For these notes, is a fixed measurable space. We shall often omit the -algebra , and simply refer to elements of as *measurable sets*. Unless otherwise indicated, all subsets of X appearing below are restricted to be measurable, and all functions on X appearing below are also restricted to be measurable.

We let denote the space of measures on X, i.e. functions which are countably additive and send to 0. For reasons that will be clearer later, we shall refer to such measures as *unsigned* measures. In this section we investigate the structure of this space, together with the closely related spaces of signed measures and finite measures.

Suppose that we have already constructed one unsigned measure on X (e.g. think of X as the real line with the Borel -algebra, and let m be Lebesgue measure). Then we can obtain many further unsigned measures on X by *multiplying* m by a function , to obtain a new unsigned measure , defined by the formula

. (1)

If is an indicator function, we write for , and refer to this measure as the *restriction* of m to A.

**Exercise 1.** Show (using the monotone convergence theorem) that is indeed a unsigned measure, and for any , we have . We will express this relationship symbolically as

. (2)

**Exercise 2.** Let m be -finite. Given two functions , show that if and only if for m-almost every x. (Hint: as usual, first do the case when m is finite. The key point is that if f and g are not equal m-almost everywhere, then either f>g on a set of positive measure, or f<g on a set of positive measure.) Give an example to show that this uniqueness statement can fail if m is not -finite. (Hint: take a very simple example, e.g. let X consist of just one point.)

In view of Exercises 1 and 2, let us temporarily call a measure *differentiable* with respect to m if (i.e. ) for some , and call f the *Radon-Nikodym derivative* of with respect to m, writing

; (3)

by Exercise 2, we see if is -finite that this derivative is defined up to m-almost everywhere equivalence.

**Exercise 3. ** (Relationship between Radon-Nikodym derivative and classical derivative) Let m be Lebesgue measure on , and let be an unsigned measure that is differentiable with respect to m. If has a continuous Radon-Nikodym derivative , show that the function is differentiable, and for all x.

**Exercise 4.** Let X be at most countable. Show that every measure on X is differentiable with respect to counting measure .

If every measure was differentiable with respect to m (as is the case in Exercise 4), then we would have completely described the space of measures of X in terms of the non-negative functions of X (modulo m-almost everywhere equivalence). Unfortunately, not every measure is differentiable with respect to every other: for instance, if x is a point in X, then the only measures that are differentiable with respect to the Dirac measure are the scalar multiples of that measure. We will explore the precise obstruction that prevents all measures from being differentiable, culminating in the *Radon-Nikodym-Lebesgue theorem* that gives a satisfactory understanding of the situation in the -finite case (which is the case of interest for most applications).

In order to establish this theorem, it will be important to first study some other basic operations on measures, notably the ability to *subtract* one measure from another. This will necessitate the study of *signed* measures, to which we now turn.

[The material here is largely based on Folland’s text, except for the last section.]

– Signed measures –

We have seen that if we fix a reference measure m, then non-negative functions (modulo m-almost everywhere equivalence) can be identified with unsigned measures . This motivates various operations on measures that are analogous to operations on functions (indeed, one could view measures as a kind of “generalised function” with respect to a fixed reference measure m). For instance, we can define the sum of two unsigned measures as

(4)

and non-negative scalar multiples for by

. (5)

We can also say that one measure is less than another if

for all . (6)

These operations are all consistent with their functional counterparts, e.g. , etc.

Next, we would like to define the difference of two unsigned measures. The obvious thing to do is to define

(7)

but we have a problem if and are both infinite: is undefined! To fix this problem, we will only define the difference of two unsigned measures if at least one of them is a finite measure. Observe that in that case, takes values in or , but not both.

