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 $\nu(A)=\int_{A}f\,d\mu$ 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 $\mathbb{Q}(A)=\mathbb{E}_{\mathbb{P}}[1_{A}X_{\mathcal{G}}]$ 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

so $X_{\mathcal{G}}\leq\tilde{X}_{\mathcal{G}}$ almost surely. Similarly, $\tilde{X}_{\mathcal{G}}\leq X_{\mathcal{G}}$ 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.

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 $\mathbb{E}_{\mathbb{P}}[|X_{{\mathcal{G}_{n}}}-X_{{\mathcal{G}_{m}}}|]\geq\epsilon$. However, letting $\mathcal{G}=\sigma(\bigcup_{n}\mathcal{G}_{n})$, Lemma 2 shows that $X_{\mathcal{G}}$ exists and

— 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,

So, $\mathbb{E}_{{\mathbb{P}}}[X_{\mathcal{G}}1_{A}]=\mathbb{Q}(A)$ as required.