Running maximal is a random variable

Question

Suppose we have a sequence $(X_n), n \in \mathbb{Z_+}$ of random variables on $(\Omega, \mathcal{F}, \mathbb{P})$ and define $X^*_{n}(\omega)=\max_{k\leq n}X_k(\omega)$. I want to prove rigorously that $X_n^*$ is a random variable.

My main issue is that I'm not sure with respect to which $\sigma$-algebra we have to check the measurability - is it the same as the one - $(\mathcal{F}_n)$, which makes the sequence $(X_n)$ adapted? It doesn't seem so, but nevertheless, I know that the maximum of $2$ measurable functions is measurable (using the identity that $\max\{f,g\}=\frac{1}{2}(|f-g|+(f+g))$ and using standard results), however this doesn't generalize directly. If this is the true $\sigma$-algebra then I want to show that for a Borel-measurable $G$ we have

$$\{\omega: X_n^*(\omega)\in G\} \in \mathcal{F}_n$$

which gets a bit messy because of the maximums.

Hence the question: Is there a specific $\sigma$-algebra with respect to which the running maximal is measurable or is it the standard one?

Thanks!

Answer 1

When we have a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and a function $Y\colon \Omega\to\mathbb R$, we say that $Y$ is a random variable if for all Borel subset $B$ of the real line, the set $Y^{-1 }(B)$ belongs to $\mathcal F$. In particular, not filtration is involved.

If you know how to prove that the maximum of two random variables still is a random variable, then you can use an induction argument because $X_n^*=\max\left\{ X_{n-1}^*,X_n\right\}$.