I am trying to prove the following, where $X_t$ is an almost surely bounded progressivley measurable process:
$$ \lim_{t\rightarrow 0}\frac 1 t \mathbb E \int_0^t X_sds = \lim_{t\rightarrow 0} \mathbb E X_t, $$
if the limit on the r.h.s exists.
Proof attempt:
$$ \lim_{t\rightarrow 0}\frac 1 t \mathbb E \int_0^t X_sds = \lim_{t\rightarrow 0}\frac 1 t \int_0^t \mathbb E(X_{s})ds= \lim_{t\rightarrow 0}\int_0^1 \mathbb E(X_{st})ds= \int_0^1 \lim_{t\rightarrow 0}\mathbb E(X_{st})ds = \lim_{t\rightarrow 0}\mathbb EX_{t}$$
Due to $X$'s boundedness, Lusin's theorem allows me to interchange the integral and the expectation. Then I use integration by substitution and bounded convergence.