Counterexample to infinite sum of expectations = sum of infinite expectations?

Question

I am looking for the counterexample to the statement $\sum E[X_n] = E[\sum X_n]$ given that $\sum E[X_n]$ and $\sum X_n$ exist. This would be also a counterexample to the dominated convergence theorem without one of its assumptions with the probability measure; I was able to create a counterexample using Lebesgue integral, but the domain is the entire real so it unfortunately fails.

Answer 1

Is this the sort of thing you had in mind?

Take Lebesgue measure on $[0,1]$. For each positive integer $n$, let $X_n = 2^n$ on an interval of measure $2^{-1-n}$, $-2^{n}$ on an interval of measure $2^{-1-n}$, $0$ everywhere else, where all these intervals are pairwise disjoint (note that the sum of the measures of the intervals is $1$). The pointwise sum $\sum_n X_n$ exists (and is finite) everywhere (in fact for any $x$ there is at most one $n$ for which $X_n(x) \ne 0$). Also each $\mathbb E X_n = 0$, so $\sum_n \mathbb E X_n = 0$. But $\mathbb E \sum_n X_n$ does not exist because $\sum_n X_n$ is not in $L^1$: $\mathbb E \left|\sum_n X_n\right| = \infty$.