Greater or equal almost surely plus equal in distribution imply equal almost surely? Suppose $X_2\geq X_1$ a.s. and $X_2\stackrel{d}{=} X_1$(i.e. equal in distribution), is it true that $X_2\stackrel{a.s.}{=} X_1$?
 A: Yes. Suppose to the contrary $X_1$ is not a.s. equal to $X_2$. Then (since $X_1 \geq X_2$ a.s.) the set $A=\{\omega:X_1(\omega)>X_2(\omega)\}$ has positive probability. Let $B=\{\omega : X_1(\omega) = X_2(\omega)\}$. By hypothesis, $P(A\cup B)=1$. Notice that for $i=1,2$:
$$E[X_i]=\int X_i (\omega) \ dP(\omega) = \int_A X_i(\omega) \ dP(\omega) + \int_B X_i(\omega) \ dP(\omega).$$
By the definition of $B$, the second term is the same for $i=1$ and $i=2$, and by the assumption that $P(A)>0$, you can show that the first term is strictly larger for $i=1$. So this implies $E[X_1]>E[X_2]$. This contradicts the fact that they are equal in distribution.
A: By $[\text{expression}]$ I mean the set $\{ \omega \in \Omega| \text{expression is true at } \omega\}$.
Let $A= [X_1 \le X_2]$, we are given that $PA = 1$.
For any $\alpha$, $[X_2 \le \alpha] \cap A \subset [X_1 \le \alpha ] \cap A \subset [X_1 \le \alpha ]$.
\begin{eqnarray}
[X_1 \le \alpha] \cap ( [ \alpha < X_2] \cup [ X_2 \le \alpha] ) &=& [X_1 \le \alpha] \\
[X_1 \le \alpha < X_2] \cup [ X_1 \le \alpha, X_2 \le \alpha]  &=& [X_1 \le \alpha] \\
[X_1 \le \alpha < X_2] \cup ([ X_1 \le \alpha, X_2 \le \alpha] \cap A)  &\subset& [X_1 \le \alpha] \\
[X_1 \le \alpha < X_2] \cup ([X_2 \le \alpha] \cap A)  &\subset& [X_1 \le \alpha] \\
\end{eqnarray}
and so
$P[X_1 \le \alpha < X_2] + P [ X_2 \le \alpha] \le  P [X_1 \le \alpha]$.
In particular, $P[X_1 \le \alpha < X_2] = 0$.
Since $[X_1 < X_2 ] = \cup_{q \in \mathbb{Q}} [X_1 \le q < X_2 ]$, we see that $P[X_1 < X_2] = 0$.
