Let
- $(\Omega,\mathcal{A},\operatorname{P})$ be a probability space
- $X=(X_t)_{t\ge 0}$ be a real-valued and continuous stochastic process on $(\Omega,\mathcal{A},\operatorname{P})$
- $0\le a<b$
I want to show, that $$\Omega\to\mathbb{R}\;,\;\;\;\omega\mapsto\int_a^bX_t(\omega)\;dt$$ is $\mathcal{A}$-measurable.
Since $X$ is continuous, for all $\omega\in\Omega$ $$X(\;\cdot\;,\omega):[0,\infty)\to\mathbb{R}\;,\;\;\;t\mapsto X_t(\omega)$$ is $\mathcal{B}\left([0,\infty)\right)$-measurable (where $\mathcal{B}(E)$ is the Borel $\sigma$-algebra on a topological space $E$). Thus, $$\mu(A):=\int_AX(\;\cdot\;,\omega)\;d\lambda\;\;\;\text{for }A\in\mathcal{B}\left([0,\infty)\right)\tag{1}$$ is a measure (measure with density $X(\;\cdot\;,\omega)$ with respect to $\left.\lambda\right|_{[0,\infty)}$, where $\lambda$ is the Lebesgue measure on $\mathcal{B}(\mathbb{R})$).
However, I don't know how I need to proceed. Maybe, it's wrong to consider $(1)$. What do we need to do?
By definition, the Riemann-integral of a function $f$ equals
$$\int_a^b f(t) \, dt = \lim_{n \to \infty} \sum_{j=1}^n f(t_j^n) (t_j^n-t_{j-1}^n)$$
where $\Pi^n = \{a=t_0^n < \ldots < t_n^n = b\}$ denotes a partition of the interval $[a,b]$ and the mesh size $|\Pi^n|$ converges to $0$ as $n \to \infty$. Consequently,
$$\int_a^b X(t)(\omega) \, dt = \lim_{n \to \infty} \underbrace{\sum_{j=1}^n X(t_j^n,\omega)(t_{j}^n-t_{j-1}^n)}_{\mathcal{A}-\text{measurable}}$$
is $\mathcal{A}$-measurable as a pointwise limit of $\mathcal{A}$-measurable functions.