Show that $\lim_n \int_A f(nx)dx =\lambda(A) \frac1T \int_0^T f(t) dt$ for a measurable, bounded, T-periodic function.

Question

I found this exercice :

Let $f : \mathbb{R} \to \mathbb{R}$ a measurable bounded and T-periodic function. Show that for every bounded borel set A, $$\lim_n \int_A f(nx)dx =\lambda(A) \frac1T \int_0^T f(t) dt$$

I can prove it when A is an bounded interval of $\mathbb{R}$. But, I can't prove the general result.

My first idea was to use that the borel set is generated by open intervals. I proved that the null set satifies the property, if $A$ and $B$ satifies the property then $A\setminus B$ also, if $(A_n)$ satifies the property and $A_n\subset A_{n+1}$ then $\cup A_n$ also. But, I wasn't able to prove that is $A_1, ..., A_N$ satisfies the property then $\cap_{i=1}^N A_n$ also.

My second idea was to use the definition of the Lebesgue measure as the infimum of lenght of covering by open intervals. But I obtained only one inequality.

Answer 1

Hint: Given $\epsilon >0$ there exists a set $B$ which is a finite disjoint union of intervals of the type $(a,b]$ with $\int |\chi_a-\chi_b| <\epsilon$. Ref: Theorem D, p. 56 of Measure Theory by P R Halmos.