Alert: I have little to no measure theory knowledge.
Assume $x,y \sim_{i.i.d} \mathbb{P}$ where $\mathbb{P}$ is Lebesgue measurable. For a function $f$ we have: \begin{align} E := \mathbb{E}[f(x,y) \ | \ x]. \end{align} I encountered a statement which says integrating the above over $x$ will return $\mathbb{E}[f(x,y)].$ How does this work?
If there exists a PDF for your joint measure, let it be $\pi(X,Y)$ then,
$$E[f(X,Y)]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(X,Y)\pi(X,Y)dxdy$$ Since $\pi(X,Y)=\pi_{Y|X}(Y|X)\pi_X(X)$ the above is equivalent to
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(X,Y)\pi_{Y|X}(Y|X)\pi_X(X)dxdy$$
Then by Fubinis Theorem,the above is equivalent to : $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(X,Y)\pi_{Y|X}(Y|X)\pi_X(X)dydx=\int_{-\infty}^{\infty}\pi_X(X)\int_{-\infty}^{\infty}f(X,Y)\pi_{Y|X}(Y|X)dydx$$
Recognizing that $\int_{-\infty}^{\infty}f(X,Y)\pi_{Y|X}(Y|X)dy=E[f(X,Y)|X]$
The above is equivalent to:
$$\int_{-\infty}^{\infty}\pi_{X}(X)E[f(X,Y)|X]dx=E[E[f(X,Y)|X]$$ where the last expectation is taken over $X$. Note this holds for the general case so it will hold for iid.