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?


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 $n \to \infty$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.