Differentiation of Measures – The Radon-Nikodym Theorem Part 1

In my last post, I ended by saying we need to discuss derivatives of measures. Now, at first you might think this sounds kind of strange, since usually to discuss derivatives we need the notion of a limit as, say x\to y in \mathbb{R}. This notion of derivative of a measure isn’t quite the same notion (though the two are related in the special case of (X,\mathcal{M})=(\mathbb{R},\mathcal{B}_{\mathcal{R}})), but before we begin to discuss it, we need to talk about signed measures. Recall that, originally, given a measurable space (X,\mathcal{M}), we defined a measure to be a function \mu:\mathcal{M}\mapsto[0,+\infty] that was countably disjointly additive and was normalized so that \mu(\emptyset)=0. We didn’t so much care about issues of convergence of the countable sums because they were sums of positive terms and so were either unbounded or convergent. Now we’re going to change the rules a little. Unfortunately this post will consist of a lot of definitions and not too many interesting results, but I’ll try to sneak in a few nice results here or there. Here’s the first definition:

Definition: Let (X,\mathcal{M}) be a measurable space. A signed measure on (X,\mathcal{M}) is a function \nu:\mathcal{M}\mapsto(-\infty,+\infty) satisfying \nu(\emptyset)=0 and for all sequences of disjoint measurable sets \{E_k\}_{k\geq1},

\displaystyle\nu\left(\bigcup_{k\geq 1}E_k\right)=\sum_{k\geq1}\nu(E_k)

where the sum converges absolutely.

A remark, I chose to force \nu(X) to be finite, though one can choose not to do this as long as the restriction is made that \nu assumes at most only one of the values +\infty,-\infty. Most of the results I’ll show are valid for \sigma-finite measures, and in the case where \nu is a positive measure (which will mostly be what we need for the determinant application), we again do not need to worry about convergence, so the measure can be infinite. But for ease (and length) or writing, for now I’ll assume all the measures are finite and I’ll try to point out what the usual extensions are.

So what are some examples of signed measures? Well, for a basic example let (X,\mathcal{M})=([-1,1],\mathcal{B}_{[0,1]}), let m be Lebesgue measure, and let \nu(E)=\int_E x^3\hspace{1mm}dm. It’s easy to check that this is indeed a signed measure, and, for example, \nu((-1,0))=\int_{-1}^0x^3\hspace{1mm}dx=-1/4. In fact, given any measure space (X,\mathcal{M},\mu) and a function f\in\mathcal{L}^1(\mu), we can always define a signed measure by \nu_f(E)=\int_E f\hspace{1mm}d\mu (the integrability assumption ensures that this is finite). In fact, once we build up some more machinery, we’ll show that this idea is incredibly useful and that, in fact, all signed measures can be written as an integral with respect to a positive “reference” measure. Now we would like to know what familiar properties of positive measures carry over to signed measures. For example, does monotonicity carry over? Well, no, not in the strictest sense. For example -1/64=\int_{(-1/2,0)}x^3\hspace{1mm}dx\geq\int_{(-1,0)}x^3\hspace{1mm}dx=-1/4. Moreover, \int_{[-1,1]} x^3\hspace{1mm}dx=0, so in fact the whole space has measure zero! Certainly this is a bit troubling. What we need is a new notion of positive, negative, and null sets.

Definition: Let \nu be a signed measure. We say a set E is positive (resp. negative, null) if for every measurable F\subset E, we have \nu(F)\geq0 (resp. \leq 0,=0).

Note that in our example, a set is positive precisely when f\geq0 (or negative, zero respectively). In our basic example of f(x)=x^3, we can write \nu(E)=\int_E f^+\hspace{1mm}d\mu-\int_Ef^-\hspace{1mm}d\mu. It turns out that this approach of writing \nu as a positive part and a negative part is a useful decomposition. I’m not going to prove it here (I’ll give a short sketch), but there is a theorem, called the Hahn Decomposition Theorem, that states for a given signed measure \nu, we can split up the space into two disjoint sets P,N on which \nu is positive and negative respectively. Moreover, this is almost unique.

Theorem (Hahn Decomposition Theorem): Let \nu be a signed measure on (X,\mathcal{M}). Then there exist measurable sets P and N such that P is positive for \nu and N is negative for \nu satisfying P\cup N=X and P\cap N=\emptyset. Moreover, if P',N' is another such pair, then P\triangle P'=N\triangle N' is null for \nu.

