# Show that $\omega\mapsto\int_a^bX_t(\omega)\;dt$ is measurable, for a real-valued and continuous stochastic process $X$

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.

• So, we should be able to conclude that $$\omega \mapsto \int_0^\infty X_t(\omega)\;dt=\lim_{b\to\infty}\int_0^bX_t(\omega)\;dt$$ is $\mathcal{A}$-measurable too, right? – 0xbadf00d May 12 '15 at 17:43
• @0xbadf00d Yeah, sure, whenever the limit exists... it is measurable as a pointwise limit of measurable functions. (Note however that in this case the integral at the left-hand side is an improper Riemann integral, not a Lebesgue integral.) – saz May 12 '15 at 17:53
• Is your note important for the measurability? – 0xbadf00d May 12 '15 at 18:14
• What I still don't understand: Why is $X(t_j^n)(t_j^n-t_{j-1}^n)$ measurable wrt $\mathcal{A}$? It's a constant, isn't it? So, it is measurable wrt any $\sigma$-algebra. – 0xbadf00d May 12 '15 at 18:19
• @0xbadf00d I see; sorry; I'm so used to it that I don't even think about it. Concerning your other question: Since $X(\cdot,\omega)$ is continuous, it doesn't matter; we have $$\int_a^b f(x) \, dx = \lim_n \sum_i (t_i^n-t_{i-1}^n) f(\xi_i^n)$$ for any intermediate point $\xi_i^n \in [t_{i-1}^n,t_i^n]$ and any continuous function $f$. – saz May 12 '15 at 18:24