I would like to show that, given two real random variables $X$ and $Y$ over the probability space $(\Omega,\mathcal{A},P)$, and with the same CDF (namely $F_X=F_Y$), then they have the same distribution, namely $\mu_X(A):=P(X \in A)=P(Y \in A)=:\mu_Y(A)$ for all $A \in \mathcal{B}(\mathbb{R})$, where $\mathcal{B}(\mathbb{R})$ is the Borel sigma algebra over $\mathbb{R}$.
I'm pretty stuck from the start. Any hint would be appreciated! Thank you!
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Post Scriptum:
I know that I have to put in my efforts, but I'm having big problems since the start.
The fact that $F_X=F_Y$ is obviously equivalent to the fact that $\forall \,x \in \mathbb{R} \quad \mu_X((-\infty,x])=\mu_Y((-\infty,x])$.
I'm learning this by myself, so maybe this is a deep result that is too hard to prove for me. I really don't know.
Again, I've searched a lot online but I didn't find any valid proof of this.
Best Answer
If $F_X = F_Y$, i.e. $F_X(z) = F_Y(z)$ for all $z\in\mathbb R$, we have that $\mu_X(I_z) = \mu_Y(I_z)$ for all $z\in\mathbb R$, where $I_z = (-\infty,z]$.
The set $\mathcal I = \{I_z:z\in\mathbb R\}$ generates the Borel $\sigma$-algebra, i,e, $\sigma(\mathcal I) = \mathcal B(\mathbb R)$, and the set $\mathcal I$ is $\pi$-stable (i.e. $I,J\in\mathcal I$ implies that $I\cap J\in\mathcal I$).
The set $\mathcal D = \{A\in\mathcal B(\mathbb R) : \mu_X(A) = \mu_Y(A)\}$ is a $\lambda$-system (https://en.wikipedia.org/wiki/Dynkin_system), and it obviously holds that $\mathcal D\subseteq\mathcal B(\mathbb R) = \sigma(\mathcal I)$. By hypothesis, $\mathcal I\subseteq\mathcal D$ also. From the $\pi$-$\lambda$-theorem then follows that $\sigma(\mathcal I) \subseteq \mathcal D$. Thus, $\mathcal D = \mathcal B(\mathbb R)$, which means $\mu_X = \mu_Y$ for all $A\in\mathcal B(\mathbb R)$.