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!

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
$$ \lim_{h\to 0} \sum_{k=-\infty}^\infty E(X\big|Y \in [kh,kh+h)) \Pr(Y \in [kh,kh+h)) = \int_{-\infty}^\infty E(X\big|Y s) f)Y(s) \, ds .$$
