# Random variables on a probability space

I need to solve the following problem.

Let $X$ be a random variable defined on the probability space $(\Omega,\mathcal{F},\textbf{P})$. Show that $X^{+} = max(0, X)$ and $X^{-} = max(0, -X)$ are random variables.

I have to work with the definition of the random variable which states that $X^{+}$ is a random variable if the set $\{\omega \in \Omega : X^{+}\in B\} \in \mathcal{F}$ for each borel set $B\in \mathcal{B}$.

Any ideas?

• Alternatively(which is what i did), we can work with : $\{\omega \in \Omega : X^{+} \leq x\} = \{\omega \in \Omega : max(0, X) \leq x\} = \underbrace{\{\omega \in \Omega : 0 \leq x\}}_{\in \mathcal{F}(\text{it's just a number})} \bigcap \underbrace{\{\omega \in \Omega : X\leq x\}}_{\in \mathcal{F}(\text{By assumption})}$ Hence the original set is $\mathcal{F}$-measurable and thus it is a random variable. Any thoughts on that? – teo theo Nov 29 '15 at 14:09
• Your solution is fine. Another way would be to consider the function $f:\mathbb R\to\mathbb R$, $f(x) = \max\{0, x\}$ and note that it's continuous. Since $X^+$ is a composition of a random variable with a continuous function, it is also measurable and so a random variable. – Milind Nov 29 '15 at 14:45

Your solution is correct. I find indicator functions helpful:

$\max(0,x) = x1_{x > 0}$

$\to \max(0,X) = X1_{X > 0}$

$X$ is a random variable in $(\Omega,\mathcal{F},\textbf{P})$.

$1_{X > 0}$ is a random variable in $(\Omega,\mathcal{F},\textbf{P})$ (Why?).

Since products of random variables in $(\Omega,\mathcal{F},\textbf{P})$ are random variables in $(\Omega,\mathcal{F},\textbf{P})$,

$X1_{X > 0}$ is a random variable in $(\Omega,\mathcal{F},\textbf{P})$. QED