# If $X_n\nearrow X$ then $E(X_n)\rightarrow E(X)$?

Let $(X_n)$ be an increasing sequence of real valued integrable rvs on a probability space $(\Omega,\mathcal{F},P)$, such that $(X_n)$ converges ae to some rv $X$. Is it true that $E(X_n)\rightarrow E(X)$?

Some thoughts: Ι guess not. If $X_n$'s were positive then we could apply the monotone convergence theorem, but now we are looking for a counterexample. I am working on $[0,1]$, equipped with the Lebesgue measure and looking for discrete rvs that do the job, but no matter how I change their formula, I seem not to be getting the desired non-convergence.

• @user159517 which could give you $\infty-\infty$ – snarfblaat Feb 10 '16 at 1:20
• @user159517, you claim that since $X_n \rightarrow X$ ae, then $X_n^{+}\rightarrow X^{+}$ and $X_{n}^{-}\rightarrow X^{-}$? (if that holds, then things are a bit straightforward, since $X_n^{+}\nearrow X^{+},~X_n^{-}\nearrow X^{-}$ and all the rvs are positive valued). – Nikolaos Skout Feb 10 '16 at 1:22
• The sequence $X_n-X_1$ of non-negative, integrable random variables increases to $X-X_1.$ – user940 Feb 10 '16 at 1:30
• Not an answer, but shouldn't this be true for weaker condition ($X_n\rightarrow X$ in probability/measure), by absolute continuity of integral w.r.t. measure? – user175968 Feb 10 '16 at 2:00
You said that the variables $X_n$ are integrable. This means that $E |X_n| < \infty$ for all $n$. Then $X_n - X_1$ make an increasing sequence of nonnegative random variables. By monotone convergence theorem, $$E(X_n - X_1) \to E \lim_{n\to\infty}(X_n - X_1) = E X - E X_1\qquad (1)$$ with finite $E X_1$. ($E X$ is either finite or is equal to $+\infty$). The desired convergence is obtained by adding $E X_1$ to both sides of (1).
If $X_n$ are measurable, but not necessarily integrable, then we can construct a counterexample. Let $U$ be a random variable uniformly distributed on $(0, 1)$. Let $$X_n = \begin{cases} -1/U & \text{if}~ U < 1/n, \\ 0 & \text{if}~ U \ge 1/n. \end{cases}$$ Then the sequence of random variables $(X_n)$ is increasing in $n$ (surely) and is converging to $0$ (surely). But $E X_n = -\infty$ for all $n$, and $E 0 = 0$. Thus, $E X_n$ do not tend to $E \lim_{n\to\infty} X_n$.