How to find $\lim_{n\rightarrow\infty} \int_{0}^{1} f(t)\phi(nt) dt$ for $1$-periodic $\phi$? How to solve the following problem:
Let $\phi\in L^{\infty}(0,1)$ be a $1$-periodic function and $f\in L^{1}(0,1)$. Find  $\lim_{n\rightarrow\infty} \int_{0}^{1} f(t)\phi(nt) dt$.
Thanks in advance.
 A: 
Claim: $$\lim_{n\to\infty}\int_0^1 f(t)\phi(nt)dt=\int_0^1f(s)ds\cdot \int_0^1\phi(s)ds.$$

Proof:
Since $\phi$ is of period $1$, for $s=nt$,
$$\int_0^1 f(t)\phi(nt)dt=\frac{1}{n}\int_0^n f(\frac{s}{n})\phi(s)ds=\frac{1}{n}\sum_{k=0}^{n-1}\int_0^1 f(\frac{s+k}{n})\phi(s)ds.\tag{1}$$
First let us assume that $f\in C[0,1]$, i.e. $f$ is continuous on $[0,1]$. Then
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{s+k}{n})=\int_0^1f(t)dt,\quad \forall s\in[0,1].\tag{2}$$
Combining $(1)$ and $(2)$, by dominated convergence theorem,
$$\lim_{n\to\infty}\int_0^1 f(t)\phi(nt)dt=\int_0^1f(s)ds\cdot \int_0^1\phi(s)ds.\tag{3}$$
Now given $f\in L^1(0,1)$, for any $\epsilon>0$, there exists $g\in C[0,1]$, such that 
$$\int_0^1|f(t)-g(t)|dt<\epsilon.\tag{4}$$
It follows that
$$\int_0^1 \big|(f(t)-g(t))\phi(nt)\big|dt\le\epsilon\|\phi\|_\infty, \quad \forall n\ge 1.\tag{5}$$
Since $(3)$ holds when $f$ is replaced with $g$, with the help of $(4)$ and $(5)$, we have: 
$$\limsup_{n\to\infty}\big|\int_0^1 f(t)\phi(nt)dt-\int_0^1f(s)ds\cdot \int_0^1\phi(s)ds\big|\le 2\epsilon\|\phi\|_\infty.\tag{6}$$
Since $\epsilon>0$ in $(6)$ is arbitrary , we can conclude that $(3)$ holds for every $f\in L^1(0,1)$.$\quad\square$