Of course, we no longer expect to be monotone. However, it is still finitely additive, and even countably additive in the sense that the sum converges to whenever are disjoint sets, and furthermore that the sum is absolutely convergent when is finite. This motivates

Definition 1.(Signed measure) A signed measure is a map such that

- ;
- can take either the value or , but not both;
- If are disjoint, then converges to , with the former sum being absolutely convergent if the latter expression is finite. [Actually, the absolute convergence is automatic from the Riemann rearrangement theorem. Another consequence of 3. is that any subset of a finite measure set is again finite measure, and the finite union of finite measure sets again has finite measure.]

Thus every unsigned measure is a signed measure, and the difference of two unsigned measures is a signed measure if at least one of the unsigned measures is finite; we will see shortly that the converse statement is also true, i.e. every signed measure is the difference of two unsigned measures (with one of the unsigned measures being finite). Another example of a signed measure are the measures defined by (1), where is now signed rather than unsigned, but with the assumption that at least one of the signed parts , of f is absolutely integrable.

We also observe that a signed measure is unsigned if and only if (where we use (6) to define order on measures).

Given a function , we can partition X into one set on which f is non-negative, and another set on which f is negative; thus and . It turns out that the same is true for signed measures:

Theorem 1.(Hahn decomposition theorem) Let be a signed measure. Then one can find a partition such that and .

**Proof. **By replacing with if necessary, we may assume that avoids the value .

Call a set E *totally positive* if , and *totally negative* if . The idea is to pick to be the totally positive set of maximal measure – a kind of “greedy algorithm”, if you will. More precisely, define to be the supremum of , where E ranges over all totally positive sets. (The supremum is non-vacuous, since the empty set is totally positive.) We claim that the supremum is actually attained. Indeed, we can always find a maximising sequence of totally positive sets with . It is not hard to see that the union is also totally positive, and as required. Since avoids , we see in particular that is finite.

Set . We claim that is totally negative. We do this as follows. Suppose for contradiction that is not totally negative, then there exists a set in of strictly positive measure. If is totally positive, then is a totally positive set having measure strictly greater than , a contradiction. Thus must contain a subset of strictly larger measure. Let us pick so that , where is the smallest integer for which such an exists. If is totally positive, then we are again done, so we can find a subset with , where is the smallest integer for whch such a exists. Continuing in this fashion, we either stop and get a contradiction, or obtain a nested sequence of sets in of increasing positive measure (with ). The intersection then also has positive measure, hence finite, which implies that the go to infinity; it is then not difficult to see that E itself cannot contain any subsets of strictly larger measure, and so E is a totally positive set of positive measure in , and we again obtain a contradiction.

**Remark 0.** A somewhat simpler proof of the Hahn decomposition theorem is available if we assume to be finite positive variation (which means that is bounded above as E varies). For each positive n, let be a set whose measure is within of . One can easily show that any subset of has measure , and in particular that has measure for any . This allows one to control the unions , and thence the lim sup of the , which one can then show to have the required properties. One can in fact show that any signed measure that avoids must have finite positive variation, but this turns out to require a certain amount of work.

Let us say that a set E is *null *for a signed measure if . (This implies that , but the converse is not true, since a set E of signed measure zero could contain subsets of non-zero measure.) It is easy to see that the sets given by the Hahn decomposition theorem are unique modulo null sets.

Let us say that a signed measure is *supported* on E if the complement of E is null (or equivalently, if . If two signed measures can be supported on disjoint sets, we say that they are *mutually singular* (or that is *singular with respect to* ) and write . If we write and , we thus soon establish

**Exercise 5. **(Jordan decomposition theorem) Every signed measure an be uniquely decomposed as , where are mutually singular unsigned measures. (The only claim not already established is the uniqueness.) We refer to as the *positive and negative parts *(or *positive and negative variations*) of .

This is of course analogous to the decomposition of a function into positive and negative parts. Inspired by this, we define the *absolute value* (or *total variation*) of a signed measure to be .

**Exercise 6.** Show that is the minimal unsigned measure such that . Furthermore, is equal to the maximum value of , where ranges over the partitions of E. (This may help explain the terminology “total variation”.)

