# 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 $$\liminf_{n\to\infty}L^n_{\infty} \ge \ell(\infty)$$.

Indeed, let $$(\Om,\Si,P)$$ be the underlying probability space. Suppose that there exists a sequence $$(\Om_n)$$ in $$\Si$$ such that $$$$P(\Om_n)\to0$$$$ (as $$n\to\infty$$) but $$$$\bigcup_{n=m}^\infty\Om_n=\Om$$$$ for all natural $$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$$ and $$$$L^n_t(\om):=1(t\ge n,\om\in\Om_n)$$$$ for all natural $$n$$, all $$\om\in\Om$$, and all real $$t\ge0$$. Then all the conditions imposed by you on $$L^n_t$$ and $$\ell$$ hold. In particular, we have $$$$L^n_\infty(\om):=1(\om\in\Om_n)$$$$ and hence $$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)$$.