$X$ is a random variable with underlying $(\Omega, \mathcal{F}, \mathbb{P})$ and so $\{\omega \in \Omega\mid X(\omega)\leq x\}\in \mathcal{F}, \hspace{3mm} \forall\, x\in \mathbb{R}$. To prove that $\max\{X, 0\}$ is a random variable, I split the max function as follows:
For $x \in \mathbb{R}$,
$$\{\max\{X, 0\} \leq x\} = \{X \leq x\} \cap \{0 \leq x\} $$
The first set on the RHS is fine, I am not sure what to do about the second set. Can I assume it to be a constant random variable on the same $(\Omega, \mathcal{F}, \mathbb{P})$ space to complete the proof?
Best Answer
We assume $x\in \mathbb{R}$ to be fixed, so $\{0\leq x\}$ is either $\Omega$ if $0\leq x$ and $\emptyset$ otherwise. In either case, it is measurable by definition of a $\sigma$-algebra.