Law of total probability in continous case According to the law of total expectation, we have
$$\mathbb{E}X=\sum_{i=1}^n \mathbb{E}(X\mid A_{i})\cdot \mathbb{P}(A_i)$$
I am wondering if the similar formula holds for the continous case.
Specifically, if $Y$ is a continous random variable with PDF $f_Y(s)$, is it true that
$$\mathbb{E}X=\int_{\mathbb{R}}\mathbb{E}(X\mid Y=s)\cdot f_Y(s) \ ds\text{ ?}$$
Moreover, is it true that
$$\mathbb{E}(X\mid Y\in A)\cdot\mathbb{P}(Y\in A)=\int_{A}\mathbb{E}(X\mid Y=s)\cdot f_Y(s) \ ds\text{ ?}$$
It seems to me that both are true, but I don't know how to prove it.
Thanks in advance!
 A: Your first question is true. Proving it uses the law of total expectation:
$$
E[X] = E[ E[X|Y]] 
 = \int_\mathbb{R} E[X|Y=s]\cdot f_y(s)\,ds
$$
The idea is this: $E[X|Y=y]$ is a function of $y$ (for each value $y$, we get a number, $E[X|Y=y]$), while $E[X|Y]$ is a random variable. Namely, if $E[X|Y=y]$ is the function $g(y)$, then $E[X|Y]$ is the random variable $g(Y)$. The right hand side of the above equation is then just $\int_\mathbb{R} g(y)\cdot f_Y(y)\,dy$, which is precisely $E[g(Y)]$, which is $E[E[X|Y]]$.
For your second question, $E[X|Y]$ is defined as the (almost-surely) unique random variable for which
$$
\int_A E[X|Y=y]\cdot f_Y(y)\,dy = \int_A\int_{\mathbb{R}}x\cdot f_{X,Y}(x,y)\,dx\,dy,
$$
for all Borel subsets $A$ of $\mathbb R$, where $f_{X,Y}$ is the joint distribution. Taking $A=\mathbb{R}$ yields the law of total expectation. A little rearranging of the right hand side of above shows
$$
\int_A E[X|Y=y]\cdot f_Y(y)\,dy =P(Y\in A)\int_{\mathbb{R}}x\cdot\frac{\int_A  f_{X,Y}(x,y)\,dy}{P(Y\in A)}\,dx=P(Y\in A)\cdot E[X|Y\in A]
$$
The second equality follows since $ \frac{\int_A  f_{X,Y}(x,y)\,dy}{P(Y\in A)}$ is precisely the conditional density of $X$ given $Y\in A$, which can be checked by integrating it over any Borel subset $B$ of $\mathbb R$ and seeing that the result is indeed $P(X\in B|Y\in A)$. 
A: Here is a non-rigorous answer.  So
$$ \mathbb EX = \sum_{k=-\infty}^\infty \mathbb E(X\big|Y \in [kh,kh+h)) \Pr(Y \in [kh,kh+h)) $$
for any $h>0$ (by the first formula you wrote).  Now as $h \to 0$, we have that $\Pr(Y \in [s,s+h)) = f_Y(s) h + o(h)$.  Let's also suppose that $\lim_{h\to 0} \mathbb E(X\big|Y \in [s,s+h))$ exists, and let's call this $\mathbb E(X\big|Y = s)$.
Then using the Riemann sum we get
$$ \lim_{h\to 0} \sum_{k=-\infty}^\infty E(X\big|Y \in [kh,kh+h)) \Pr(Y \in [kh,kh+h)) = \int_{-\infty}^\infty E(X\big|Y s) f)Y(s) \, ds .$$
