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 $f_X(x)=f_Y(x)$ ?