Show that if $Y$ is $\sigma(X)$-measurable, $X\sim\operatorname B(p)$ then $Y=aX+b$

Question

Let $X$ a Bernoulli random variable with parameter $p\in [0,1]$ : $$ P(X=1)=p,\quad P(X=0)=1-p.$$ Let $Y$ a real r.v. $\sigma(X)$-measurable. Then there exists $a,b\in \mathbb{R}$ such that almost surely: $$Y=aX+b.$$ I would like to prove this; any hint/help would be appreciated.

Answer 1

Since $Y$ is $\sigma(X)$-measurable, $Y$ is constant on the atoms of $\sigma(X)$, which are $\{X=1\}$ and $\{X=0\}$. Let $y_1 = Y(\omega)$ if $\omega \in \{X=1\}$ and $b = Y(\omega)$ if $\omega \in \{X=0\}$. Let $a = y_1 - b$. Then, $$Y(\omega) = aX(\omega) + b$$ for almost all $\omega$.