Is $\mathbb P((X,Y)\in A)=\int_{\mathbb R}\mathbb P(X\in A^y\mid Y=y)\mu_Y(dy)$ always true?

Question

I know that if $\mu_1$ and $\mu_2$ are two measures on $\mathbb R$, then if $E\subset \mathbb R^2$, then $$\mu_1\otimes\mu_2(E)=\int_{\mathbb R}\mu_1(E^y)\mu_2(dy),$$ where $E^y=\{x\mid (x,y)\in E\}$. Now, if $X$ and $Y$ are independents, then $\nu(E)=\mu_X\otimes\mu_Y(E):=\mathbb P((X,Y)\in E)$ defines a product measure on $\mathbb R^2$, where $\mu_X$ is the measure on $\mathbb R$ generate by $X$. But if there are not independent, then $$\mu_X\otimes \mu_Y=\mathbb P((X,Y)\in E)$$ doesn't hold anymore. So, $$\mu_X\otimes \mu_Y(E)=\int_{\mathbb R}\mu_X(E^y))\mu_Y(dy),$$ will be obviously, but does $$\mathbb P((X,Y)\in E)=\int_{\mathbb R}\mathbb P(X\in E^y\mid Y=y)\mu_Y(dy),$$ will hold ?

Answer 1

Answer: Yes!

Proof. Let's fix the probability space $(\Omega,\mathcal F,\mathbb P)$. I'm wondering how you define conditional probability on the event $\{Y=y\}$. A natural way to do so is through the conditional expectation (definition with $\sigma$-algebra etc). Let $g$ be a Borel-measurable function such that $$g(Y)=\mathbb E[\mathbf{1}_{\{(X,Y)\in E\}}\mid \sigma(Y)]\ \ \ \text{a.s. }$$ Then we set $$\mathbb P(X\in E^y\mid Y=y):=g(y)$$ You ask whether $$\mathbb P((X,Y)\in E)=\int_\mathbb Rg(y)\,d\mu_Y(y)$$ is true or not.

Let's see, we have using change of measure formula and the definition of conditional expectation $$\int_\mathbb Rg(y)\,d\mu_Y(y)=\int_\Omega g(Y)\,d\mathbb P=\int_\Omega \mathbf{1}_{\{(X,Y)\in E\}}\,d\mathbb P=\mathbb P((X,Y)\in E)$$ Formula proved!