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Let $S,T\in\mathbb R$ such that $0\leq S<T<\infty$. For a function $X:\mathbb R\times \Omega\to\mathbb R$ I know that $$ \lim\limits_{t\to s}\int_\Omega (X(t,\omega)-X(s,\omega))^2\mathsf P(d\omega) = 0 \quad (1) $$ for all $s\in \mathbb R$. Here $\mathsf P$ is a probability measure on $\Omega$ but the probabilistic part of the question is not the most important one. I have to show, that given a family of partitions of $[S,T]$: $$ \mathscr T^n = \{S = t_0^n<t_1^n<\dots<t_n^n = T\} $$ the following holds: $$ \lim\limits_{\mu_n\to 0}\int\limits_S^T\left(\int_\Omega (X(t,\omega) - X^n(t,\omega))^2\mathsf P(d\omega)\right)dt = 0 $$ where $\mu_n = \max\limits_{j=0,\dots,n-1}\Delta t^n_j = \max\limits_{j=0,\dots,n-1}(t^n_{j+1} - t_j^n)$ and $$ X^n_t = \sum\limits_{j=0}^{n-1}X(t_j,\omega)1_{[t_j^n,t_{j+1}^n)}(t) $$ where $1_A(t)$ is a characteristic function of $A$. I've obtain the following bound: $$ \int\limits_S^T\left(\int_\Omega (X(t,\omega) - X^n(t,\omega))^2\mathsf P(d\omega)\right)dt\leq (T-S)\cdot \sup\limits_j\sup\limits_{t\in [t_j^n,t_{j+1}^n)}\int_\Omega (X(t,\omega)-X(t^n_j,\omega))^2\mathsf P(d\omega) $$ but I am not sure that the latter $\sup$ can be made as small as I want using $(1)$ only.

If I am not mistaken, it is equivalent to consider $$ m(t,s) = \int_\Omega (X(t,\omega)-X(s,\omega))^2\mathsf P(d\omega) $$ such that $\lim\limits_{t\to s}m(t,s) = 0$ for all $s\in \mathbb R$, and to ask if $$ \sup\limits_j\sup\limits_{t\in [t_j^n,t_{j+1}^n)}m(t,t_j) $$ can be made small by taking $\mu_n$ small enough.

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@Elias: in first thing the integral $\|X-X^n\|_{L^2([S,T]\times\Omega)}$ depends on the partition and so we want it to go to zero with partition becoming smaller and smaller. What is unclear with the second part? –  Ilya Feb 25 '12 at 8:57

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