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


$\newcommand{\ep}{\varepsilon}\newcommand{\om}{\omega}\newcommand{\Om}{\Omega}\newcommand{\Si}{\Sigma}$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 $\ell(\infty)$.

  • $\begingroup$ I find this solution fantastic. Thanks for your help $\endgroup$
    – Neymar
    Jun 4 '21 at 7:38
  • $\begingroup$ @Neymar : I am glad this was of help. $\endgroup$ Jun 4 '21 at 11:46

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