Does convergence under expectation implies convergence almost surely? 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  $\liminf_{n\to\infty}L^n_{\infty} \ge \ell(\infty)$.
 A: $\newcommand{\ep}{\varepsilon}\newcommand{\om}{\omega}\newcommand{\Om}{\Omega}\newcommand{\Si}{\Sigma}$The answer is: not in general.
Indeed, let $(\Om,\Si,P)$ be the underlying probability space. Suppose that there exists a sequence $(\Om_n)$ in $\Si$ such that
\begin{equation}
    P(\Om_n)\to0
\end{equation}
(as $n\to\infty$) but
\begin{equation}
    \bigcup_{n=m}^\infty\Om_n=\Om
\end{equation}
for all natural $m$; such a sequence exists if e.g. $(\Om,\Si,P)$ is the standard probability space with $\Om=[0,1]$.
Let then $\ell(t):=0$ for all real $t\ge0$ and
\begin{equation}
    L^n_t(\om):=1(t\ge n,\om\in\Om_n)
\end{equation}
for all natural $n$, all $\om\in\Om$, and all real $t\ge0$. Then all the conditions imposed by you on $L^n_t$ and $\ell$ hold. In particular, we have
\begin{equation}
    L^n_\infty(\om):=1(\om\in\Om_n)
\end{equation}
and hence $EL^n_\infty=P(\Om_n)\to0=\ell(\infty)$.
However, for all $\om\in\Om$
\begin{align}
    \limsup_{n\to\infty}L^n_\infty(\om)&=\lim_{m\to\infty}\sup_{n=m}^\infty L^n_\infty(\om) \\ 
    &=\lim_{m\to\infty}\sup_{n=m}^\infty 1(\om\in\Om_n) \\ 
    &=\lim_{m\to\infty}1(\om\in\bigcup_{n=m}^\infty\Om_n) \\ 
    &=\lim_{m\to\infty}1(\om\in\Om)=1\ne0=\ell(\infty). 
\end{align}
So, $L^n_\infty$ does not converge almost surely to $\ell(\infty)$.
