Let $\Omega$ be a probability space with $\sigma$-algebra $\mathcal{A}$, and let $\mathcal{B}$ be the Borel $\sigma$-algebra.
Let $X:(\Omega,\mathcal{A}) \to (\mathbf{R},\mathcal{B})$ and $Y:(\Omega,\mathcal{A}) \to (\mathbf{R},\mathcal{B})$ be two measurable functions, and let $B\in \mathcal{B}$.
If $X$=$Y$ almost surely, then is $P(\omega\in \Omega : X(\omega)\in B)$=$P(\omega\in \Omega : Y(\omega)\in B)$? If so please provide a proof.
Background: I am doing some self-study on probability and measure theory and am having difficulty with the concept of two random variables being almost surely equal. I want to know if two random variables that are almost surely equal necessarily have the same probability. I suspect this is true but would like a formal proof.
Best Answer
This elaborates on Ian's comment.
Using the convention where $P(\omega\in \Omega:X(\omega)\in B)$ is written as $P(X\in B)$, you want to show $$ X=Y\quad a.s.\implies P(X\in B)=P(Y\in B) $$ for all $B\in \mathcal B$. We start by "conditioning on" $\{X=Y\}$: $$ P(X\in B)=P(\{X\in B\}\cap \{X=Y\})+P(\{X\in B\}\cap \{X\not=Y\})=P(\{X\in B\}\cap \{X=Y\})+0 $$ since $P(\{X\neq Y\})=0$. Similarly, $$ P(Y\in B)=P(\{Y\in B\}\cap \{X=Y\}) $$ The final step is to realize that $$ P(\{X\in B\}\cap \{X=Y\})=P(\{Y\in B\}\cap \{X=Y\}) $$ because the sets on either side of the above equality are equal. Specifically, if $\omega\in \{X\in B\}\cap \{X=Y\}$, then this means that $X(\omega)\in B$ and $X(\omega)=Y(\omega)$, implying $Y(\omega)\in B$ so that $\omega\in \{Y\in B\}\cap \{X=Y\}$. This shows that $\{Y\in B\}\cap \{X=Y\}\subset \{Y\in B\}\cap \{X=Y\}$, and the reverse inclusion holds by an identical argument.
Anyway, combining the last three displayed equations proves what you want.