How to show that $\lim_{n\to \infty} f_n(x) = 0$. I hate to admit it, but I don't know how to begin solving this problem: 
Let $f\in L^1(\mathbb R)$. Let $\displaystyle f_n(x) = \frac{f(nx)}{n},~n\geqslant 1$. Then $\displaystyle \lim_{n\to \infty} f_n(x) = 0$ for almost every $x\in \mathbb{R}$. 
Please this is not a home problem, so hints and full solutions will be very much appreciated.
 A: First method: we use Fubini's theorem, applied to Lebesgue measure and counting measure, which are both $\sigma$-finite. Using a substitution
$$\sum_{n=0}^{+\infty}\int_{\mathbb R}\frac{|f(nx)|}n\lambda(x)=\sum_{n=0}^{+\infty}\frac 1{n^2}\int_{\mathbb R}|f(x)|\lambda(x)$$ so $\sum_{n=0}^{+\infty}\frac{|f(nx)|}n$ is convergent for almost every $x$ and we are done.

Second method, using only basis facts of measures and sequences.
Fix $\delta>0$ and an integer $n$ and put $A_{\delta}^n:=\left\{x\in\mathbb R,\left|\frac{f(nx)}n\right|\geq \delta\right\}$. Then
$$\delta \mu(A_{\delta}^n)\leq \int_{A_{\delta}^n}\left|\frac{f(nx)}n\right|d\lambda(x)=\int_{nA_{\delta}^n}\frac{f(t)}{n^2}d\lambda(t)\leq \frac{||f||_{L_1}}{n^2}$$
so $\lambda(A_{\delta}^n)\leq \frac{||f||_{L_1}}{\delta n^2}$ and therefore $\lambda\left(\bigcap_{k\geq 1}\bigcup_{n\geq k}A_{\delta}^n\right)=0$. Denote $B_{\delta}=\bigcap_{k\geq 1}\bigcup_{n\geq k}A_{\delta}^n$. Then $\lambda\left(\bigcup_{p\geq 1}B_{p^{-1}}\right)=0$ and $N:=\bigcup_{p\geq 1}B_{p^{-1}}$ is exactly the set of the $x$ such that $\frac{f(nx)}x$ doesn't converge to $0$. Indeed, if $x\in N$ then $x\in B_{p^{-1}}$ for some $p$ so there are infinitely many $n$ such that $x\in A_{p^{-1}}^n$ i.e. $\left|\frac{f(nx)}n\right|\geq \frac 1p$ so in particular $\frac{f(nx)}n$ doesn't converge to $0$. Conversely, if $\frac{f(nx)}n$ doens't converge to $0$, then we can find an integer $n_0$ and $J\subset \mathbb N$ infinite such that for all $n\in J$, $\left|\frac{f(nx)}n\right|\geq n_0^{-1}$, so $x\in B_{n_0^{-1}}$.
