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 $\liminf\_{n\to\infty}L^n\_{\infty} \ge \ell(\infty)$.



Does convergence under expectation implies convergence almost surely?
$\newcommand{\ep}{\varepsilon}\newcommand{\om}{\omega}\newcommand{\Om}{\Omega}\newcommand{\Si}{\Sigma}$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 $\ell(\infty)$.


