We apply the martingale convergence theorem to prove the Radon-Nikodym theorem, which states that if $\mu$ and $\nu$ are $\sigma$ -finite measures on a measurable space $(\Omega,\mathcal{F})$ and $\nu$ is absolutely continuous with respect to $\mu$ then there exists a non-negative and measurable $f\colon\Omega\rightarrow\mathbb{R}$ such that $\nu(A)=\int_Af\,d\mu$ for all measurable sets $A$ .

As any $\sigma$ -finite measure is equivalent to a probability measure, it is enough to prove the result in the case where $\mu$ and $\nu$ are probability measures. Furthermore, by the Jordan decomposition, the result generalizes to the case where $\nu$ is a signed measure. So, we just need to prove the following.

Here, $X$ is called the Radon-Nikodym derivative of $\mathbb{Q}$ with respect to $\mathbb{P}$ .

More generally, for any sub-$\sigma$ -algebra $\mathcal{G}$ of $\mathcal{F}$ we can restrict the measures $\mathbb{P}$ and $\mathbb{Q}$ to $\mathcal{G}$ and ask if the Radon-Nikodym derivative of $\mathbb{Q}|_\mathcal{G}$ with respect to $\mathbb{P}|_\mathcal{G}$ exists. If it does we shall denote it by $X_\mathcal{G}$ , which by definition is a non-negative $\mathcal{G}$ -measurable random variable satisfying $\mathbb{Q}(A)=\mathbb{E}_\mathbb{P}[1_AX_\mathcal{G}]$ for all $A\in\mathcal{G}$ .

We note that if $X_\mathcal{G}$ does exist, then it is uniquely defined ($\mathbb{P}$ -almost everywhere). Suppose that $\tilde X_{\mathcal{G}}$ also satisfied the required properties, then \begin{equation*}\begin{split} \mathbb{E}_{\mathbb{P}}[\max(X_{\mathcal{G}}-\tilde X_{\mathcal{G}},0)] &=\mathbb{E}_{\mathbb{P}}[X_\mathcal{G}1_{\left\{X_{\mathcal{G}}>\tilde X_{\mathcal{G}}\right\}}]-\mathbb{E}_{\mathbb{P}}[\tilde X_\mathcal{G}1_{\left\{X_{\mathcal{G}}>\tilde X_{\mathcal{G}}\right\}}]\\ &=\mathbb{Q}(X_{\mathcal{G}}>\tilde X_{\mathcal{G}})-\mathbb{Q}(X_{\mathcal{G}}>\tilde X_{\mathcal{G}})=0 \end{split}\end{equation*}so $X_\mathcal{G}\le \tilde X_\mathcal{G}$ almost surely. Similarly, $\tilde X_\mathcal{G}\le X_\mathcal{G}$ and therefore $\tilde X_\mathcal{G}= X_\mathcal{G}$ (almost surely).

First, the easy case. For a finite $\sigma$ -algebra, the Radon-Nikodym derivative can be written out explicitly.

Proof. Let $A_1,A_2,\ldots,A_n$ be the minimal non-empty elements of $\mathcal{G}$ . These are pairwise disjoint subsets of $\Omega$ such that every set in $\mathcal{G}$ is a union of a subcollection of the $A_k$ . Set \begin{equation*} X_\mathcal{G}=\sum_{k=1}^n \frac{\mathbb{Q}(A_k)}{\mathbb{P}(A_k)}1_{A_k} \end{equation*}Note that whenever $\mathbb{P}(A_k)=0$ then $\mathbb{Q}(A_k)=0$ , and we adopt the convention that $\frac{0}{0}=0$ . Clearly, $X_\mathcal{G}$ is $\mathcal{G}$ -measurable, and \begin{equation*}\begin{split} \mathbb{E}_{\mathbb{P}}[1_{A_k}X_\mathcal{G}] &=\frac{\mathbb{Q}(A_k)}{\mathbb{P}(A_k)}\mathbb{E}_{\mathbb{P}}[1_{A_k}]+\sum_{j\not=k}\frac{\mathbb{Q}(A_j)}{\mathbb{P}(A_j)}\mathbb{E}_{\mathbb{P}}[1_{A_k\cap A_j}]\\ &=\mathbb{Q}(A_k). \end{split}\end{equation*}Here, we have used $\mathbb{E}_{\mathbb{P}}[1_{A_k}]=\mathbb{P}(A_k)$ and $1_{A_k\cap A_j}=0$ . By linearity, this equality remains true if both sides are replaced by any union of the $A_k$ , and therefore $X_\mathcal{G}$ is the required Radon-Nikodym derivative.