**Exercise 7.** Show that is finite for every E if and only if is a finite unsigned measure, if and only if are finite unsigned measures. If any of these properties hold, we call a *finite measure*. (In a similar spirit, we call a signed measure *-finite* if is -finite.)

The space of finite measures on X is clearly a real vector space, and is denoted .

– The Lebesgue-Radon-Nikodym theorem –

Let m be a reference unsigned measure. We saw in the introduction that the map is an embedding of the space of non-negative functions (modulo m-almost everywhere equivalence) into the space of unsigned measures. The same map is also an embedding of the space of absolutely integrable functions (again modulo m-almost everywhere equivalence) into the space of finite measures. (To verify this, one first makes the easy observation that the Jordan decomposition of a measure given by an absolutely integrable function f is simply .)

In the converse direction, one can ask if every finite measure in can be expressed as for some absolutely integrable f. Unfortunately, there are some obstructions to this. Firstly, from (1) we see that if , then any set that has measure zero with respect to , must also have measure zero with respect to . In particular, this implies that a non-trivial measure that is singular with respect to m cannot be expressed in the form .

In the -finite case, this turns out to be the only obstruction:

Theorem 2.(Lebesgue-Radon-Nikodym theorem) Let m be an unsigned -finite measure, and let be a signed -finite measure. Then there exists a unique decomposition , where is measurable and . If is unsigned, then f and are also. If is finite, lies in and is finite.

**Proof. ** We prove this only for the case when are finite rather than -finite, and leave the general case as an exercise. The uniqueness follows from Exercise 2 and the previous observation that cannot be mutually singular with m for any non-zero f, so it suffices to prove existence. By the Jordan decomposition theorem, we may assume that is unsigned as well. (In this case, we expect f and to be unsigned also.)

The idea is to select f “greedily“. More precisely, let M be the supremum of the quantity , where f ranges over all non-negative functions such that . Since is finite, M is finite. We claim that the supremum is actually attained for some f. Indeed, if we let be a maximising sequence, thus and , one easily checks that the function attains the supremum.

The measure is a non-negative finite measure by construction. To finish the theorem, it suffices to show that .

It will suffice to show that for all , as the claim then easily follows by letting be a countable sequence going to zero. But if were not singular with respect to m, we see from the Hahn decomposition theorem that there is a set E with such that , and thus . But then one could add to f, contradicting the construction of f.

**Exercise 8.** Complete the proof of Theorem 2 for the -finite case.

We have the following corollary:

Corollary 1.(Radon-Nikodym theorem) Let m be an unsigned -finite measure, and let be a signed -finite measure. Then the following are equivalent.

- for some measurable .
- whenever .
- For every , there exists such that whenever .
When any of these statements occur, we say that is

absolutely continuouswith respect to m, and write . As in the introduction, we call f theRadon-Nikodym derivativeof with respect to m, and write .

**Proof. ** The implication of 3. from 1. is Exercise 11 from Notes 0. The implication of 2. from 3. is trivial. To deduce 1. from 2., apply Theorem 1 to and observe that is supported on a set of m-measure zero E by hypothesis. Since E is null for m, it is null for and also, and so is trivial, giving 1.

Corollary 2.(Lebesgue decomposition theorem) Let m be an unsigned -finite measure, and let be a signed -finite measure. Then there is a unique decomposition , where and . (We refer to and as theabsolutely continuousandsingular componentsof with respect to m.) If is unsigned, then and are also.

**Exercise 9.** If every point in X is measurable, we call a signed measure *continuous* if for all x. Let the hypotheses be as in Corollary 2, but suppose also that every point is measurable and m is continuous. Show that there is a unique decomposition , where , is supported on an at most countable set, and is both singular with respect to m and continuous. Furthermore, if is unsigned, then are also. We call and the *singular continuous *and *pure point* components of respectively.

**Example 1.** A Cantor measure is singular continuous with respect to Lebesgue measure, while Dirac measures are pure point. Lebesgue measure on a line is singular continuous with respect to Lebesgue measure on a plane containing that line.

