# $\lim_{n \to \infty} P(A_n) = 0$ implies $\lim_{n\to\infty}\int_{A_n}{X}dP = 0$

Let $(\Omega, \mathcal{A}, P)$ be a probability space and let $X$ be a non-negative, real random variable with $\int{X}dP < \infty$. If $A_n \in \mathcal{A}$ is a sequence such that $\lim_{n \to \infty} P(A_n) = 0$, is it true then that $$\lim_{n\to\infty}\int_{A_n}{X}dP = 0$$ If yes, I would appreciate a hint on how to prove this.

• $\mu(A_n) := \int_{A_n} XdP$ is a measure that is absolutely continuous with respect to $P$. Commented Feb 14, 2017 at 22:47
• $\mu(A_n)=\int_{A_n}XdP$ is a finite measure that is absolutely continuous with respect to $P$. Commented Feb 15, 2017 at 2:10
• Thanks, I got it now! Would you like to post this as an answer, so I can accept it? Commented Feb 15, 2017 at 9:55
• You could also just apply the dominated convergence theorem.
– saz
Commented Feb 15, 2017 at 14:36
• @saz In which way? $1_{A_n} X$ does not appear to converge almost surely to $X$ Commented Feb 15, 2017 at 15:37

First observe that you can define $B_n:=\{X<n\}\uparrow \Omega$ ( because $X$ is integrable). Then, by dominated/monotone convergence, you have $$\int_{B_n}{X\ dP}\rightarrow\ \int_{\Omega}{X\ dP}$$ Now, let $\varepsilon >0$. There is an $n_1\in N$ such that $\int_{B_{n_1}^{\ c}}{X\ dP}< \varepsilon/2$.

So $$\begin{split} \int_{A_n}{X\ dP} &=\int_{A_n \cap B_{n_1}}{X\ dP} + \int_{A_n \cap B_{n_1}^{\ c}}{X\ dP} \\ & < n_1 P(A_n)+ \varepsilon /2 \end{split}$$ And there exists $n_0$ large enough such that $P(A_n) < \varepsilon /2n_1$ for any $\ n>n_0$ .

• I guess you meant to say "for any $n>n_0$" instead of "for any $n>n_1$" at the end.
– user940
Commented Feb 16, 2017 at 2:01
• That is right, I've corrected it already, thanks.
– Max
Commented Feb 17, 2017 at 18:26

Here's another way, using only non-negativity:

Show this using an argument building from simple non-negative functions to non-negative measurable ones. Using the setup you have above,

1. Take $X = \sum_1^n a_i 1_{B_i}$ where $a_i \geq 0$ are real and $\cup_1^n B_i = \Omega$, $B_i$ are disjoint measurable sets.

By your assumption, for every $\epsilon >0$ there is an $N(\epsilon)$ so that $P(A_k) < \epsilon / \sum_1^n a_i$ for all $k \geq N$.

$$\int_{A_k} X \, dP = \sum_1^n a_i P(A_k \cap B_i) < \epsilon$$ for all $k \geq N$. So $\int_{A_k} X \, dP \to 0$

1. For every non-negative measurable function $X$, there is an increasing sequence of non-negative simple functions $f_n$ converging to $X$ almost everywhere.

Monotone convergence says also $\int_\Omega |X - f_n| \, dP \to 0$.

Pick an $\epsilon > 0$ and $N, M \geq 1$ so that $\int_\Omega |X - f_n| \, dP < \epsilon /2$ for all $n \geq M$ and $\int_{A_k} f_M \, dP < \epsilon /2$ for $k \geq N$ using part 1.

$$\int_{A_k} X \, dP \leq \int_\Omega |X - f_M| \, dP + \int_{A_k} f_M \, dP < \epsilon$$

for any $k \geq N$