Next, martingale convergence is used to prove the existence of the Radon-Nikodym derivative in the case where the $\sigma$ -algebra $\mathcal{G}$ is separable. By separable, we mean that there is a countable sequence of sets $A_1,A_2,\ldots$ generating $\mathcal{G}$ . Note that if we let $\mathcal{G}_n$ be the $\sigma$ -algebra generated by $A_1,A_2,\ldots,A_n$ , then $\mathcal{G}_n$ is an increasing sequence of finite sub-$\sigma$ -algebras such that $\bigcup_n\mathcal{G}_n$ generates $\mathcal{G}$ . The following result is general enough to apply in many useful cases, such as with the Boral $\sigma$ -algebra on $\mathbb{R}^n$ .

Finally, by approximating by finite $\sigma$ -algebras we can prove the Radon-Nikodym theorem for arbitrary inseparable $\sigma$ -algebras $\mathcal{F}$ .

Proof of the Radon-Nikodym theorem:

First, we use contradiction to show that for any $\epsilon>0$ there exists a finite $\sigma$ -algebra $\mathcal{G}\subseteq\mathcal{F}$ satisfying $\mathbb{E}_\mathbb{P}[|X_\mathcal{G}-X_\mathcal{H}|]<\epsilon$ for every finite $\sigma$ -algebra $\mathcal{H}$ with $\mathcal{G}\subseteq\mathcal{H}\subseteq{F}$ . If this were not the case, then by induction we could find an increasing sequence of finite sub-$\sigma$ -algebras of $\mathcal{F}$ satisfying $\mathbb{E}_\mathbb{P}[|X_{\mathcal{G}_n}-X_{\mathcal{G}_m}|]\ge\epsilon$ . However, letting $\mathcal{G}=\sigma(\bigcup_n\mathcal{G}_n)$ , Lemma 2 shows that $X_\mathcal{G}$ exists and \begin{equation*} \epsilon\le\lim_{n\rightarrow\infty}\mathbb{E}_\mathbb{P}[|X_{\mathcal{G}_n}-X_{\mathcal{G}_{n+1}}|] \le\lim_{n\rightarrow\infty}\mathbb{E}_\mathbb{P}[|X_{\mathcal{G}_n}-X_{\mathcal{G}}|]+\lim_{n\rightarrow\infty}\mathbb{E}_\mathbb{P}[|X_{\mathcal{G}_{n+1}}-X_{\mathcal{G}}|]=0 \end{equation*}-- a contradiction.

So, there exists a sequence of finite sub-$\sigma$ -algebras $\mathcal{G}_n$ of $\mathcal{F}$ such that $\mathbb{E}_\mathbb{P}[|X_{\mathcal{G}_n}-X_\mathcal{H}|]<2^{-n}$ for every finite sub-$\sigma$ -algebra $\mathcal{H}$ of $\mathcal{F}$ containing $\mathcal{G}_n$ . Let $\mathcal{G}$ be the (separable) $\sigma$ -algebra generated by $\bigcup_n\mathcal{G}_n$ , and set $\mathcal{\tilde G}_n=\sigma(\bigcup_{k=1}^n\mathcal{G}_k)$ . By Lemma 2, the Radon-Nikodym derivative restricted to $\mathcal{G}$ , $X_\mathcal{G}$ , exists, and we show that it is the required derivative of $\mathbb{Q}$ with respect to $\mathbb{P}$ .

Choose any set $A\in\mathcal{F}$ and let $\mathcal{H}_n$ be the (finite) $\sigma$ -algebra generated by $\mathcal{G}_n\cup\{A\}$ . Then, $X_{\mathcal{H}_n}$ exists and satisfies $\mathbb{E}_{\mathbb{P}}[X_{\mathcal{H}_n}1_A]=\mathbb{Q}(A)$ and, \begin{equation*}\begin{split} \left|\mathbb{E}_\mathbb{P}[X_\mathcal{G}1_A]-\mathbb{Q}(A)\right| &=\lim_{n\rightarrow\infty}\left|\mathbb{E}_\mathbb{P}[X_{\mathcal{\tilde G}_n}1_A]-\mathbb{Q}(A)\right|\\ &=\lim_{n\rightarrow\infty}\left|\mathbb{E}_\mathbb{P}[X_{\mathcal{\tilde G}_n}1_A]-\mathbb{E}_\mathbb{P}[X_{\mathcal{H}_n}1_A]\right|\\ &\le\lim_{n\rightarrow\infty}\mathbb{E}_\mathbb{P}[|X_{\mathcal{\tilde G}_n}-X_{\mathcal{G}_n}|]+\lim_{n\rightarrow\infty}\mathbb{E}_\mathbb{P}[|X_{\mathcal{H}_n}-X_{\mathcal{G}_n}|]\\ &\le\lim_{n\rightarrow\infty}(2^{-n}+2^{-n})=0. \end{split}\end{equation*}So, $\mathbb{E}_{\mathbb{P}}[X_\mathcal{G}1_A]=\mathbb{Q}(A)$ as required.

Bibliography

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David Williams, Probability with martingales, Cambridge Mathematical Textbooks, Cambridge University Press, 1991.