**Remark 1.** Suppose one is decomposing a measure on a Euclidean space with respect to Lebesgue measure m on that space. Very roughly speaking, a measure is pure point if it is supported on a 0-dimensional subset of , it is absolutely continuous if its support is spread out on a full dimensional subset, and is singular continuous if it is supported on some set of dimension intermediate between 0 and d. For instance, if is the sum of a Dirac mass at , one-dimensional Lebesgue measure on the x-axis, and two-dimensional Lebesgue measure on , then these are the pure point, singular continuous, and absolutely continuous components of respectively. This heuristic is not completely accurate (in part because I have left the definition of “dimension” vague) but is not a bad rule of thumb for a first approximation.

To motivate the terminology “continuous” and “singular continuous”, we recall two definitions on an interval , and make a third:

- A function is
*continuous*if for every and every , there exists such that whenever is such that . - A function is
*uniformly continuous*if for every , there exists such that whenever has length at most . - A function is
*absolutely continuous*if for every , there exists such that whenever are disjoint intervals in I of total length at most .

Clearly, absolute continuity implies uniform continuity, which in turn implies continuity. The significance of absolute continuity is that it is the largest class of functions for which the fundamental theorem of calculus holds (using the classical derivative, and the Lebesgue integral), as was shown in the previous course.

**Exercise 10. **Let m be Lebesgue measure on the interval , and let be a finite unsigned measure.

- Show that is a continuous measure if and only if the function is continuous.
- Show that is an absolutely continuous measure with respect to m if and only if the function is absolutely continuous.

– A finitary analogue of the Lebesgue decomposition (optional) –

At first glance, the above theory is only non-trivial when the underlying set X is infinite. For instance, if X is finite, and m is the uniform distribution on X, then every other measure on X will be absolutely continuous with respect to m, making the Lebesgue decomposition trivial. Nevertheless, there is a non-trivial version of the above theory that can be applied to finite sets (cf. my blog post on the relationship between soft analysis and hard analysis). The cleanest formulation is to apply it to a sequence of (increasingly large) sets, rather than to a single set:

Theorem 3.(Finitary analogue of the Lebesgue-Radon-Nikodym theorem) Let be a sequence of finite sets (and with the discrete -algebra), and for each n, let be the uniform distribution on , and let be another probability measure on . Then, after passing to a subsequence, one has a decomposition(9)

where

- (Uniform absolute continuity) For every , there exists (independent of n) such that whenever , for all n and all .
- (Asymptotic singular continuity) is supported on a set of -measure , and we have uniformly for all , where denotes an error that goes to zero as .
- (Uniform pure point) For every there exists (independent of n) such that for each n, there exists a set of cardinality at most N such that .

**Proof. ** Using the Radon-Nikodym theorem (or just working by hand, since everything is finite), we can write for some with average value 1.

For each positive integer k, the sequence is bounded between 0 and 1, so by the Bolzano-Weierstrass theorem, it has a convergent subsequence. Applying the usual diagonalisation argument (as in the proof of the Arzelà-Ascoli theorem), we may thus assume (after passing to a subsequence, and relabeling) that converges for positive k to some limit .

Clearly, the are decreasing and range between 0 and 1, and so converge as to some limit .

Since , we can find a sequence going to infinity such that as . We now set to be the restriction of to the set . We claim the absolute continuithy property 1. Indeed, for any , we can find a k such that . For n sufficiently large, we thus have

(10)

and

(11)

and hence

. (12)

If we take , we thus see (for n sufficiently large) that 1. holds. (For the remaining n, one simply shrinks delta as much as is necessary.)

Write , thus is supported on a set of size by Markov’s inequality. It remains to extract out the pure point components. This we do by a similar procedure as above. Indeed, by arguing as before we may assume (after passing to a subsequence as necessary) that the quantities converge to a limit for each positive integer j, that the themselves converge to a limit d, and that there exists a sequence such that converges to d. If one sets and to be the restrictions of to the sets and respectively, one can verify the remaining claims by arguments similar to those already given.

