Is $(T,\omega) \mapsto \int_0^T f(t,\omega)\ dt$ measurable?

Question

Let $(\Omega, \mathcal{A}, P)$ be a probability space. Denote by $\mathcal{B}$ the Borel space on the real line and denote by $\mathcal{B}_{[0, \infty)}$ the Borel space on the interval $[0, \infty)$. Denote by $\mu$ the Lebesgue measure on $\mathcal{B}_{[0,\infty)}$.

Let $f:[0,\infty)\times\Omega\rightarrow\mathbb{R}$ be a $(\mathcal{B}_{[0,\infty)}\otimes\mathcal{A})/\mathcal{B}$-measurable function such that for every $T \in [0,\infty)$, $f\mathbb{1}_{[0, T]\times\Omega} \in L_1(\mu \otimes P)$. For every $T \in [0,\infty)$ define $X_T : \Omega \rightarrow \mathbb{R}$ as follows: $$ X_T(\omega) := \int_0^T f(t, \omega)\ dt $$ as long as the (Lebesgue) integral on the right is defined and finite. Otherwise, set $X_T(\omega) := 0$. From Fubini's theorem we know that, w.l.g., we may assume that $X_T$ is $\mathcal{A}/\mathcal{B}$-measurable.

Consider the stochastic process $X := (X_T)_{T \in [0,\infty)}$. As a function from $[0,\infty)\times \Omega$ to $\mathbb{R}$, is $X$ $(\mathcal{B}_{[0,\infty)}\otimes \mathcal{A})/\mathcal{B}$-measurable?

Answer 1

Yes, it is. This follows from the fact that $(X_t)_{t \geq 0}$ has continuous sample paths.

Lemma: Let $(X_t)_{t \geq 0}$ be a stochastic process with continuous sample paths such that $X_t$ is $\mathcal{A}$-measurable for each $t \geq 0$. Then $X: [0,\infty) \times \Omega \to \mathbb{R}$ is measurable.

For the proof show first that

$$X^n(t,\omega) := \sum_{j \geq 0} X_{\frac{j}{n}}(\omega) 1_{[\frac{j}{n},\frac{j+1}{n})}(t)$$

is $\mathcal{B}[0,\infty) \otimes \mathcal{A}$-measurable for each $n$ and then use that $X^n(t,\omega) \to X(t,\omega)$ as $n \to \infty$.