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Show that if $f$ is a Lebesgue integrable function on $A\subset\mathbb R$ and $$A_n=\{x\in A:|f(x)|\geq n\}$$ for $n\in\mathbb N$, then $\lim_{n\to\infty} n\cdot m(A_n)=0$.

My solution which is different from the answer provided is as follows:

Assume to the contrary that $\lim_{n\to\infty} n\cdot m(A_n)\neq0$. Then $\exists\varepsilon>0$ s.t. $\forall n\in\mathbb N,n\cdot m(A_n)>\varepsilon$.

$A=\bigcup_{n\in\mathbb N}A_n$, so $m(A)=\sum_{n=1}^{\infty}m(A_n)$

So $\int_A|f|d\mu=\sum_{n=1}^{\infty}\int_{A_n}|f|d\mu\geq\sum_{n=1}^{\infty}n\cdot m(A_n)\geq\sum_{n=1}^{\infty}\epsilon\rightarrow\infty$. So we get a contradiction and hence $\lim_{n\to\infty} n\cdot m(A_n)\neq0$.

Could somebody check if this argument works? Thanks!

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  • $\begingroup$ You probably know that, but in case: this is very easy if $f^2$ is integrable as $m(A_n)\leq \frac{1}{n^2}\int |f|^2dm$ by Chebyshev's inequality. If $f$ is merely, integrable, we only get $m(A_n)\leq\frac{1}{n}\int |f|dm$. This is not enough to porve the result, but it shows $\lim m(A_n)=0$ which you may use now. $\endgroup$
    – Julien
    Apr 12, 2013 at 20:46

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It does not work. For $m>n$, we have $A_m\supset A_n$. Thus though we do have $A=\bigcup_{n=0}^\infty A_n$, because this union is not disjoint we do not have $m(A)=\sum_{n=0}^\infty m(A_n)$ -- we have $m(A)\leq \sum_{n=0}^\infty m(A_n)$. Thus the first equality in your penultimate line is not valid.

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  • $\begingroup$ Thanks for pointing that out! :) $\endgroup$
    – Haikal Yeo
    Apr 12, 2013 at 20:12
  • $\begingroup$ No problem! Do you have a different solution in mind? $\endgroup$
    – Ian Coley
    Apr 12, 2013 at 20:30
  • $\begingroup$ Basically show that $m(A_n)\rightarrow0$ first? And then I think everything works out nicely I think.. $\endgroup$
    – Haikal Yeo
    Apr 12, 2013 at 20:34

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