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Suppose $(\Omega, \mathcal F, P)$ is our probability space. $X, Y$ are two real-valued random variables (Borel measurable) defined on this space. In the book "Probability: Theory and Examples", (1) $X$ and $Y$ have the same distribution if they induce the same measure on $\mathbb R$, that is $P(X \le x) = P(Y \le x)$ for all $x \in \mathbb R$. (2)In another source, two random variables have the same distribution if for all $B \in \mathcal B(\mathbb R)$, $P(X \in B) = P(Y \in B)$.

The two definition should be the same. (2) implies (1) trivially. I am having trouble to see why (1) implies (2). I know set with the form $(-\infty, x]$ generates the Borel $\sigma$-algebra on $\mathbb R$. But in my mind, for any Borel set, in general we can only approximate it by closed (open) sets. How could we conclude the general case just by the equality on the closed sets?

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  • $\begingroup$ Maybe show $\{ B : P(X \in B) = P(Y \in B) \}$ is a monotone class and also contains the algebra generated by $\{ (-\infty, x] \}$, then apply the monotone class theorem? $\endgroup$ – Daniel Schepler Oct 13 '17 at 23:13
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This answer uses Dynkin's theorem. Suppose that $P_{1}$ and $P_{2}$ are probability measures that agree on rays of the form $(-\infty, x]$. Let $$ \mathcal{D}=\{B\in\mathcal{B}\colon\, P_{1}(B)=P_{1}(B)\} $$ which is a Dynkin system. Let $\mathcal{P}$ be the collection of these rays of the form $(-\infty, x]$. They are a $\pi$ system. Further $$ \mathcal{P}\subseteq \mathcal{D}\implies\mathcal{B}=\sigma{(P)}\subseteq \mathcal{D}. $$ by Dynkin's theorem. Hence the probability measures agree on Borel sets.

Alternatively there is a bijection between CDFs and the measures $F\to \mu_F$ induced by them by Caratheodory-Hahn extension theorem where $\mu_F$ is the unique probability measure such that $\mu_F(a,b]=F(b)-F(a)$. Hence if two probability measures agree on rays of the form $(-\infty, x]$, then they are induced by the same CDF and hence equal on all Borel sets.

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