Differentiation of Measures – The Radon-Nikodym Theorem Part 2

Hello everyone! My apologies for the long delay between posts in this series, these last few weeks have been a bit crazy with the NSF fellowship application due as well as an exam and a boatload of homework. Anyways, I’m at home for Thanksgiving, and I’ve got enough homework done that I can take a bit of a break and write about a very cool theorem in measure theory: the Radon-Nikodym theorem. In the last post I talked a lot about writing one measure as an integral with respect to some positive “reference” measure. The Radon-Nikodym theorem tells us precisely how to do this for any (reasonably well behaved) signed measure \nu and positive measure \mu. I’ll present the theorem in its full generality (as stated in Folland) and give a sketch of the proof, but I want to spend more time discussing the meaning of this result.

Lemma: Let \mu,\nu be finite measures. Either \nu\perp\mu or there is \epsilon>0 and E measurable so that \mu(E)>0 and \nu\geq\epsilon\mu on E, i.e. E is a positive set for the measure \nu-\epsilon\mu.

The proof is an easy exercise using Hahn decompositions, so I’m going to omit it. Before I present the main theorem, I want to present one piece of fairly common notation: If \nu is a measure defined by the equation \nu(E)=\int_E f\hspace{1mm}d\mu, we often abbreviate this as d\nu=f\hspace{1mm}d\mu. And now the main theorem:

Theorem (Lebesgue-Radon-Nikodym): Let \nu be a \sigma-finite signed measure and \mu a \sigma-finite positive measure on (X,\mathcal{M}). There exist unique \sigma-finite signed measures \rho,\lambda on (X,\mathcal{M}) such that

\lambda\perp \mu\hspace{5mm}\rho\ll\mu\hspace{5mm}\nu=\lambda+\rho

Moreover, there is an extended \mu-integrable function f:X\to\mathbb{R} such that d\rho=f\hspace{1mm}d\mu, and any two such functions are equal almost everywhere.

Remark: To prove this, start with the case \mu,\nu finite and positive. If \nu is signed, then simply apply the result for positive measures to the positive and negative variations of \nu and subtract the results. Going to the \sigma-finite case is a bit more difficult, but is essentially what you would expect: get countably many measures and functions, then sum them up. Thus I’m only going to concern myself with the case of positive, finite measures. Moreover, the singular measure \lambda is easy to construct once we have \rho, just take \lambda=\nu-\rho and show it works (this uses the above lemma, which is why I included it if you want to work out the details), and showing uniqueness is a relatively standard proof and is not very interesting, so I leave these two parts to the reader. The interesting part of this proof is the construction of the measure d\rho=f\hspace{1mm}d\mu. You will see it’s similar to the proof of the Hahn Decomposition theorem in that we will produce a “large” set of functions and we will want to take a “maximal” function. As before, we will only be able to do this up to a supremum in some sense, but it will turn out to be good enough. One last thing, the decomposition \nu=\lambda+f\hspace{1mm}d\mu is called the Lebesgue decomposition of \nu with respect to \nu, hence his name is on the theorem. If \nu\ll\mu, then this theorem states that d\nu=f\mu for some function and this result is what is usually known as the Radon-Nikodym theorem. The function f is often called the Radon-Nikodym derivative of \nu with respect to \nu and is often denoted f=d\nu/d\mu.

Proof of Part of Theorem: Let \mu,\nu be finite, positive measures. Define

\mathcal{F}=\{f:X\to[0,\infty]:\int f\hspace{1mm}d\mu\leq \nu(E)\text{ for all }E\in\mathcal{M}\}

Note that \mathcal{F} is non-empty because the zero function is in \mathcal{F}. Moreover, it is easily verified that if f,g\in\mathcal{F}, then so is the maximum of f and g. Let a=\sup\{\int f\hspace{1mm}d\mu:f\in\mathcal{F}\}. Since \mu is finite, we have a\leq \mu(X)<\infty. Choose a sequence \{f_n\}\subset\mathcal{F} such that \int f_n\hspace{1mm}d\mu\to a and define g_n=\max\{f_1,f_2,\dots,f_n\}. Also, define f=\sup_n f_n, and note that g_n\in \mathcal{F} and g_n\uparrow f pointwise. Moreover, a\geq\int g_n\geq\int f_n, so taking n\to\infty gives that \int g_n\to a. By the monotone convergence theorem, f\in\mathcal{F} and \int f\hspace{1mm}d\mu=a. Note that f<\infty a.e., so we can assume f is real valued everywhere. We lose this when going to the \sigma-finite case.

