probability theory - If $Z=X$ on $A$ and $Z=Y$ on $A^c$ then $Z$ is a random variable

Let $X$ and $Y$ be random random variables and let $A \in \mathcal{B}$. Prove that the function $Z$ defined by

$$Z(\omega)=\begin{cases} X(\omega),& \text{ if } \omega \in A \\ Y(\omega),& \text{ if } \omega \in A^{c} \end{cases}$$
is a random variable

Proof so far:

$$Z^{-1}((-\infty,a])=\{\omega:Z(\omega)\geq a\}=\{\omega: Z(\omega)\geq a, \omega \in A\}\cup\{\omega: Z(\omega)\leq a, \omega \in A^{c}\}=Y^{-1}[a,\infty) \cup X^{-1}([a,\infty))$$
So $Z$ is measurable

Answer

Let $X, Y$ be random variables in $(\Omega, \mathcal B, \mathbb P)$.

If $A \in \mathcal B$, then $1_A$ and $1_{A^C}$ are random variables.

Note that

$$Z = X1_A + Y1_{A^C}$$

Since sums or products of random variables in $(\Omega, \mathcal B, \mathbb P)$ are random variables in $(\Omega, \mathcal B, \mathbb P)$, $Z$ is a random variable in $(\Omega, \mathcal B, \mathbb P)$.

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As for your proof, I think you should say:

1. $\forall a \in \mathbb R$




2. have $Z \ge a$ instead of $Z \le a$




3. $Z$ is $\mathcal B$-measurable