Suppose $X, Y$ are real valued random variables defined in a probability space $(\Omega, \sigma, P)$, with joint density function $f(x,y)$. I want to prove that $E(Y|X=x) = \int \frac{yf(x,y)}{f(x)}dy$, where $f(x) = \int f(x,y)dy$. In order to do so, I have to prove that for each measurable set $B \subseteq R$,
$$\int_B \left(\int \frac{yf(x,y)}{f(x)}dy \right)d P_X = \int_{X \in B} Y dP$$
Where $P_X = P \circ X^{-1}$ is the induced probability on $\mathbb{R}$. Using that $dP_X = f(x) dx$, we have that
$$\int_B \left(\int \frac{yf(x,y)}{f(x)}dy \right)d P_X =\int_B \left(\int \frac{yf(x,y)}{f(x)}dy \right)f(x)dx = \int_B \int yf(x,y) dy dx$$
Now, I suppose that $\int_{X \in B} Y dP = \int_B yf(y) dy$, where $f(y) = \int f(x,y)dx$, because of properties of density functions (I am not sure if this is indeed the case, as I am quite inexperienced in probability theory).
Then, is it true that $\int_B \int yf(x,y) dy dx = \int_B y f(y) dy$? If it is, how do I show it? Using Fubini's, I can conclude only that $$\int_B \int yf(x,y) dy dx = \int \int_Byf(x,y)dxdy = \int y \int_Bf(x,y)dx dy$$
Is there something wrong conceptually? Edit: For the record, this is exercise 4.2 of Breiman's book on probability.
You are nearly done, but there is mistake in the following:
Note that $\int_{X\in B} Y\,\mathrm{d}P=E[Y1_B(X)]$. Generally this is not the same as $\int_B yf(y)\,\mathrm{d}y=E[Y1_B(Y)]$. Instead you have $$ \int_B\int yf(x,y)\,\mathrm{d}y\,\mathrm{d}x=\int\int1_B(x)yf(x,y)\,\mathrm{d}y\,\mathrm{d}x=\int1_B(X)Y\,\mathrm{d}P=\int_{\{X\in B\}}Y\,\mathrm{d}P. $$