**Exercise 11.** Generalise Theorem 3 to the setting where the can be infinite and non-discrete (but we still require every point to be measurable), the are arbitrary probability measures, and the are arbitrary finite measures of uniformly bounded total variation.

**Remark 2.** This result is still not fully “finitary” because it deals with a sequence of finite structures, rather than with a single finite structure. It appears in fact to be quite difficult (and perhaps even impossible) to make a fully finitary version of the Lebesgue decomposition (in the same way that the finite convergence principle in this blog post of mine was a fully finitary analogue of the infinite convergence principle), though one can certainly form some weaker finitary statements that capture a portion of the strength of this theorem. For instance, one very cheap thing to do, given two probability measures , is to introduce a threshold parameter k, and partition , where , and is supported on a set of m-measure at most ; such a decomposition is automatic from Theorem 2 and Markov’s inequality, and has meaningful content even when the underlying space X is finite, but this type of decomposition is not as powerful as the full Lebesgue decompositions (mainly because the size of the support for is relatively large compared to the threshold k). Using the finite convergence principle, one can do a bit better, writing for any function F and any , where , , is supported on a set of m-measure at most , and has total mass at most , but this is still fails to capture the full strength of the infinitary decomposition, because needs to be fixed in advance. I have not been able to find a fully finitary statement that is equivalent to, say, Theorem 3; I suspect that if it does exist, it will have quite a messy formulation.

[*Update*, Jan 5: Exercise added.]

[*Update*, Jan 7: Proof of Hahn decomposition theorem altered; my original proof works, but one of the steps was much trickier than I had anticipated, so I am reverting to Folland’s proof.]

## 41 comments

Comments feed for this article

5 January, 2009 at 7:38 am

liuxiaochuanDear Professor Tao:

Here are several typos,

1, In exercise 1, should be .

2, In Theorem 1, there is a “latex” missing.

[Corrected, thanks – T.]6 January, 2009 at 5:39 am

AnonymousFor the function f in the statement of Theorem 2 to be in L^1, one needs to

assume that \mu is finite. Similarly for the implication of 3. from 2. in Corollary 1.

6 January, 2009 at 6:13 am

liuxiaochuanDear Professor:

The existence of conditional expectation comes from R-N derivative, which I think is worth mentioning, as an exercise or something. ( maybe you have other plans. :-))

Thanks for the last theorem! I spent several hours to go over your previous post on soft analysis and hard analysis and learned a lot, again!

6 January, 2009 at 7:53 am

liuxiaochuanAnd two more corrections( if I am right):

1,Before (10) it should be , and (10) and (11) should be and

2,In the last paragraph of the last proof, , the denominator of lefthand side should be

6 January, 2009 at 10:58 am

Terence TaoThanks for the corrections and suggestions!

6 January, 2009 at 11:47 am

AnonymousI believe there is a typo in Remark 2 line 9: We need that is supported on a set of –measure at most , not .

[Corrected, thanks – T.]7 January, 2009 at 3:01 am

BobDear Terry,

Off topic, but do you generally write your posts in some text editor first. and then copy and paste into wordpress?

Thanks

7 January, 2009 at 11:38 am

Max BaroiI could be wrong, but in Corollary 2, you write is mutually singular to when I tihnk it should be is mutually singular to m.

[Corrected, thanks – T.]9 January, 2009 at 9:43 pm

An Exercise of Real Analysis « Liu Xiaochuan’s Weblog[…] 10, 2009 at 1:42 下午 | In Maths（数学）, Real Analysis(实分析) | The last exercise of Notes 1 of Terence Tao’s lecture notes of the course “Real Analysis” is relavent with his […]

10 January, 2009 at 8:39 am

245B, notes 3: L^p spaces « What’s new[…] result should be compared with the Radon-Nikodym theorem (Corollary 1 from Notes 1). Both theorems start with an abstract function or , and create a function out of it. Indeed, […]

30 January, 2009 at 10:50 am

CristiHello,

I think that in remark 0 :

“One can in fact show that any signed measure that avoids +inf must have finite positive variation, but this turns out to require a certain amount of work”

you should add that it must avoid -(infinity) too.

