Measurability of the integral of Brownian motion

Question

Hi once again a measurability question about Brownian motion. I'm not that familiar in how to prove that a function is measurable rigorously, so I'm stuck at the following exercise: Show that for $T>0$ the mapping $$\omega \rightarrow X_T(\omega) = \int_{0}^{T} B_t(\omega) dt$$ is measurable. The Brownian motion is therefore assumed to be jointly measurable with continuous paths. How can I prove this? I know the definition of measurability that the preimage of measurable sets are measureable. But what are here the measurable sets in the image space. And how to proceed from there? Any help is really appreciated.

Answer 1

For each $n\ge 1$ and $0\le k <n$, define $\Delta t=T/n$ and $t_k=k\Delta t$. Then $$ X_T^{(n)}=\sum_{k=0}^{n-1}B(t_k)\Delta t\to X_T $$ as $n\to\infty$, and each $X_T^{(n)}$ is measurable.