Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space. I know that if $X$ and $Y$ are two independent r.v., then $$\mu(A\times B):=\mathbb P\{X\in A, Y\in B\}=\mu_X\otimes \mu_Y(A\times B)=\mu_X(A)\mu_Y(B)$$ is a product measure on $\mathbb R^2$. And moreover $$\mu(A\times B)=\int_{A}\mu_Y(B^x)\mu_X(dx)=\int_{B}\mu_X(A^y)\mu_Y(dy),$$ where $A^y=\{x\mid (x,y)\in A\times B\}$ and $B^x=\{y\mid (x,y)\in A\times B\}$
Question : Now, suppose $X$ and $Y$ are not independent. The measure $\mu_X\otimes\mu_Y$ is a measure on $\mathbb R^2$ but is not a product measure, i.e. $\mu_X\otimes\mu_Y(A\times B)=\mu_X(A)\mu_Y(B)$ doesn't hold anymore. But, does the formula $$\mu_X\otimes \mu_Y(A\times B)=\int_{A}\mu_Y(B^x)\mu_X(dx)=\int_{B}\mu_X(A^y)\mu_Y(dy)$$ still hold ? I'm a bit in truble since on wikipedia they elaborate this formula for $\mu_X\otimes \mu_Y$ being the product measure (i.e. $\mu_X\otimes\mu_Y(A\times B)=\mu_X(A)\mu_Y(B)$), but it looks that in proposition 3.1 page 278 of "Real-Analysis" of Stein and Shakarchi they make the proof of this formula, but they don't look to use this property. So is this formula still true ?