# martingale proof of the Radon-Nikodym theorem

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 (http://planetmath.org/AnySigmaFiniteMeasureIsEquivalentToAProbabilityMeasure), 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.

Let $\mathbb{P}$ and $\mathbb{Q}$ be probability measures on the measurable space $(\Omega,\mathcal{F})$, such that $\mathbb{Q}$ is absolutely continuous with respect to $\mathbb{P}$. Then, there exists a non-negative random variable  $X$ such that $\mathbb{E}_{\mathbb{P}}[X]=1$ and $\mathbb{Q}(A)=\mathbb{E}_{\mathbb{P}}[1_{A}X]$ for every $A\in\mathcal{F}$.

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

 $\begin{split}\displaystyle\mathbb{E}_{\mathbb{P}}[\max(X_{\mathcal{G}}-\tilde{% X}_{\mathcal{G}},0)]&\displaystyle=\mathbb{E}_{\mathbb{P}}[X_{\mathcal{G}}1_{% \left\{X_{\mathcal{G}}>\tilde{X}_{\mathcal{G}}\right\}}]-\mathbb{E}_{\mathbb{P% }}[\tilde{X}_{\mathcal{G}}1_{\left\{X_{\mathcal{G}}>\tilde{X}_{\mathcal{G}}% \right\}}]\\ &\displaystyle=\mathbb{Q}(X_{\mathcal{G}}>\tilde{X}_{\mathcal{G}})-\mathbb{Q}(% X_{\mathcal{G}}>\tilde{X}_{\mathcal{G}})=0\end{split}$

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.

###### Lemma 1.

If $\mathcal{G}$ is a finite sub-$\sigma$-algebra of $\mathcal{F}$ then the Radon-Nikodym derivative $X_{\mathcal{G}}$ exists.

###### 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

 $X_{\mathcal{G}}=\sum_{k=1}^{n}\frac{\mathbb{Q}(A_{k})}{\mathbb{P}(A_{k})}1_{A_% {k}}$

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

 $\begin{split}\displaystyle\mathbb{E}_{\mathbb{P}}[1_{A_{k}}X_{\mathcal{G}}]&% \displaystyle=\frac{\mathbb{Q}(A_{k})}{\mathbb{P}(A_{k})}\mathbb{E}_{\mathbb{P% }}[1_{A_{k}}]+\sum_{j\not=k}\frac{\mathbb{Q}(A_{j})}{\mathbb{P}(A_{j})}\mathbb% {E}_{\mathbb{P}}[1_{A_{k}\cap A_{j}}]\\ &\displaystyle=\mathbb{Q}(A_{k}).\end{split}$

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 $\mathbb{R}^{n}$.

###### Lemma 2.

Let $\mathcal{G}$ be a separable sub-$\sigma$-algebra of $\mathcal{F}$. Then, the Radon-Nikodym derivative $X_{\mathcal{G}}$ exists. If furthermore, $\mathcal{G}_{n}$ is an increasing sequence of finite $\sigma$-algebras satisfying $\mathcal{G}=\sigma(\bigcup_{n}\mathcal{G}_{n})$ then $\mathbb{E}_{\mathbb{P}}[|X_{\mathcal{G}}-X_{\mathcal{G}_{n}}|]\rightarrow 0$ as $n\rightarrow\infty$.

###### Proof.

Let us set $X_{n}\equiv X_{\mathcal{G}_{n}}$. If $m then the conditional expectation $\mathbb{E}_{\mathbb{P}}[X_{n}\mid\mathcal{G}_{m}]$ is $\mathcal{G}_{m}$-measurable, and for every $A\in\mathcal{G}_{m}$,

 $\mathbb{E}_{\mathbb{P}}\left[1_{A}\mathbb{E}_{\mathbb{P}}[X_{n}\mid\mathcal{G}% _{m}]\right]=\mathbb{E}_{\mathbb{P}}\left[1_{A}X_{n}\right]=\mathbb{Q}(A).$

This equality just uses the definition of the conditional expectation and then the definition of $X_{n}$ as the Radon-Nikodym derivative restricted to $\mathcal{G}_{n}$. So, $\mathbb{E}_{\mathbb{P}}[X_{n}\mid\mathcal{G}_{m}]$ is the Radon-Nikodym derivative restricted to $\mathcal{G}_{m}$, and equals $X_{m}$ (almost-surely).

