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I'm confuse on some probability formula with conditional probability. I know that $$\mathbb P(A\cap B)=\mathbb P(A\mid B)\mathbb P(B).$$

Let $X,Y$ two (continuous) random variable. Then, $$\mathbb P\{X\in A,Y\in B\}=\mathbb P\{X\in A\mid Y\in B\}\mathbb P\{Y\in B\},\tag{*}$$ should be true regarding the above formula. I have in my lecture that $$\mathbb P\{X\in A,Y\in B\}=\int_B \mathbb P\{X\in A\mid Y=t\}f_Y(t)dt,\tag{**}$$ where $f_Y$ is the pdf of $Y$.

Q1) If it's true how can I prove that $$\int_B \mathbb P\{X\in A\mid Y=t\}f_Y(t)dt=\mathbb P\{X\in A\mid Y\in B\}\mathbb P\{Y\in B\} \ \ ?$$

For me $$\mathbb P\{X\in A\mid Y=t\}=\frac{\mathbb P\{X\in A, Y=t\}}{\mathbb P\{Y=t\}}$$ is not well defined since $\mathbb P\{Y=t\}=0$.

Q2) Since $$\mathbb P\{X\in A\mid Y=t\}=\int_A f_{X,Y}(x,t)dx,$$ and since by (**), the formula $$\mathbb P\{X\in A,Y\in B\}=\int_A \mathbb P\{Y\in B\mid X=x\}f_X(x)dx$$ should be true, one should have $$\int_A\int_B f_{X,Y}(x,t)dtf_X(x)dx=\mathbb P\{X\in A, Y\in B\}=\int_B\int_A f_{X,Y}(x,t)dxf_Y(t)dt\underset{fubini}{=}\int_A\int_B f_{X,Y}(x,t)f_Y(t)dtdx.$$

Wouldn't this implies that $f_X(x)=f_Y(x)$ ?

user621345
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    $(**)$ is just the total probability theorem for random variables. We have $P(Y\in B)=\int_B f_Y(t),dt$ and the conditional probability $P(X\mid Y=t)$ is in a limiting sense for a continuous random variable $Y$, i.e. something like $P(X\mid Y=t)=\lim_{h\to 0}P(X\mid t-h<Y<t+h)$. – StubbornAtom Dec 27 '18 at 14:09
  • For Q2), your formula is not correct : $$\mathbb P{X\in A\mid Y=t}=\int_A\frac{f_{X,Y}(x,t)}{f_Y(t)}dx.$$ In fact, $$\int_A f_{X,Y}(x,t)dx=\mathbb P{X\in A, Y=t}=0.$$ – Surb Dec 27 '18 at 14:19

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