Let $(\Omega, \mathcal F, P)$ be a probability space and denote by $L^0(\Omega, \mathcal F, P)$ the space of all random variables $X : \Omega \to \mathbb R$ (i.e. measurable functions between $(\Omega, \mathcal F)$ and $(\mathcal R, \mathcal B)$). Let this space equipped with the topology given by convergence in probability, i.e. $X_n \to X$ iff for each $\varepsilon > 0$ $$ \lim_{n\to \infty} P(|X_n - X| \ge \varepsilon) = 0. $$ Also let $L^1(\Omega, \mathcal F, P)$ be the usual space with $L^1$-norm and the topology induced by it.
Is every set closed in $L^0(\Omega,\mathcal F, P)$ also closed in $L^1(\Omega, \mathcal F, P)$ with respect to the $L^1$-norm?
I think this does it: if a sequence converges in $L^1$, then a subsequence converges a.s. to the same limit. That subsequence converges in probability by Egorov's theorem. So the $L^1$ limit of an $L^1$-convergent sequence must be in an $L^0$-closed set.
The only catch here is that this assumes that you are dealing with a subset of $L^1$ which is closed in $L^0$. This won't work if you start with a set which isn't even contained in $L^1$, such as all of $L^0$.