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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.

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    $\mu(A_n) := \int_{A_n} XdP$ is a measure that is absolutely continuous with respect to $P$. – 2017-02-14
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    $\mu(A_n)=\int_{A_n}XdP$ is a **finite** measure that is absolutely continuous with respect to $P$. – 2017-02-15
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    Thanks, I got it now! Would you like to post this as an answer, so I can accept it? – 2017-02-15
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    You could also just apply the dominated convergence theorem. – 2017-02-15
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    @saz In which way? $1_{A_n} X$ does not appear to converge almost surely to $X$ – 2017-02-15
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    @AndreiKh The dominated convergence theorem only needs convergence in measure. – 2017-02-15
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    a similar question was asked here: http://math.stackexchange.com/questions/421176/convergence-of-int-a-n-f-to-0?rq=1 – 2017-02-15
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    @AndreiKh See here: http://math.stackexchange.com/questions/206851/generalisation-of-dominated-convergence-theorem – 2017-02-15

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First observe that you can define $B_n:=\{X0 $. There is an $n_1\in N $ such that $\int_{B_{n_1}^{\ c}}{X\ dP}< \varepsilon/2 $.

So \begin{equation} \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} \end{equation} And there exists $n_0$ large enough such that $P(A_n) < \varepsilon /2n_1$ for any $\ n>n_0$ .

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    I guess you meant to say "for any $n>n_0$" instead of "for any $n>n_1$" at the end. – 2017-02-16
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    That is right, I've corrected it already, thanks. – 2017-02-17
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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 $k \geq N$