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 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 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 By the lemma, each $Z_{k,k+1,m}$ is of measure 0, so is B_m and so is B.
Q.E.D