Therefore, $X_{n}$ is a martingale and the martingale convergence theorem implies that the limit

 $X_{\mathcal{G}}=\lim_{n\rightarrow\infty}X_{n}$ (1)

exists almost surely. We now show that the sequence $X_{n}$ is uniformly integrable. Choose any $\epsilon>0$. As $\mathbb{Q}$ is absolutely continuous with respect to $\mathbb{P}$, there exists a $\delta>0$ such that $\mathbb{Q}(A)<\epsilon$ whenever $\mathbb{P}(A)<\delta$. Using

 $\mathbb{P}(X_{n}>K)=\mathbb{E}_{\mathbb{P}}[1_{\{X_{n}>K\}}]\leq\mathbb{E}_{% \mathbb{P}}\left[\frac{X_{n}}{K}\right]=\frac{1}{K}$

we see that $\mathbb{P}(X_{n}>K)<\delta$ whenever $K>\delta^{-1}$ and, therefore, $\mathbb{Q}(X_{n}>K)<\epsilon$. So

 $\mathbb{E}_{\mathbb{P}}[X_{n}1_{\{X_{n}>K\}}]=\mathbb{Q}(X_{n}>K)<\epsilon$

for every $n$, showing that $X_{n}$ is a uniformly integrable sequence with respect to $\mathbb{P}$. Therefore, convergence in (1) is in $L^{1}$, and $\mathbb{E}_{\mathbb{P}}[|X_{n}-X_{\mathcal{G}}|]\rightarrow 0$ as $n\rightarrow\infty$. So, for any $A\in\bigcup_{n}\mathcal{G}_{n}$,

 $\mathbb{E}_{\mathbb{P}}[X_{\mathcal{G}}1_{A}]=\lim_{m\rightarrow\infty}\mathbb% {E}_{\mathbb{P}}[X_{m}1_{A}]=\mathbb{Q}(A).$ (2)

By linearity and the monotone convergence theorem  , the collection  of sets $A$ satisfying (2) is a Dynkin system containing the $\pi$-system $\bigcup_{n}\mathcal{G}_{n}$ so, by Dynkin’s lemma, is satisfied for every $A\in\sigma(\bigcup_{n}\mathcal{G}_{n})=\mathcal{G}$ and, by definition, $X_{\mathcal{G}}$ is the Radon-Nikodym derivative restricted to $\mathcal{G}$. ∎

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

 $\epsilon\leq\lim_{n\rightarrow\infty}\mathbb{E}_{\mathbb{P}}[|X_{\mathcal{G}_{% n}}-X_{\mathcal{G}_{n+1}}|]\leq\lim_{n\rightarrow\infty}\mathbb{E}_{\mathbb{P}% }[|X_{\mathcal{G}_{n}}-X_{\mathcal{G}}|]+\lim_{n\rightarrow\infty}\mathbb{E}_{% \mathbb{P}}[|X_{\mathcal{G}_{n+1}}-X_{\mathcal{G}}|]=0$

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,
 $\begin{split}\displaystyle\left|\mathbb{E}_{\mathbb{P}}[X_{\mathcal{G}}1_{A}]-% \mathbb{Q}(A)\right|&\displaystyle=\lim_{n\rightarrow\infty}\left|\mathbb{E}_{% \mathbb{P}}[X_{\mathcal{\tilde{G}}_{n}}1_{A}]-\mathbb{Q}(A)\right|\\ &\displaystyle=\lim_{n\rightarrow\infty}\left|\mathbb{E}_{\mathbb{P}}[X_{% \mathcal{\tilde{G}}_{n}}1_{A}]-\mathbb{E}_{\mathbb{P}}[X_{\mathcal{H}_{n}}1_{A% }]\right|\\ &\displaystyle\leq\lim_{n\rightarrow\infty}\mathbb{E}_{\mathbb{P}}[|X_{% \mathcal{\tilde{G}}_{n}}-X_{\mathcal{G}_{n}}|]+\lim_{n\rightarrow\infty}% \mathbb{E}_{\mathbb{P}}[|X_{\mathcal{H}_{n}}-X_{\mathcal{G}_{n}}|]\\ &\displaystyle\leq\lim_{n\rightarrow\infty}(2^{-n}+2^{-n})=0.\end{split}$
So, $\mathbb{E}_{\mathbb{P}}[X_{\mathcal{G}}1_{A}]=\mathbb{Q}(A)$ as required.