Let $(L^n)_{n\ge 1}$ be a sequence of non-decreasing and right-continuous stochastic processes s.t. $0\le L^n_t\le 1$ for all $t\ge 0$. Let $\ell:[0,\infty)\to [0,1]$ be a non-decreasing and right-continuous function. Assume that one has almost surely

$$\lim_{n\to\infty}L^n_t=\ell(t),\quad \mbox{for all the points of continuity $t$ of } \ell.$$

My question is whether $\lim_{n\to\infty}\mathbb E[L^n_{\infty}]=\ell(\infty)$ implies that $\lim_{n\to\infty}L^n_{\infty}=\ell(\infty)$ (almost surely)? Here we set $L^n_{\infty}:=\lim_{t\to\infty}L^n_t$ and $\ell(\infty):=\lim_{t\to\infty}\ell(t)$.

**Personal thoughts :** Take a sequence $(t_m)_{m\ge 1}$ s.t. $t_m\uparrow \infty$ and $\ell$ is continuous at $t_m$. Then

$$L^n_{\infty}\ge L^n_{t_m}~~\Longrightarrow~~ \limsup_{n\to\infty}L^n_{\infty}\ge \liminf_{n\to\infty}L^n_{\infty} \ge \ell(t_m).$$

Letting $m\to\infty$, it holds $\limsup_{n\to\infty}L^n_{\infty}\ge \liminf_{n\to\infty}L^n_{\infty} \ge \ell(\infty).$ On the other hand, one obtains by Fatou's lemma

$$\ell(\infty)=\liminf_{n\to\infty}\mathbb E[L^n_{\infty}] \ge \mathbb E[\liminf_{n\to\infty}L^n_{\infty}] \ge \ell(\infty),$$

which yields $\liminf_{n\to\infty}L^n_{\infty} = \ell(\infty)$ as $\liminf_{n\to\infty}L^n_{\infty} \ge \ell(\infty)$.