Existence of limit of increasing sequence of random variables.

Question

Let $X_1 \leq X_2 \leq \cdots$ be an increasing sequence of integrable random variables. Suppose there exists $M$ such that $$\mathbb{E}(X_n)\leq M \quad \forall n$$ I am trying to prove that $X := \lim_{n \rightarrow \infty} X_n$ exist and is a.s. finite. Can someone give me a hint regarding this?

Answer 1

Well first of all $X = \lim_{n \to \infty} X_n$ exists (just do it pointwise for each $\omega$ and use the fact that for each $\omega$, $(X_n(\omega))_n$ is an increasing sequence thus has a limit which might be finite or $\infty$).

The next part consists of showing Beppo-Levi's theorem: Let

$$Y_i = X_i - X_1 \geq 0$$

$$Y = X - X_1 \geq 0$$

Then $Y_n$ is also a non-decreasing non-negative sequence with $Y_n \to Y$. Thus by monotone convergence we get that:

$$ \int Y d\mu = \lim_{n\to\infty} \int Y_n d\mu $$

note that $\int Y_n d\mu \leq 2M$, hence $Y_n$ and $Y$ are also integrable and additionally $|X| \leq Y + |X_1|$, thus $X$ is integrable ($X_1$ integrable by assumption). But since it is integrable, $X$ must be almost surely finite. Thus we are done.

Remark: By adding $\int X_1 d\mu$ (which we know is finite) to both sides of the above limit we also get:

$$ \int X d\mu = \lim_{n\to\infty} \int X_n d\mu $$

This is the actual result of Beppo Levi's theorem.