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How can we show that: if $x$ is an integrable function (i.e., $\int_X|x|<\infty$), and $\{S_m\}$ is a sequence in $\Sigma$ ($\Sigma$ is a $\sigma$-algebra on $X$) in $X$ such that $\mu(S_m)\to 0$ as $m\to\infty$, where $\mu$ is a finite measure defined over $\Sigma$, then $$\int_{S_m}|x|\,\mathrm{d}\mu\to 0.$$ Can anyone give me some ideas? Many thanks. (I think the core is to show that $|x|\cdot 1_{S_m}$ converges to $0$ almost surely, but I don't know how to show that.)

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up vote 1 down vote accepted

Given the conditions, it's not necessarily true that $|x| 1_{S_m}$ converges to $0$ almost surely.

Here is one way to solve the problem: Fix $N>0$. You can partition $X$ into $X_N$, the set where $|x|\le N$, and $X'_N$, the set where $|x|>N$. The integral of $|x|$ over $S_m\cap X_N$ is bounded by $N \mu(S_m)$, and the integral of $|x|$ over $X'_N$ must approach $0$ as $N\to\infty$ because $x$ is integrable. Then, let $N\to\infty$ as $m\to\infty$ at an appropriate rate.

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What does Then, let $N\to\infty$ as $m\to\infty$ at an appropriate rate mean? – Did Mar 4 '13 at 23:52
It means to construct a function $N(m)$ such that $\lim_{m\to\infty} N(m)=\infty$ and such that $\lim_{m\to\infty} {\cal B}(N(m),m)=0$, where ${\cal B}(N,m)$ is the bound outlined above. – David Moews Mar 4 '13 at 23:57
Such as $N(m)=$ the integer part of $1/\sqrt{\mu(S_m)}$? I see, thanks. – Did Mar 5 '13 at 0:00
One more question: How can we show that if x is integrable, then the integral of |x| over $X'_N$ goes to zero?(intuitive, but I don't know how to put it rigorously) Thanks! – user65018 Mar 5 '13 at 1:34
Use the dominated convergence theorem. – David Moews Mar 5 '13 at 1:35

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