Assume $\mathbb{E}|X_1| < \infty$ and $X_n \uparrow X$ a.s., then Monotone Convergence Theorem either provide $\mathbb{E}X_n \uparrow \mathbb{E} X < \infty$ or else $\mathbb{E}X \uparrow \infty$ and $\mathbb{E}|X| = \infty.$

($\mathbb{E}$ = expected value, equivalently, it is the same as an integration with respect to probability measure)

Here is my approach. First assume that $\mathbb{E}|X| < \infty.$ Define $$Y_n = X - X_n.$$ Then $Y_n$ is well-defined a.s. (almost surely) because $\mathbb{E}|X| < \infty$ implies $X$ is bounded a.s. Moreover, $Y_n \geq 0$ and $Y_n \downarrow 0.$ Since $Y_n \leq |X|$, $\mathbb{E}Y_n < \infty$ for all $n$. By Monotone convergence theorem, $$\mathbb{E}Y_n \downarrow \mathbb{E}0 = 0$$ which yields $$\mathbb{E}X_n \uparrow \mathbb{E}X < \infty.$$

Next, assume that $\mathbb{E}|X| = \infty.$ Suppose that $X$ is bounded a.s. Then it is clear that $\mathbb{E}|X| < \mathbb{E}M < \infty$ where $M > 0$ constant, a contradiction. So $X$ is not bounded a.s. Since $X_n \leq X_{n+1}$, $\mathbb{E}X_n \leq \mathbb{E}X_{n+1}$, that is, $(\mathbb{E}X_n)$ is an increasing sequence. Suppose that there exist $N > 0$ such that $\mathbb{E}X_n \leq N$ for all $n$. Then $\lim \mathbb{E}X_n < \infty.$

I think that this will lead to a contradiction with $X$ is not bounded a.s., but somehow, I cannot produce the contradiction. Anyway, I also do not use $\mathbb{E}|X_1| < \infty$. So I doubt that my proof is not somewhere correct.

Any help please ?

  • $\begingroup$ Any hints or suggestions ????? $\endgroup$ – Both Htob Feb 3 '16 at 13:04
  • $\begingroup$ It doesn't have to be $X_1$. It can be that some random variable $X_k$ is integrable. We just want to avoid the case where they all have $E[|X_i|] = \infty$ $\endgroup$ – user198044 Feb 3 '16 at 14:34

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