I have a question concerning point 3 of the following problem:

Let $(\Omega,\mathfrak{F},\mathbb{P},\mathfrak{(F_n})_{n\in\mathbb{N}})$ be a filtered probability space and $\mu$ a finite measure on $\mathfrak{F_\infty}:=\sigma{(\cup_{n=1}^{\infty}\mathfrak{F_n}})$. Assume that, for every $n\geq0$, the measure $\mu$ is absolutely continuous with respect to $\mathbb{P}$ on $\mathfrak{F_n}$, and denote by $M_n$ the Radon-Nikodym density: $M_n$ is thus non-negative, $\mathfrak{F_n}$-measurable and for every $A \in \mathfrak{F_n}$,$$\mu(A)=\int_{A}M_n \;d\mathbb{P}$$

1. Prove that $(M_n)_{n\geq0}$ is a martingale. (OK, I showed it)
2. Prove that $(M_n)_{n\geq0}$ converges a.s. to an integrable random variable $M_{\infty}$. (Ok, I showed it using a corollary of Doob's martingale convergence theorem)
3. Prove that $\mu$ is absolutely continuous with respect to $\mathbb{P}$ on $\mathfrak{F_\infty}$ if and only if the martingale $(M_n)$ is closed and, that, in this case $M_{\infty}$ is the Radon-Nikodym density.

I have a problem with point 3. I do not know how to start the proof knowing that $\mu$ is absolutely continuous with respect to $\mathbb{P}$ on $\mathfrak{F_\infty}$. I would like to show that $M_n$ is uniformly integrable (bounded in $L^1$ is already good, but $\mathbb{P}$-continuous?) because then $M_n$ converges in $L^1$ and this would help a lot. But this is perhaps the wrong way to attack the problem. Perhaps somebody knows the solution.

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I don't know the complete solution, but if $\mu \ll \Bbb P$ on $\mathfrak F_\infty$ wouldn't it help if you just take $M:= \frac{\mathrm d\mu}{\mathrm d\Bbb P}$ and show that in fact $M_n = \Bbb E[M|\mathfrak F_\infty]$ since this it a Levy martingale? –  Ilya Jan 25 at 10:47

• $\Leftarrow$: Assume that $(M_n)_n$ is closed, i.e. $\mathbb{E}(Y|\mathcal{F}_n)=M_n$ for some (integrable) random variable $Y$. Hence $$\mu(F) \stackrel{\text{Def}}{=} \mathbb{E}(M_n \cdot 1_F) = \mathbb{E}(Y \cdot 1_F)$$ for all $F \in \mathcal{F}_n$, i.e. the measure $F \mapsto \mu(F)$, $F \mapsto \mathbb{E}(Y \cdot 1_F)$ coincide on $\bigcup_n \mathcal{F}_n$. This is a generator of $\mathcal{F}_{\infty}$!
• $\Rightarrow$: By the Radon-Nikodym theorem there exists a $\mathcal{F}_{\infty}$ measurable (posiitve) random variable $Y$ such that $$\mu(F) = \mathbb{E}(Y \cdot 1_F)$$ for $F \in \mathcal{F}_{\infty}$. Hence $$\mathbb{E}(M_n \cdot 1_F)=\mu(F) = \mathbb{E}(Y \cdot 1_F)$$ for $F \in \mathcal{F}_n$, i.e. $\mathbb{E}(Y|\mathcal{F}_n)=M_n$.

Moreover, we know that $(M_n)_n$ is equi-integrable (since $M_n = \mathbb{E}(Y|\mathcal{F}_n)$), so there exists $Z \in L^1(\mathcal{F}_{\infty})$ such that $M_n \to Z$ in $L^1$, almost surely. Since $M_n \to M_{\infty}$ almost surely, we conclude $Z=M_{\infty}$. By the $L^1$-convergence we obtain

$$\mathbb{E}(M_{\infty} \cdot 1_F) = \lim_{n \to \infty} \mathbb{E}(M_n \cdot 1_F) = \mathbb{E}(Y \cdot 1_F)$$

for all $F \in \bigcup_n \mathcal{F}_n$ and therefore $M_{\infty} = Y$ almost surely (since the random variables are $\mathcal{F}_\infty$-measurable).

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Thank you! I am just wondering why you have that $M_n$ converges to $M_\infty$ in $L^1$ –  Mathoman Jan 25 at 17:33
@Mathoman I added some more hints concerning the $L^1$-convergence. –  saz Jan 25 at 22:51
Thank you for your perfect answer! It helps me a lot! –  Mathoman Jan 26 at 9:26