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I've tried an idea similar to this post. But that idea does not work here, since $\sum_{n=1}^\infty \frac{1}{n^2}<+\infty$ but here $\sum_{n=1}^\infty \frac{1}{n}$ does not converge. (This is required by Borel–Cantelli lemma used in that idea.)

And now I have no new idea about this problem.

Could you please help? Thank you.

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I don't understand...what if $f(x)=x$ ? –  Belgi Sep 23 '12 at 22:32
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'Integrable' includes that the integral is finite –  Berci Sep 23 '12 at 22:33
    
Did you perhaps post an incorrect link? I expected to find a use of Borel-Cantelli, but did not. –  Quinn Culver Sep 24 '12 at 2:24
    
@QuinnCulver: In the second method, the author concludes that $\lambda\left(\bigcap_{k\geq 1}\bigcup_{n\geq k}A_{\delta}^n\right)=0$. I think this conclusion needs Borel-Cantelli lemma. –  Roun Sep 24 '12 at 2:49
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An answer is given here: hal.archives-ouvertes.fr/.../noteEL_nov08_v2.pdf –  Davide Giraudo Sep 26 '12 at 21:39

2 Answers 2

Note that it suffices to show that $\lim_{n \to \infty}f(nx) = 0$ for a.e. $x$ with $1 \leq x \leq 2$, as the general case follows by applying this case to $f(ax)$ for some $a$. Consider the quantity $\int_1^2 \sum_{n=1}^{\infty} |f(nx)|\,dx = \sum_{n=1}^{\infty}\int_1^2 |f(nx)|\,dx $. Changing variables in each term here from $nx$ to $x$, we get the following, where as usual $\chi_A(x)$ denotes the characteristic function of $A$: $$\sum_{n=1}^{\infty}\int_1^2 |f(nx)|\,dx = \sum_{n=1}^{\infty}\int_n^{2n} {1 \over n}|f(x)|\,dx $$ $$= \sum_{n=1}^{\infty}\int_0^{\infty}{1 \over n}\chi_{[n,2n]}(x) |f(x)|\,dx $$ $$= \int_0^{\infty}\sum_{n=1}^{\infty}{1 \over n}\chi_{[n,2n]}(x) |f(x)|\,dx $$

If $x$ is between $n$ and $2n$, then $n$ is between $x/2$ and $x$. So for a given $x$ there are at most ${x \over 2}$ terms for which ${1 \over n}\chi_{[n,2n]}(x)$ is nonzero, and each such ${1 \over n}\chi_{[n,2n]}(x)$ is at most ${2 \over x}$. Thus the sum of all these ${1 \over n}\chi_{[n,2n]}(x)$ is at most $1$, and the above is at most the finite number $$\int_0^{\infty} |f(x)|\,dx $$ We conclude that $\int_1^2 \sum_{n=1}^{\infty} |f(nx)|\,dx $ is finite. So the integrand $\sum_{n=1}^{\infty} |f(nx)|$ is finite a.e. on $[1,2]$. So the terms go to zero a.e. and $\lim_{n \rightarrow \infty} f(nx) = 0$ a.e. as needed.

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We asked our professor today. He gave us a proof, which used the similar technique (mentioned in the post above), plus a new technique.

Here's the proof:

Lemma: Let $H_{a,b,m}^{(1)}=\{x\in[a/m,b/m]:f(nx)>1 \text{ for infinite many } n\}$, then $\mu(H_{a,b,m}^{(1)})=0$. ($a$ and $b$ are positive integers)
Proof of Lemma:
We can easily see $H_{a,b,m}^{(1)}=\limsup_{n\to\infty} E_n$, where $E_n=\{x\in[a/m,b/m]:f(nx)>1\}$. By Borel-Cantelli Lemma, it suffices to show that $$ \sum_{n=1}^{\infty}\mu(E_n)<+\infty $$ We have $$ \begin{aligned} \mu(E_n)&=\int\limits_{\substack{x\in[a/m,b/m]\\f(nx)>1}}1\,dx\le\int_{a/m}^{b/m}f(nx)\,dx=\frac{1}{n}\int_{an/m}^{bn/m}f(x)\,dx\\ &=\frac{1}{n}\sum_{an\le k<bn}\int_{k/m}^{(k+1)/m}f(x)\,dx \end{aligned} $$ Then, $$ \begin{aligned} \sum_{n=1}^{\infty}\mu(E_n)&=\sum_{n\ge 1}\frac{1}{n}\sum_{an\le k< bn}\int_{k/m}^{(k+1)/m}f(x)\,dx\\ &=\sum_{a\le an\le k < bn}\frac{1}{n}\int_{k/m}^{(k+1)/m}f(x)\,dx\\ &=\sum_{\substack{k\ge a\\k/b<n\le k/a}}\frac{1}{n}\int_{k/m}^{(k+1)/m}f(x)\,dx\\ &=\sum_{k\ge a}\int_{k/m}^{(k+1)/m}f(x)\,dx\sum_{k/b<n\le k/a}\frac{1}{n}\\ &\le \frac{b-a}{a}\sum_{k\ge 1}\int_{k/m}^{(k+1)/m}f(x)\,dx\\ &=\frac{b-a}{a}\int_{1/m}^{+\infty}f(x)\,dx<+\infty \end{aligned} $$ Therefore, $\mu(H_{a,b,m}^{(1)})=0$.

Proof of the problem:
Now, denote $H_{a,b,m}^{(j)}=\{x\in[a/m,b/m]:f(nx)>\frac{1}{j} \text{ for infinite many } n\}$. Then, $$ H_{a,b,m}^{(j)}=\{x\in[a/m,b/m]:jf(nx)>1 \text{ for infinite many } n\} $$ where we $jf$ is another integrable function, therefore by the Lemma, $\mu(H_{a,b,m}^{(j)})=0,\,\forall m$
Therefore, $$ Z_{a,b,m}=\{x\in[a/m,b/m]:\limsup_{n\to\infty}f(nx)>0\}=\bigcup_{j}H_{a,b,m}^{(j)} $$ and $\mu(Z_{a,b,m})=0$ Therefore, $\mu(A)=0$, where $$A=\{x\in[1,+\infty):\limsup_{n\to\infty}f(nx)>0\}=\bigcup_{n=1}^{+\infty} Z_{n,n+1,1}$$.

Now it suffices to show $\mu(B)=0$, where $$B=\{x\in(0,1]:\limsup_{n\to\infty}f(nx)>0\}=\bigcup_m B_m $$ where
$$ B_m=\{x\in[1/m,1]:\limsup_{n\to\infty}f(nx)>0\}=\bigcup_{1\le k<m}Z_{k,k+1,m} $$ By the lemma, each $Z_{k,k+1,m}$ is of measure 0, so is B_m and so is B.

Q.E.D

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