu A is on A and nu S is on A complement. So, nu is nothing but nu A plus nu S and the measure nu S on A complement is equal to 0. So, this satisfies the required properties only we have to check that nu of A is given by the integral and that is obvious because nu A of E is nu of E intersection x 2 union x 3 by the definition because A is x 2 union x 3 and on x 2 it is given by E intersection x 2 integral g d mu because nu of x 3 is equal to 0. So, nu of A is given by the integral. So, that proves the theorem except for the fact that at present g is defined only on x 2. So, there is not a issue we can extend it to the whole by putting it equal to 0 then it has the required property that f is defined is non-negative measurable function defined on the whole space x and nu A of E is given by the integral. So, that proves the Lebesgue decomposition theorem and the uniqueness is only a manipulation of the measures. So, which will really leave it as a reading exercise for the reader to verify. So, uniqueness we will assume the uniqueness part of it. So, Lebesgue decomposition. So, let me go back and state understand what is Lebesgue decomposition theorem. It says that given two sigma finite measures on a measure space the measure one of the measures say let us nu can be written as a sum of two measures nu A and nu S where nu A is given by integration and nu S sits on a part which is complement to the part of mu of A. So, this is nu of A is given as a integral and mu of A complement is same as nu S of A is equal to 0. So, this theorem is called Lebesgue decomposition theorem. So, this measure nu S that we have defined just now gives a let us make it as a definition. So, two measures mu and nu are said to be singular with respect to each other or we say mu is singular with respect to nu if there is a set say that mu of the set E is 0 and nu of the E complement is equal to 0. That essentially says we can decompose the space x into two parts E and E complement and mu sits on one part and nu sits on the other part. So, such measures are called singular and it is obvious that if mu is singular with respect to nu then nu is singular with respect to mu. So, it is a commutative relation of singularity while absolute continuity was not. So, nu absolutely continuous with respect to mu need not imply mu is absolutely continuous while the singularity is true that namely if mu is and this singular is also written as mu perpendicular. So, this is also read as singularity is also said to mu is orthogonal or mu is perpendicular to nu and written as mu perpendicular to nu. So, Lebesgue decomposition theorem and now can be stated in terms of this singularity that given two measures which are sigma finite on a measure space x s there exists sigma finite measures nu a and nu s such that the following properties hold namely nu is decomposed into two parts nu a plus nu s where nu of a is absolutely continuous with respect to mu and nu s is orthogonal with respect to mu. So, this is the decomposition that essentially it says that the measure nu has got absolutely continuous part and absolute continuous part says you can obtain nu of a via integration and singular part says that this is the other part which is completely orthogonal to mu. So, they says disjoint says you cannot do anything there they are disjoint sets they set on what one says they are their sports are disjoint essentially. So, this is what is called Lebesgue decomposition theorem and as a consequence of this we will get what is called the radon recursive theorem which characterizes absolutely continuous measures and it says that if two measures mu and nu which are sigma finite on a measurable space x s be such that nu is absolutely continuous with respect to mu then there is a non-negative measurable function f such that nu of e is given via integration of f over the set e. So, that completely characterizes absolutely continuous measures recall if we define nu of e by this then we have already shown that was the beginning of our analysis saying that any measure nu defined in terms of integral with respect to mu is absolutely continuous and this is the converse part of it namely if nu is absolutely continuous with respect to mu then there must be a non-negative measurable function f such that nu of e is equal to integral of f over e with respect to the measure mu. And proof is obvious from the Lebesgue decomposition theorem and the uniqueness part of it because we know that nu is absolutely continuous with respect to mu. So, when we apply Lebesgue decomposition theorem to the measures nu and mu nu will be decomposed into two parts the absolute continuous part and the singular part but there is no singular part. So, there is only absolutely continuous part and absolutely continuous part we have already seen in Lebesgue decomposition theorem is given by integrals. So, this is a direct application of Lebesgue decomposition theorem and this function f is also unique in the sense that if there are two functions then f must be equal to g almost everywhere and that happens because if there is another function g with the same property then integral of f over every set e is equal to integral of g over every set e and that implies that f must be equal to g almost everywhere that we have seen earlier. So, Radon-Degardim theorem the proof is an application of Lebesgue decomposition theorem and Lebesgue decomposition theorem is an application of von Neumann's theorem. So, once again saying that nu is absolutely continuous with respect to mu by Lebesgue decomposition theorem nu of a must be equal to nu because that is absolutely continuous and nu of s is equal to 0. So, that says that nu of e is given by integral over f by the Lebesgue decomposition theorem and as I said we said uniqueness is obvious because integral f is equal to integral g must imply that f is equal to g almost everywhere. So, this is so Radon-Degardim theorem is one of the most important and subtle theorems of our subject because just from the existence of same null sets mu of e equal to 0 implies nu of e equal to 0. This property simple property about null set says that nu must be obtainable from mu via integration. It is really a deep and amazing theorem of our subject and let me also point out there are many proofs available of this theorem. We have given a proof which is via von Neumann's theorem and von Neumann's theorem uses the fact that the dual of L 2 is L 2. There is another purely measure theoretic proof of this, but that goes into the realm of what is called signed measures. So, one looks at signed measures and then one decomposes a signed measure. There is something called a Han decomposition theorem that every signed measure is a difference of two measures and then from there one disuses Lebesgue decomposition theorem and then comes to Radon-Degardim theorem. That is another route possible for proving this theorem. Both the proofs are available in the text and we have outlined here only one proof which is more function theoretic using that dual of L 2 is L 2. This gives the notion of absolute continuity also gives rise to the notion of what is called the derivative. Let us define that. Whenever two measures mu and nu are sigma finite and nu is absolutely continuous with respect to mu, Radon Degardim theorem says that nu of E must be given by a function, a unique function f. This unique function f is called the Radon Degardim derivative of the measure nu with respect to mu and is denoted by d nu by d mu. So, here are some simple properties of this Radon Degardim derivative which are very much similar to the derivative of functions namely if mu 1, mu 2 and nu are sigma finite measures and if mu i each mu 1 and mu 2 is absolutely continuous with respect to mu then mu 1 plus mu 2 is also absolutely continuous and the Radon Degardim derivative of the sum is equal to sum of the Radon Degardim derivatives of course almost everywhere. So, very much similar to the derivative of the sum is equal to sum of the derivatives for functions and similarly if mu 1 is absolutely continuous with respect to mu 2, this is a mistake here and mu 2 is absolutely continuous with respect to mu 1 that means both are absolutely continuous with respect to each other then the product of the derivative is equal to 1. So, that is the second property. So, here it should be mu 2 absolutely continuous with respect to mu and finally if nu is absolutely continuous with respect to mu 1 and mu 1 is absolutely continuous with respect to mu 2 then there is a kind of associative property namely this implies that nu must be absolutely continuous with respect to mu 2 and the derivative. So, d mu 2 over d nu is computable as d mu 2 over d mu 1 multiplied by the derivative d mu 2 over this should be d nu here, this should be d nu. So, it is like a chain rule for the derivative functions. So, with that idea of this set of ideas namely absolutely continuous functions and absolutely continuous measures is complete and in the remaining next lecture we will look at some special properties of sequences of measurable functions and the various ways they can converge to a function f. Thank you.