Let
- $\mathbb{F}$ be a filtration
- $(X_t)_{t\ge 0}$ be a right-continuous $\mathbb{F}$-supermartingale
- $\sigma,\tau$ be bounded $\mathbb{F}$-stopping times with $\sigma\le \tau$ and $\sigma^n:=2^{-n}\lceil 2^n\sigma\rceil$, $\tau^n:=2^{-n}\lceil 2^n\tau\rceil$
I want to show, that we've got $$\operatorname{E}\left[X_{\tau^m}\mid\mathcal{F}_{\sigma^n}\right]\stackrel{n\to\infty}{\to}\operatorname{E}\left[X_{\tau^m}\mid\mathcal{F}_\sigma\right]\;\;\;\text{and}\;\;\;X_{\sigma^n}\stackrel{n\to\infty}{\to}X_\sigma\tag{1}$$ almost surely and in $L^1$.
Maybe it's not that hard to prove $(1)$, but I absolutely don't find a starting point. So, what do we need to do to prove $(1)$?
So let's see that $X_{\sigma^n} \to X_{\sigma}$ in $L^1$. Since $\sigma^n \downarrow \sigma$ right continuity gives us that $X_{\sigma^n} \to X_{\sigma}$ almost surely. Now call $X(n) = X_{\sigma^n}$ and $\mathcal{F}_n = \mathcal{F}_\sigma$.
Note that
1) $X(n)$ is $\mathcal{F}_n$ adapted
2) $X(n) \in L^1$ for every $n$
3)$\Bbb{E}[X(n+1) - X(n)\vert \mathcal{F}_n] \geq 0$
4) $\Bbb{E}[X_n]\leq\Bbb{E}[X(n+1)]\leq \Bbb{E}[X_\sigma] <\infty$
Let $\Delta_{n} = \Bbb{E}[X(n) - X(n-1)\vert \mathcal{F}_{n-1}] \geq 0$.
Let $A_{n} = \sum_{i=2}^{n} \Delta_i$
$$\Bbb{E}[A_n] = \Bbb{E}[X(n) - X(1)]\leq \Bbb{E}[X_\sigma - X(1)]< \infty$$
$A_{\infty}:= \sum_{i=2}^\infty \Delta_i$.By the monotone convergence theorem $\Bbb{E}[A_\infty]< \infty$.
Define now, $M(n) = X(n) - A(n)$. Note that $M_n$ is a martingale. Indeed $$\Bbb{E}[M(n+1) - M(n)\, \vert \mathcal{F}_n] = \Bbb{E}[X(n+1) - X(n) - \Delta_{n+1}\, \vert \mathcal{F}_n] = 0 $$
Since $\mathcal{F}_n \subset \mathcal{F}_{n-1}$, $$\Bbb{E}[M(1)- M(n)\vert \mathcal {F}_n] = \Bbb{E}[\Bbb{E}[M(1)- M(n)\vert \mathcal {F}_1]\vert\mathcal {F}_n]=0$$ $M(n)$ is uniformly integrable.
Moreover since $A_n \leq A_\infty \in L^1$ $A_n$ is also uniformly integrable. Therefore $X_n = M_n + A_n$ is also uniformly integrable.
Thus, almost sure convergence (from right continuity) implies convergence in law which combined with Uniform integrability yields convergence in $L^1$.
we now want to prove $$\operatorname{E}\left[X_{\tau^m}\mid\mathcal{F}_{\sigma^n}\right]\stackrel{n\to\infty}{\to}\operatorname{E}\left[X_{\tau^m}\mid\mathcal{F}_\sigma\right]$$ for $T = \tau^m$ $$f_n(\omega)=\Bbb{E}[X_T \vert \mathcal{F}_{\sigma^n}](\omega)$$ Is almost an uniformly integrable backward martingale therefore it converges almost surely and in $L^1$.
To see that where is the backward martingale remember that $\mathcal{F}_{\sigma^n} \supset \mathcal{F}_{\sigma^m}$ when $m>n$ $$\Bbb{E}[f_n \vert \mathcal{F}_{\sigma^m}] = f_m $$
Then consider $Y_{n} = f_{-n}$ and $\mathcal{F}_{n} = \mathcal{F}_{\sigma^{-n}}$ now this sequence is a an uniformly integrable backward martingale.
Then the limit $Y_{-\infty}= \lim_{n\to -\infty} X_n$ exists almost surely and in $L^1$ define $f = Y_{-\omega}$ and note that $\lim_{n\to \infty} f_n(\omega) = f(\omega)$ is our natural candidate for $\Bbb{E}[X_T \vert \mathcal{F}_{\sigma}]$ since $\sigma^n \downarrow \sigma$.
We need to check two properties
A) $\int_A f(\omega)\,d\Bbb{P}(\omega) =\int_A X_T(\omega)\,d\Bbb{P}(\omega) $ for every $A \in \mathcal{F}_\sigma$
Since $A \in \mathcal{F}_\sigma \Rightarrow A \in \mathcal{F}_{\sigma^n}$ we have for each $n > 0$: $$\int_A f_n(\omega)\,d\Bbb{P}(\omega) =\int_A X_{T}\,d\Bbb{P}(\omega) $$
Therefore since $X_{\sigma^n} \xrightarrow[]{L^1} X_{\sigma} $ taking limits in both sides we obtain (use monotone convergence for the left side) $$\int_A f(\omega)\,d\Bbb{P}(\omega) =\int_A X_{T}(\omega)\,d\Bbb{P}(\omega) $$
B) the last condition we need to check is that $f(\omega)$ is $\mathcal{F}_{\sigma}$ measurable. Since $f_n$ is $\mathcal{F}_{\sigma^n}$ measurable and $\mathcal{F}_{\sigma^n}\supset\mathcal{F}_{\sigma^{n+1}}$we conclude that $f$ is $\mathcal{F}_{\sigma^n}$ measurable for every $n$. since $\sigma^n \downarrow \sigma$ $forall \delta>0$ we choose $n$ such that $\mathcal{F}_{\sigma^n} \subset \mathcal{F}_{\sigma + \delta}$ This implies that $f is \mathcal{F}_{\sigma}^+$ measurable. If your filtration satisfies the usual conditions (see https://almostsure.wordpress.com/2009/11/08/filtrations-and-adapted-processes/) then you conclude that $f$ is $\mathcal{F}_{\sigma}$ measurable and therefore that
$$\lim_n \Bbb{E}[X_T \vert \mathcal{F}_{\sigma^n}] = \lim_n f_n(\omega) = f(\omega) = \Bbb{E}[X_T \vert \mathcal{F}_{\sigma}]$$