Let $S,T\in\mathbb R$ such that $0\leq S. 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 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.
real-analysis
stochastic-processes