When $ t\mapsto X_t(\omega)$ is continuous for almost all omega then does it imply that $t \mapsto E[X_t]$ continuous? If this is true how can go about proving it? If not what condition do I need?
Obviously if $X$ is a martingale then $t \mapsto E[X_t]$ continuous. How could I show this in the general case. I am just not sure how can i start to approach this problem. Any hints would be appreciated
Take as your probability space $[0;1]$ with the Lebesgue measure and the Borel-sigma-algebra. Pick $f:\mathbb{R} \rightarrow \mathbb{R}$ continuous with $supp(f)\subseteq [0;1]$ and $E[f]\neq 0$ and define $$ X_t(\omega) = \begin{cases} t^{-1} f(\omega\cdot \vert t\vert^{-1}),& t\neq 0,\\ 0,& t=0 \end{cases} $$ Then $[-1;1] \rightarrow \mathbb{R}, t \mapsto X_t(\omega)$ is continuous for every $\omega\in [0;1]$, but $$ E[X_t] = sgn(t) E[f] $$ which is not continuous. Indeed, for $0< \vert t \vert \leq 1$ we have $$ E[X_t] = \int_0^1 X_t(\omega) d\omega = \int_0^1 t^{-1} f(\omega \cdot \vert t \vert^{-1}) d\omega \stackrel{s=\omega \cdot \vert t \vert^{-1}}{=} \frac{\vert t \vert}{t}\int_0^{\vert t \vert^{-1}} f(s) ds = sgn(t) E[f] $$ where we used $\vert t \vert^{-1}\geq 1$ (as we assumed $\vert t \vert \leq 1$) and $supp(f) \subseteq [0,1]$.