TITLE: Topology of almost sure convergence
QUESTION [5 upvotes]: If $(\Omega ,\mathcal{A},\mathbb{P})$ is a probability space and $(Y,\tau)$ a topological space, we usually say that a sequence $X_n$ of measurable functions : $\Omega \rightarrow Y$ converges almost surely (a.s.) to $L$ measurable if $\mathbb{P}(X_n \rightarrow L)=1$.
I wonder from which topology this convergence is. If $\tau^\Omega$ is the product topology on $Y^\Omega$, $\tau'$ the induced topology on the measurable functions $\mathcal{M}(\Omega,Y)$ and $\mathcal{R}$ the equivalence relation of equality a.s., I thought that maybe the right topology would be $\tau'/\mathcal{R}$, the quotient topology on $\mathcal{M}(\Omega,Y)/\mathcal{R}$. It is easy to show that if $X_n$ converges to $L$ a.s., then $\tilde{X_n}$ (the equivalence class of $X_n$) converges to $\tilde{L}$ for $\tau'/\mathcal{R}$ (one can take a representant of $\tilde{X_n}$ which converges to $L$ for $\tau'$, and use the continuity of the projection).
Though I do not manage to prove the necessary condition. Can someone prove that $\tilde{X_n} \rightarrow \tilde{L}$ for $\tau'/\mathcal{R}$ implies that $X_n \rightarrow L$ a.s. or give me a counterexample ?

REPLY [3 votes]: Suppose $\tau'/\mathcal{R}$ were a topology such that $\tilde{X_n} \rightarrow \tilde{L}$ for $\tau'/\mathcal{R}$ if and only if $X_n \rightarrow L$ almost surely. Let $p:\mathcal{M}(\Omega,Y)\rightarrow \mathcal{M}(\Omega,Y)/\mathcal{R}$ be the canonical projection. Let $\tau_{\mathcal{R}}$ be the coarsest topology which makes $p$ continuous. We have $X_n \rightarrow L$ for $\tau_{\mathcal{R}}$ if and only if $\tilde{X_n} \rightarrow \tilde{L}$ for $\tau'/\mathcal{R}$. Then $X_n \rightarrow L$ almost surely if and only if $X_n \rightarrow L$ for $\tau_{\mathcal{R}}$, which is impossible as explained in the answer to this question. Thanks to Niels Diepeveen for his comments.