Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. Consider the stochastic process $X = \{X(t) \, | \, t \geq 0\}$. Choose arbitrary $t_1, t_2 \in [0, \infty)$ and denote the increment
$$ \Delta X := X(t_2) - X(t_1). $$
I want to prove that $\Delta X$ is a random variable. That is, I want to prove that the mapping $$ \Delta X : \Omega \rightarrow \mathbb{R} $$ is $\mathcal{F}$-measurable, i.e. the inverse image of any Borel set is measurable:
$$ \Delta X^{-1}(B) = \{\omega \in \Omega \, | \, X(t_2)(\omega) - X(t_1)(\omega) \in B\} \in \mathcal{F} \quad \text{for all} \ B \in \mathbb{B}(\mathbb{R}). $$
I'm stuck trying to prove this. My best guess is that given any Borel set $B$, I somehow construct a Borel set $B'$ such that $X(t_2)(\omega) \in B'$ implies $\Delta X = X(t_2)(\omega)-X(t_1)(\omega) \in B$. Then $\Delta X^{-1}(B) = X(t_2)^{-1}(B') \in \mathcal{F}$ since $X(t_2)$ is a random variable. However, I'm stuck in constructing the Borel set $B'$, and/or there may be a better way to prove it.
Thanks!
Observe that the function $g:\mathbb R^2\to\mathbb R$ prescribed by $(x,y)\mapsto x-y$ is continuous hence is measurable if domain $\mathbb R^2$ is equipped with the $\sigma$-algebra of Borel subsets of $\mathbb R^2$ (denoted as $\mathcal B(\mathbb R^2)$) and codomain $\mathbb R$ is equipped with the $\sigma$-algebra of Borel subsets of $\mathbb R$ (denoted as $\mathcal B(\mathbb R)$).
Since $X(t_2)$ and $X(t_1)$ are both random variables the map $Y:\Omega\to\mathbb R^2$ prescribed by $$\omega\mapsto(X(t_2)(\omega),X(t_1)(\omega))$$is measurable wrt product $\sigma$-algebra $\mathcal B(\mathbb R)\otimes\mathcal B(\mathbb R)$ and it can be shown that:$$\mathcal B(\mathbb R^2)=\mathcal B(\mathbb R)\otimes\mathcal B(\mathbb R)$$
A composition of measurable functions is again measurable so on base of the facts above we conclude that $$g\circ Y:\Omega\to\mathbb R$$is measurable, hence is a random variable.
Now observe that this function is prescribed by:$$\omega\mapsto X(t_2)(\omega)-X(t_1)(\omega)$$So actually we have $g\circ Y=\Delta$ and it has been proved above that $\Delta$ is a random variable.