The Unapologetic Mathematician

Mathematics for the interested outsider

The Radon-Nikodym Theorem (Proof)

Today we set about the proof of the Radon-Nikodym theorem. We assumed that (X,\mathcal{S},\mu) is a \sigma-finite measure space, and that \nu is a \sigma-finite signed measure. Thus we can write X as the countable union of subsets on which both \mu and \nu are finite, and so without loss of generality we may as well assume that they’re finite to begin with.

Now, if we assume for the moment that we’re correct and an f does exist so that \nu is its indefinite integral, then the fact that \nu is finite means that f is integrable, and then if g is any other such function we can calculate


for every measurable E\in\mathcal{S}. Now we know that this implies f-g=0 a.e., and thus the uniqueness condition we asserted will hold.

Back to the general case, we know that the absolute continuity \nu\ll\mu is equivalent to the conjunction of \nu^+\ll\mu and \nu^-\ll\mu, and so we can reduce to the case where \nu is a finite measure, not just a finite signed measure.

Now we define the collection \mathcal{K} of all nonnegative functions f which are integrable with respect to \mu, and for which we have


for every measurable E. We define

\displaystyle\alpha=\sup\limits_{f\in\mathcal{K}}\int f\,d\mu

Since \alpha is the supremum, we can find a sequence \{f_n\} of functions in \mathcal{K} so that

\displaystyle\lim\limits_{n\to\infty}\int f\,d\mu=\alpha

For each n we define

\displaystyle g_n(x)=\max\limits_{1\leq i\leq n}f_i(x)

Now if E is some measurable set we can break it into the finite disjoint union of n sets E_i so that g_n=f_i on E_i. Thus we have


and so g_n\in\mathcal{K}.

We can write g_n=g_{n-1}\cup f_n, which tells us that the sequence \{g_n\} is increasing. We define f_0 to be the limit of the g_nf_0(x) is the maximum of all the f_i(x) — and use the monotone convergence theorem to tell us that


Since all of the integrals on the right are bounded above by \nu(E), their limit is as well, and f_0\in\mathcal{K}. Further, we can tell that the integral of f_0 over all of X must be \alpha. Since f_0 is integrable, it must be equal \mu-a.e. to some finite-valued function f. What we must now show is that if we define


then \nu_0 is identically zero.

If it’s not identically zero, then by the lemma from yesterday there is a positive number \epsilon and a set A so that \mu(A)>0 and so that

\displaystyle\epsilon\mu(E\cap A)\leq\nu_0(E\cap A)=\nu(E\cap A)-\int\limits_{E\cap A}f\,d\mu

for every measurable set E. If we define g=f+\epsilon\chi_A, then

\displaystyle\int\limits_Eg\,d\mu=\int\limits_Ef\,d\mu+\epsilon\mu(E\cap A)\leq\int\limits_{E\setminus A}+\nu(E\cap A)\leq\nu(E)

for every measurable set E, which means that g\in\mathcal{K}. But

\displaystyle\int g\,d\mu=\int f\,d\mu+\epsilon\mu(A)>\alpha

which contradicts the maximality of the integral of f. Thus \nu_0 must be identically zero, and the proof is complete.

July 7, 2010 Posted by | Analysis, Measure Theory | 3 Comments