$L_1$ convergence implies a.s. convergence for a sequence that is monotonically decreasing in $n$?

Question

Basically my question is how to prove (if true) that $L_1$ convergence implies a.s. convergence for a sequence that is monotonically decreasing in $n$. (not just for a subsequence)

More formally, for a sequence of measurable functions $\{X_n\}_{n=1}^\infty$,

If $0\leq X_{n+1}(\omega) \leq X_{n}(\omega)$ for each $n$ and each $\omega$, and $\mathbb{E}[X_n]\to0$ then $X_n \to 0$ almost surely.

It was inspired by this post When does the $L_1$ convergence imply almost everywhere convergence? but I have not been able to find a reference or proof for it.

Answer 1

Suppose for a contradiction that $X_n$ does not converge to $0$ almost surely. Then there exists a set $E$ of positive measure and $\varepsilon > 0$ such that for each $\omega \in E$ there exists a sequence $n_k(\omega)$ such that $X_{n_k(\omega)}(\omega) > \varepsilon$ for all $k \geq 1$.

By Monotonicity, we obtain that $X_n(\omega) > \varepsilon$ for all $n \geq 1$.

This implies that $\mathbb{E}[X_n] \geq \mathbb{P}(E) \varepsilon > 0$ which is in contradiction with $L^1$ convergence to $0$.

Answer 2

Let $X(\omega)=\inf_nX_n(\omega)=\lim_nX_n(\omega)$. Then by the Monotone Convergence Theorem, one has that $X_n\to X$ in $L^1$.

By assumption $X_n\to0$ in $L^1$ as well, so the uniqueness of the limit implies that $X=0$ a.s.