Cumulative distribution function implication How can I prove the following: Let $X$ and $Y$ be two random variables. Suppose that their cumulative distribution functions satisfies $F_X(x)=F_Y(x)$ for all $x$. How can I show that $X$ and $Y$ are identically distributed?
I got a hint that this is not easy for beginner and it requires heavy use of $\sigma$-algebras.
 A: "$X$ and $Y$ are identically distributed" means that $P(X \in B)=P(Y \in B)$ for each Borel set $B$. Knowing that $F_X=F_Y$ means you know $P(X \in (-\infty,a])=P(Y \in (-\infty,a])$ for every real number $a$. So you prove the desired result in three steps:


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*$P(X \in U)=P(Y \in U)$ for each open $U$.

*If $P(X \in U)=P(Y \in U)$ for each open $U$, then $P(X \in F)=P(Y \in F)$ for each closed $F$.

*If $P(X \in U)=P(Y \in U)$ for each open $U$ and $P(X \in F)=P(Y \in F)$ for each closed $F$, then $P(X \in B)=P(Y \in B)$ for each Borel $B$.


To prove the first part, start from proving the result for open intervals, and then extend it to general open sets by using the fact that any open set is a countable union of open intervals.
To prove the second part you need only note that a closed set is the complement of an open set, so $P(X \in F)=1-P(X \in F^c)$, and we already have agreement for open sets.
To prove the third part, you show that the set $E=\{ A : P(X \in A)=P(Y \in A) \}$ is a $\sigma$-algebra. This takes some work. First, the above implies that this set contains an algebra (no $\sigma$ here), namely the algebra consisting of open and closed sets. Next, you can check $E$ is closed under increasing unions and decreasing intersections, using continuity of measure. (Here the fact that a probability space has finite measure is required). Then the monotone class theorem implies that $E$ is a $\sigma$-algebra. Now by definition, any $\sigma$-algebra containing all open sets must contain the Borel $\sigma$-algebra.