So this is the main idea. If you get caught up on details and want to know what’s going on, let me know and I’ll fill in some spots for you in the comments. So what did we do? We looked at functions whose integrals with respect to \mu got close the the \nu measure of E for all E measurable. Then we found functions whose integrals converged correctly to the measure we wanted, and we took the “best part” of each of those functions, namely the supremum at each point. Then after that it’s a routine monotone convergence theorem check to get the result. One note: in my previous post I said we can generalize signed measures to complex measures, and there is an obvious generalization for this theorem as well. We can also weaken the hypotheses a little bit, which I’ll do below.

So what can we do with this theorem? Well, for one thing there is a lot that can be done with functions of one variable. For example, it turns out that in the case of one variable, the familiar fundamental theorem of calculus connection between integrals and derivatives holds in the case of Lebesgue integrals and R/N derivatives, but that’s for a different post. What we are more interested in is that we now have a way to somehow compare a given measure on \mathbb{R}^n to the standard Lebesgue measure. Our push forward measures are measures on \mathbb{R}^n, so perhaps we should compare these to Lebesgue measure! We’ll do this next post. For now, I’ll close by stating a slightly stronger version of the Radon-Nikodym theorem, and do one example to show how these derivatives might be used.

Proposition: If \nu is an arbitrary signed measure and \mu is a \sigma-finite positive measure such that \nu\ll\mu, then there exists and extended \mu-integrable function f such that d\nu=f\hspace{1mm}d\mu.

I won’t prove this, one essentially chooses clever set E so that \nu is \sigma-finite on E, and then apply R/N there and extend f to the rest of X in a fairly canonical way. Okay, one last little example for this post! This result is an exercise in Folland’s book and is very useful in probability theory:

Proposition Let (X,\mathcal{M},\mu) be a finite measure space and let \mathcal{N} be a sub \sigma-algebra of \mathcal{M} and \nu=\mu|_{\mathcal{N}}. If f\in L^1(\mu), then there exists a \nu-a.e. unique g\in L^1(\nu) (so g is \mathcal{N}-measurable) so that \int_E f\hspace{1mm}d\mu=\int_E g\hspace{1mm}d\nu for all E\in\mathcal{N}. The function g is often called theconditional expectation of f on \mathcal{N}.

Proof: Define a measure \lambda on \mathcal{N} by \lambda(E)=\int_E f\hspace{1mm}d\mu. Note that \lambda is finite because f\in L^1, and clearly \nu is a finite measure (because \mu is). Moreover, for E\in\mathcal{N}, if \nu(E)=0, then \mu(E)=0, so \lambda(E)=\int_E f\hspace{1mm}d\mu=0 as well, thus \lambda\ll\nu, so there is a function g so that \int g\hspace{1mm}d\nu=\lambda(E)=\int f\hspace{1mm}d\mu. Moreover this function is \nu-a.e. unique. To check that g\in L^1, note that if f is positive, then g will also be positive (by construction in the proof of R/N), and in this case int_X |g|\hspace{1mm}d\nu=\int_X g\hspace{1mm}d\nu=\int_X f\hspace{1mm}d\mu<\infty by assumption. Taking positive and negative parts of an arbitrary f gives the final result.

So that’s the Radon-Nikodym theorem… pretty cool stuff! Next time we’ll talk about push-forward measures and wrap up my first series on determinants. Have a good last few weeks of semester everyone!


About Ryan

I'm a software developer at Hudl where I work on awesome software. Before that, I was a grad student in mathematics, interested in probability theory as well as analysis, more on the side of functional analysis and less on the side of PDEs. Apart from that I'm pretty lame. Though I do enjoy watching football, playing golf, and playing the trumpet.
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3 Responses to Differentiation of Measures – The Radon-Nikodym Theorem Part 2

  1. Z Norwood says:

    Gawd Hotovy, you’d better apologise! It’s been DAYS since you last posted!

  2. Pingback: RT Part 7; Van Der Waerden’s Theorem | whateversuitsyourboat

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