30 January, 2009 at 11:48 am

Terence TaoIt is certainly easier to prove finite positive variation if one also avoids , but it is not necessary. (Of course, signed measures that attain will have infinite negative variation, and thus infinite total variation as well, but will still have finite positive variation.)

30 January, 2009 at 12:37 pm

CristiOops, I thought, you meant finite total variation (because of the “measure is finite iff the total variation is finite” corollary) instead of positive variation.

Sorry.

4 February, 2009 at 1:25 pm

245B, Notes 9: The Baire category theorem and its Banach space consequences « What’s new[…] finite and are all absolutely continuous with respect to ; now use the Radon-Nikodym theorem (see Notes 1) and the fact that .) On the other hand, is closed, hence weakly closed (by Hahn-Banach), and so […]

16 February, 2009 at 7:28 am

实分析0-10 « Liu Xiaochuan’s Weblog[…] 第一节是关于关于符号测度和Radon-Nikodym-Lebesgue定理。本节从正常测度开始讲起，事实上为符号测度做铺垫，为对符号测度有一个更清楚的了解而写的一个引子。我更看重陶教授在这方面做的事情。而且整个课程他一直是此种风格。本节主体内容跟常见的教材差不多，我更喜欢夹杂在证明过程中的一些解释。比如在证明Radon-Nikodym-Lebesgue定理的时候，正文中讲述整个过程中“贪婪”的思想，很有意思。结尾附上了一个选学的内容，即Lebesgue分解定理的有限形式。这个跟陶教授以前的一篇帖子有很大关系，我跟人很喜欢这个想法，我把最后的习题写了下来。 […]

9 March, 2009 at 12:08 am

KeRegarding

Remark 0, I don’t think works, If the first one is then the union is also . Limsup doesn’t work either. There is an nontrivial example where converges to while and .9 March, 2009 at 8:51 am

Terence TaoDear Ke,

Yes, you’re right; but I think it can be fixed by making converge to the supremum sufficiently rapidly. I’ve amended the remark to reflect this.

9 March, 2009 at 6:28 pm

studentDear Prof. Tao,

in a finite measure space if a function is in all spaces for p greater than or equal to 1, what can we say about it?just it is a.e. bounded?

what happens, in sigma finite and infinite measure case?

thanks,

9 March, 2009 at 8:58 pm

Terence TaoDear student,

In the finite measure case, is included in for all , and so is the same as .

In general, this inclusion does not hold, but instead we have whenever , and so .

One can get some intuition for this by looking at the special case of step functions , in which .

9 March, 2009 at 9:46 pm

studentThank you very much Prof. Tao.

16 April, 2009 at 3:56 am

Anonymous“(indeed, one could view measures as a kind of “generalised function” with respect to a fixed reference measure m).”

Do you mean to say that we can view unsigned measure as a kind of “generalised function” with respect to a fixed reference measure m?

21 August, 2009 at 2:13 pm

Anush TserunyanDear Prof. Tao,

I think there is a typo in Corollary 1 (Radon-Nikodym theorem). \mu should be finite, otherwise 2 doesn’t imply 1 or 3: let m be the Lebesgue measure on (0,1) and \mu = m_f with f(x) = 1/x.

Thanks

[Corrected, thanks – T.]7 October, 2009 at 2:48 am

AnonymousIf

\mu is unsigned, thenm_f and\mu_s are also.29 May, 2010 at 12:49 pm

Kestutis CesnaviciusI think in Exercise 4 one should add that $X$ is $\sigma$-finite. Otherwise, if $X$ is countable and $\{ \emptyset, X\}$ is the $\sigma$-algebra, then no finite nonzero measure is differentiable wrt the counting measure.

