Do two almost surely equal random variables necessarily have the same probability? 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. 
 A: 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. 
A: It is true, as
$$\begin{align}
P(X \in B) &= P(X\in B, X=Y) + P(X\in B, X\ne Y) \\
&\le P(Y\in B, X=Y) + P(X\ne Y) \\
&\le P(Y\in B) + P(X\ne Y) \\
&= P(Y\in B).
\end{align}$$
By symmetry, it also follows $P(Y\in B) \le P(X\in B)$ and thus $P(X\in B) = P(Y\in B)$.
