A question on measurability of stochastic process Let $(\Omega,\mathcal{F},P)$ be a probability space and $\{X_t:t\geq0\}$ be a collection of real-valued random variables with index set $[0,\infty)$. Show that the mapping $t\mapsto X_t$ is Borel-measurable and furthermore if $X_t\in L^1(\Omega,\mathcal{F},P)$ for all $t$ then the mapping $t\mapsto E[X_t]$ is Borel-measurable as well.
I am a bit stuck on this so any help is greatly needed and appreciated. Thanks in advance.
 A: Your added assumption is basically saying that the map $\;X: [0,\infty) \times \Omega \to \mathbb{R}$ is jointly measurable in $t$ and $\omega$.
To prove the first claim, notice that for any fixed $\omega \in \Omega$, the map $t \mapsto X(t,\omega)$ is the composition of two measurable maps: $$t \stackrel{i_{\omega}}{\longmapsto} (t,\omega) \stackrel{X}{\longmapsto} X(t,\omega)$$ where the inclusion map $i_{\omega}:[0,\infty) \to [0,\infty) \times \Omega$ is measurable because its components are measurable, and the map $X:[0,\infty) \times \Omega \to \mathbb{R}$ is given to be measurable. Since compositions of measurable maps are measurable, it follows that the map $t \mapsto X(t,\omega)$ is measurable.
For the second part, you want to show that the map $t \mapsto \int_{\Omega}X(t,\omega)\; dP(\omega)$ is measurable. This is the special case of a more general statement in measure theory:
Claim: Let $(X,\mathcal{A})$ be a measurable space, and let $(Y, \mathcal{B},\mu)$ be a finite measure space. Then for any map $\;f: X \times Y \to \mathbb{R}$ such that $f$ is $(\mathcal{A} \times \mathcal{B})$-measurable and $\;f(x,\cdot) \in L^1(Y, \mathcal{B},\mu)$ for all $x\in X$, the map $\;I_f:X \to \mathbb{R}$ given by $x \mapsto \int_Y f(x,y)\;d\mu(y)$ is measurable.
Proof sketch of claim: Prove it first when $\;f=1_{A \times B}$ for some $A \in \mathcal{A}$ and $B \in \mathcal{B}$ (this should be easy). Then the set $\{E \in \mathcal{A} \times \mathcal{B}: I_{1_E}$ is measurable$\}$ is a Dynkin-system which contains a $\pi$-system which generates $\mathcal{A} \times \mathcal{B}$, and thus $\{E \in \mathcal{A} \times \mathcal{B}: I_{1_E}$ is measurable$\} = \mathcal{A} \times \mathcal{B}$, so the claim is proved when $\;f=1_E$ for some $E \in \mathcal{A} \times \mathcal{B}$. Now extend to all non-negative measurable $f$ by approximating from below by simple functions and using MCT; and then extend to all $f$ such that $f(x,\cdot) \in L^1$ by considering positive and negative parts. $\Box$
A: This can't be true. Let $\Omega$ be a one-point probability space and $D\subseteq[0,\infty)$ be non-measurable.  For each $t\ge 0$, put
$X_t(\omega) = 1$ if $t\in D$ and $0$ otherwise.  All of the $X_t$ are measurable, but $t\mapsto X_t(\omega)$ is not and neither is $t\mapsto E(X_t)$. 
