Limit of $\int_E f(nx) dx$ for a $1$-periodic function $f$ on $[0,2\pi]$ Let $E$ be a measurable subset of $[0, 2\pi]$. Assume that $f \in C(\mathbb R)$ is $1$-periodic, i.e. $f(x + 1) = f(x)$. Compute $$\lim_{n\to\infty} \int_{E} f(nx) dx$$.
Since $f$ is continuous on $\mathbb R$ so it is continuous in each measurable subset of $[0, 2\pi]$ and so $f(nx)$ is Lebesgue integrable for each $n$. But how can I use $1$ periodicity of $f$?
 A: Define $c=\int_{0}^{1}f(x)dx$, $M=\max|f(x)|$ exist by the given conditions, try to prove the answer is $c\mu(E)$


*

*First we prove it's true for any interval, $$\displaystyle \lim_{n\to \infty}\int_{a}^{b}f(nx)dx\\=\lim_{n\to \infty}\frac{1}{n}\int_{na}^{nb}f(y)dy\\=\lim_{n\to \infty}\frac{1}{n}\int_{na}^{na+[nb-na]}f(y)dy+\frac{1}{n}\int_{na+[nb-na]}^{nb}f(y)dy\\=\lim_{n\to \infty}\frac{[nb-na]}{n}\int_{0}^{1}f(y)dy+\frac{1}{n}\int_{na+[nb-na]}^{nb}f(y)dy\\=(b-a)c$$


(Substitution $y=nx$ for the first equality, $[]$ denote floor function, in the last step the first part is straight foward limit evaluation, the second part has to be $0$ because regardless of $a,b,n$ the integral is bounded by $\int_{0}^1|f(x)|dx\leq \max{|f(x)|}$, a finite number)


*Now we prove it's true for $E=$ countable union of intervals and finite measure
Let $E=\displaystyle \cup_{j=1}^{\infty}I_j$, each $I_j$ is interval (can make it disjoint). Define $E_k=\cup_{j=1}^{k}I_j$, not hard to see $\displaystyle \lim_{n\to \infty}\mu(I_n)=0$ and $\displaystyle \lim_{n\to \infty}\mu(E_n)=\mu(E)$


So for every $\epsilon>0$, we can pick $k$ large enough such that $|\int_{E}f(nx)dx-\int_{E_k}f(nx)dx|\leq M\mu(E-E_k)<\epsilon/3$
By part 1. pick $n$ large enough we have
$|\int_{E_k}f(nx)dx-c\mu(E_k)|<\epsilon/3$
Once again, pick $k$ large enough, we have 
$|c\mu(E)-c\mu(E_k)|<\epsilon/3$
Combine all of them together by triangle inequality we have $|\int_{E}f(nx)dx-c\mu(E)|<\epsilon$, which is the conclusion.


*Now for arbitrary measurable set $E$, by a famous result for any $\epsilon $we can find an open set $U$(which in $\mathbb{R}$ is countable disjoint union of open interval) such that $E\subset U$, and $\mu(U)-\mu(E)<\epsilon$. Similar to argument in part 2, because $|f|$ is bounded for any $\epsilon>0$ we can make $|\int_{E}f(nx)dx-\int_{U}f(nx)dx|$,$|\int_{U}f(nx)dx-c\mu(U)|$, and $|c\mu(U)-c\mu(E)|$ all $<\epsilon/3$ and reach the conclusion.