29 May, 2010 at 2:22 pm

Terence TaoAh, I forgot to add the assumption that X is endowed with the discrete -algebra. Thanks for the correction!

25 September, 2010 at 10:59 pm

245A, Notes 3: Integration on abstract measure spaces, and the convergence theorems « What’s new[…] is a measure. (We will study such measures in greater detail in 245B.) […]

16 October, 2010 at 8:30 pm

245A, Notes 5: Differentiation theorems « What’s new[…] If is bounded, then can be expressed as the sum of a continuous monotone non-decreasing function and a jump function . Remark 14 This decomposition is part of the more general Lebesgue decomposition, which we will discuss later in this course. […]

30 October, 2010 at 6:54 pm

245A, Notes 6: Outer measures, pre-measures, and product measures « What’s new[…] This is the final set of notes in this sequence. If time permits, the course will then begin covering the 245B notes, starting with the material on signed measures and the Radon-Nikodym-Lebesgue theorem. […]

23 December, 2010 at 8:23 pm

QiangHi, Prof Tao

I’m a bit confused in the Hahn decomposition theorem. For a signed measure , there is a decomposition such that or just . Cause both in the process of proof and wikipedia entry it’s the latter ?

Thanks, Qiang

23 December, 2010 at 8:36 pm

Terence Taois equivalent to being totally positive (or a positive set, as per the Wikipedia notation), by definition. (I assume you mean here “ is positive” rather than , since of course a signed measure need not be positive in general.)

25 December, 2010 at 12:22 am

Hahn decomposition theory « Mcknight0219's Blog[…] am confused about the proof of Prof. Tao for a while, mainly due to notation mistaken. Here I try to write down the proof following […]

18 March, 2011 at 8:13 am

O Teorema de Radon-Nikodym | renanfinder[…] não é a original. Outra linha de ataque pode ser encontrada no link da primeira linha, bem como aqui (mas acho que elas são essencialmente equivalentes). De toda forma, me parece que as ideias mais […]

11 February, 2012 at 5:28 am

RexIn the statement of the Lebesgue-Radon-Nikodym theorem, I think the assertion that the Radon-Nikodym derivative $f$ is absolutely integrable is not true.

For example, let $\mu = dx$ be the Lebesgue measure on the real line and let $\nu = 2dx$ so that $f$ is the constant function 2.

[Corrected, thanks – T.]11 February, 2012 at 12:50 pm

RexYou’ll probably want to update this in your statement of Corollary 1 as well. The LaTeX in the statement of Theorem 2 now doesn’t seem to be parsing (at least on my computer).

In the general case, what can be said about the function f? I have heard that it should be locally integrable, but I didn’t see anything in the proof to indicate this.

11 February, 2012 at 2:38 pm

Terence TaoCorrected, thanks. With the assumptions given, f need not be locally integrable; consider for instance on when . However, if one makes the further assumption that is a Radon measure (and is in particular locally finite), then will be locally integrable.

11 February, 2012 at 12:03 pm

Anonymousf ought to be *locally* integrable, though

29 June, 2014 at 11:40 am

AnonymousWhy is instead of used in this note? Are they the same?

2 November, 2014 at 7:07 am

Anonymous“Indeed, if we let be a maximising sequence, thus and , one easily checks that the function attains the supremum (and that is in the set of all non-negative functions such that ).”

I don’t see how easily one can check this.. If is monotone, then one can use MCT. If not, let . increases to and . It follows that . How can one show that to complete the argument?

2 November, 2014 at 8:53 am

Terence TaoOne has , rather than .

2 November, 2014 at 11:58 am

AnonymousAh… I meant and . Still this only gives one half of the desired equality.

2 November, 2014 at 3:35 pm

Terence TaoSince , we have , so by definition of .