$X$ is a non-negative r.v. with finite expectation. Prove $E[\min(X,n)]\to E[X]$ as $n\to \infty$ The original problem given to me is to prove
$$E[X\wedge n]\to E[X] ~~\mbox{as}~~ n\to\infty$$
but I thought the $\wedge$ here means taking the smaller one out of the two. Am I correct?
Moreover, can any one give a counter example that $$E[\min(X,n)-X]\nrightarrow 0 ~~\mbox{if}~~ E[X]=\infty$$?
 A: Non-negativity was added later to the question but my answer is still valid. 
Suppose $E|X| <\infty$. Then $EXI_{X>n} \to 0$ by DCT (with dominating function $|X|)$. This gives $E(\min \{X, n\}) \to EX$.
For the second part take a random variable $X$ which takes the value $n$ with probability $\frac c {n^{2}}$, $n=1,2...$ where $c$ is such that these probabilities add up to $1$. Can you check that $E(\min \{X,n \}-X)=-\infty$ for a,ll $n$?.
A: Yes your interpretation of the symbol $\wedge$ is correct, and the first question you asked is a direct application of the dominated convergence theorem.
Your second question is more interesting.
Let $Y_n=X-X\wedge n$, and note that $Y_n\geq 0$. You are asking for an example of a random variable $X$ such that $\mathbb EY_n$ does not converge to $0$. Suppose that
$$
\mathbb P(X\geq n)=\frac{1}{n},\qquad n\in\mathbb N.
$$
Then for all $n\in\mathbb N$, we have that
$$\mathbb EY_n=\sum _{m=1}^{\infty} \mathbb P(Y_n\geq m)=\sum _{m=1}^{\infty} \mathbb P(X\geq m+n)=\sum_{m=1}^{\infty}\frac{1}{m+n}=\infty.$$
