Expected value of a random function Assume $r>0$ is a variable (not random) but $X(r)$ is a random variable (like the total number of points randomly distributed over a plane in a circle of radius $r$). We know that for all $r$ (no matter how large):
$$E[X(r)]\leq 2$$
Under what conditions we can conclude?:
$$E\left[\lim\limits_{r \to \infty}X(r)\right]\leq 2$$
 A: If $X(r)$ satisfies the requirements of any of the following three theorems then you can exchange the limit and expectation:


*

*Dominated Convergence Theorem

*Bounded Convergence Theorem

*Monotone Convergence Theorem
You have that $E[X(r)]\leq 2 \hspace{2mm} \forall r \in \mathbb{R}^+$, then it's true of the limit too: $\lim_{r\rightarrow\infty}E[X(r)]\leq2$
To complete the proof you just need to use one of those previous three theorems to permit yourself to exchange the limit and expectation.
EDIT
A usually pretty convenient way to do this is to re-express the expectation as the integral of probabilities:
\begin{equation}
E[X(r)] = \int_{-\infty}^0(1-P[X(r)\geq x])dx +  \int_{0}^\infty P[X(r)\geq x]dx
\end{equation}
Note that the two functions $(1-P[X(r)\geq x])$ and $P[X(r)\geq x]$ are bounded above by 2 for all $x$, thus you can use the Dominated Convergence Theorem to exchange the integrals and limits:
\begin{equation}
\lim_{r\rightarrow \infty} E[X(r)] = \int_{-\infty}^0 \lim_{r\rightarrow \infty}(1-P[X(r)\geq x])dx +  \int_{0}^\infty \lim_{r\rightarrow \infty}P[X(r)\geq x]dx
\end{equation}
The previous functions are almost surely continuous in $X(r)$, so using the Continuous Mapping Theorem:
\begin{equation}
\lim_{r\rightarrow \infty} E[X(r)] = \int_{-\infty}^0 (1-P[X^*\geq x])dx +  \int_{0}^\infty P[X^*\geq x]dx = E[X^*]
\end{equation}
Where $X^* = \lim_{r\rightarrow \infty}X(r)$
