Right continuity of Expectation of the Limit of a Right-Continuous Supermartingale Sequence It is known that if $\left\{ X_{t},\mathscr{F}_{t}:0\le t<\infty\right\}  $ be a right-continuous supermartingale, then the function $t\mapsto EX_{t} $ is right-continuous. How about the limit of a Right-Continuous Supermartingale Sequence. To be specific, 
Suppose that the filtration $\left\{ \mathscr{F}_{t}\right\}$ satisfies the usual conditions and let $X^{\left(n\right)}=\left\{ X_{t}^{\left(n\right)},\mathscr{F}_{t}:0\le t<\infty\right\} $, $n\geq 1$ be an increasing sequence of right-continuous supermartingales, such that the random variable $\xi_{t}={\displaystyle \lim_{n\rightarrow\infty}X_{t}^{\left(n\right)}}$ is nonnegative and integrable for every $0\le t<\infty$.
I can show that $\left\{ \xi_{t},\mathscr{F}_{t}:0\le t<\infty\right\} $ is a supermartingale, but it may not be right-continuous. To show it has  a right-continuous modification, I need to show that $t\rightarrow E\xi_{t}$ is right-continuous. But how? Can anyone give me a hint? Thanks very much in advance!
 A: Following the comment @John Dawkins, I completed this proof. Here is the complete proof.
Firstly let's show an lemma:
Lemma: if $\left\{ a_{m,n}\right\}$  is a collection of numbers with $m\mapsto a_{m,n}$ non-decreasing ($n$ fixed) and $n\mapsto a_{m,n}$ non-decreasing ($m$ fixed), then 
$\lim_{m\rightarrow\infty}\lim_{n\rightarrow\infty}a_{m,n}=\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}a_{m,n}$
Proof of Lemma: Denote $b_{n}={\displaystyle \lim_{m\rightarrow\infty}a_{m,n}} $and $c_{m}={\displaystyle \lim_{n\rightarrow\infty}a_{m,n}}$. Monotonicity implies that these limits exist (they might be $+\infty$) and the sequences $\left\{ b_{n}\right\}$ , $\left\{ c_{m}\right\}$  are non-decreasing. Hence let $b={\displaystyle \lim_{n\rightarrow\infty}b_{n}}$ and $c{\displaystyle =\lim_{m\rightarrow\infty}c_{m}}$. We have $a_{m,n}\le c_{m}\le c\Longrightarrow b_{n}\le c\Longrightarrow b\le c $. The proof that $c\le b$ is analogous. Hence $\lim_{m\rightarrow\infty}\lim_{n\rightarrow\infty}a_{m,n}=\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}a_{m,n}$
End of Lemma Proof.
Hence for any $\left\{ t_{m}\right\}$  with $t_{m}\searrow t$, we have $\lim_{m\rightarrow\infty}E\xi_{t_{m}}=\lim_{m\rightarrow\infty}E\left(\lim_{n\rightarrow\infty}X_{t_{m}}^{\left(n\right)}\right)$. Since $X^{\left(n\right)}$ is an increasing sequence, we have $E\left(\lim_{n\rightarrow\infty}X_{t_{m}}^{\left(n\right)}\right)=\lim_{n\rightarrow\infty}E\left(X_{t_{m}}^{\left(n\right)}\right) $ by monotone convergence. Note that $\left\{ X_{t}^{\left(n\right)},\mathscr{F}_{t}:0\le t<\infty\right\}$  is right-continuous supermartingale. We have $E\left(X_{t_{m}}^{\left(n\right)}\right)\le E\left(X_{t_{m+1}}^{\left(n\right)}\right)and X^{\left(n\right)}$ is an increasing sequence implies $E\left(X_{t_{m}}^{\left(n\right)}\right)\le E\left(X_{t_{m}}^{\left(n+1\right)}\right)$ hence by lemma we have
$\lim_{m\rightarrow\infty}E\xi_{t_{m}} =\lim_{m\rightarrow\infty}E\left(\lim_{n\rightarrow\infty}X_{t_{m}}^{\left(n\right)}\right)
 =\lim_{m\rightarrow\infty}\lim_{n\rightarrow\infty}E\left(X_{t_{m}}^{\left(n\right)}\right)
 =\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}E\left(X_{t_{m}}^{\left(n\right)}\right)
 =\lim_{n\rightarrow\infty}E\left(\lim_{m\rightarrow\infty}X_{t_{m}}^{\left(n\right)}\right) 
=\lim_{n\rightarrow\infty}E\left(X_{t}^{\left(n\right)}\right)=E\left(\lim_{n\rightarrow\infty}X_{t}^{\left(n\right)}\right)=E\left(\xi_{t}\right) $ 
by using monotone convergence theorems twice. Hence $t\rightarrow E\xi_{t}$ is right-continuous. 
