Why does the indicator function fulfill the random variable definition?

Question

Def. of random variable: Pre-image of random variable

It's intuitive that the pre-image of $1_A=1, \omega \in A$ is in $F$. Since $A \in F$.

However, since $1_A=0$, when $\omega \not \in A$, then the pre-image of this would not belong to $F$? But it would need to, because of the def. of r.v.?

You probably mean $A\in F$ (not $A\subset F$). – drhab May 03 '19 at 11:40 — drhab, May 03 '19 at 11:40

score 2 · Answer 1 · answered May 03 '19 at 11:38

Function $\mathsf1_A$ can have at most $4$ distinct preimages:$$\varnothing, A, A^{\complement},\Omega$$where $\Omega$ denotes the whole space.

If $A\in\mathcal F$ then also $A^{\complement}\in\mathcal F$ since $\sigma$-algebra $\mathcal F$ is closed under complements.

In all cases we have $\varnothing,\Omega\in\mathcal F$.

Why does the indicator function fulfill the random variable definition?

1 Answers1