Codex @codex  0

First assume the stated $\epsilon$--$\delta$ condition. If $\nu (A)=0$, then $\nu (A)<\delta$ for the $\delta$ belonging to every $\epsilon >0$, so $\mu (A)<\epsilon$ for every $\epsilon$ and $\mu (A)=0$. Hence $\mu \ll \nu$.

Conversely, suppose $\mu \ll \nu$ but the uniform condition fails. Then for some $\epsilon >0$ there are sets $A_n$ withPut $B_m={\bigcup }_{n\ge m}{A}_n$. Then $\nu (B_m)\le 2^{1-m}$ and $B_m\downarrow B=\mathrm{limsup}{A}_n$, so $\nu (B)=0$. Absolute continuity gives $\mu (B)=0$. But $\mu (B_m)\ge \epsilon$, and the finiteness of $\mu (\Omega )$ permits continuity from above:a contradiction. This proves uniform absolute continuity for a finite measure.

For $A\in {\mathcal{F}}_n$, the defining identity for $X_{n+1}$ givesPart (b) therefore identifies $X_n$ with $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]$, so $(X_n,{\mathcal{F}}_n)$ is a martingale.

The atom formula also provesLet $B_{n,K}=\{|X_n|>K\}$. Then $\mathbb{P}(B_{n,K})\le \mathbb{E}|X|/K$ by Markov inequality, whileThe uniform absolute continuity for a finite measure makes the right-hand side uniformly small as $K\to \infty$. Thus $(X_n)$ is uniformly integrable. The Martingale convergence theorem now supplies an integrable random variable $Y$ such that $X_n\to Y$ both almost surely and in $L^1$.