The interesting idea in this proof, to me, is how the sets P and N are constructed. The rest of the argument is a contradiction proof to show that the sets have the desired property. To construct P, one considers \sup\{\nu(E):E\text{ positive}\}. The set is non-empty because the empty set is positive. Then, using a typical measure theory argument, one can construct a sequence of positive sets whose measures converge to this supremum. Taking the countable union of these sets preserves positivity (this is easy to show), and so we let P be this union. Then, taking N to be the complement, we get the desired sets. Of course this decomposition is not unique, because we can move null sets freely between the two sets. However, the sets are “unique enough” as we will show in the next theorem. First, one more definition.

Definition: Suppose \mu,\nu are two signed measures on the same \sigma-algebra. If there exist E,F measurable such that X=E\cup F, E\cap F=\emptyset, E is null for \mu, and F is null for \nu, then we say E and F are mutually singular and write \mu\perp\nu.

Theorem (Jordan Decomposition Theorem): If \nu is a signed measure, there exist unique positive measures \nu^+ and \nu^- so that \nu=\nu^+-\nu^- and \nu^+\perp\nu^-.
Proof: Let P,N be a Hahn decomposition for X and define \nu^+(E)=\nu(E\cap P) and \nu^-(E)=\nu(E\cap N). Clearly \nu=\nu^+-\nu^-. To show uniqueness, suppose \lambda^+,\lambda^- are another such pair of measures. Then there are A and B such that A and B form another Hahn decomposition for \nu, hence \nu(P\triangle E)=0, and using this fact it is easily verified that \nu^+(E)=\lambda^+(E) for all E, and similarly for \nu^-.

So basically, what this theorem tells us is that every signed measure can be written as two positive measures, which lets us use all our familiar properties of such measures that we know and love! Moreover, we even get a new positive measure out of the deal. These measures turn out to be so useful that they get their own names. We call the measures \nu^+ and \nu^- the positive and negative variations of \nu, and we define the positive measure |\nu|:=\nu^++\nu^-, called the total variation of \nu. There are two interesting things to note about this positive measure |\nu|. First, we can write \nu(E)=\int_Ef\hspace{1mm}d|\nu|, where f=\chi_P-\chi_N (where P,N are a Hahn decomposition for \nu), so we can now always write a signed measure as i) a difference of two positive measures and ii) an integral with respect to a positive reference measure. The second thing to notice is that |\nu| dominates \nu in the sense that if |\nu|(E)=0, then we must have \nu(E)=0. This idea has a name, and it will turn out that whenever a positive measure dominates a signed measure in this sense, we will always be able to write the signed measure as an integral with respect to the positive measure (provided both are reasonably well behaved). I won’t prove this fact today, but it will be in the next post. For now, I’ll close with the definition of absolute continuity and prove a characterization of it that justifies the name.

Definition: Let \nu be a signed measure and \mu be a positive measure on (X,\mathcal{M}). If \mu(E)=0\implies \nu(E)=0 whenever \mu(E)=0, then we say \nu is absolutely continuous with respect to \mu and write \nu\ll\mu.

Proposition: Let \nu be a signed measure and \mu be a positive measure on (X,\mathcal{M}). Then \nu\ll\mu if and only if for all \epsilon>0, there is \delta>0 so that \mu(E)<\delta implies |\nu(E)|<\epsilon.
Proof: It is clear that \nu\ll\mu iff |\nu|\ll\mu, so it suffices to assume \nu is a positive measure. The epsilon-delta condition clearly implies \nu\ll\mu, so suppose that the \epsilon-\delta condition does not hold. Then there is \epsilon>0 so that for each n\in\mathbb{N}, there is E_n measurable so that \mu(E_n)<1/n but \nu(E_n)\geq \epsilon. Set F_k=\bigcup_{n\geq k} E_n and let F=\bigcap_{k\in\mathbb{N}} F_k. Then for each k, we have \mu(F_k)\leq\sum_{n=k}^\infty 1/2^n=2^{1-k}. Thus by continuity of measures we have \mu(F)=0, but (since \nu is finite), we have \nu(F_k)\geq\epsilon, so \nu(F)=\lim \nu(F_k)\geq\epsilon, hence \nu is not absolutely continuous with respect to \mu.

I don't think I'll use this proposition much, but it's a nice characterization (for finite signed measures, this one does not need to hold even if \nu is \sigma-finite) of absolute continuity. In the next post I'll present the Radon-Nikodym theorem, though I'm not sure I'll prove it (I haven't decided yet), and then discuss a few results about the Radon-Nikodym derivative. After that, I'll move back to push-forward measures and ultimately connect back to determinants, and I think then I'll finally be done with the determinant series. Enjoy!


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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1 Response to Differentiation of Measures – The Radon-Nikodym Theorem Part 1

  1. JCummings says:

    I also read “\nu \ll \mu” as “\mu dominates \nu“!

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