I am working on the following problem from some old exam:
For $\ n>0, \ $ a set $\ A_n \subset [0,1] \ $ is defined as $$A_n \ = \ \bigcup_{k=1}^{2^{n-1}}\Bigg[\frac{2k-2}{2^n},\frac{2k-1}{2^n}\Bigg].$$ Prove that for any Lebesgue integrable function $\ u:[0,1]\to \mathbb{R} \ $ we have $$\lim_{n\to \infty}\int_{A_n}u(t)\mathrm{d}t \ = \ \frac{1}{2}\int_{0}^{1}u(t)\mathrm{d}t. \ \ \ \ \ \ (\star) $$
Thus $\ A_n \ $ is a union of odd intervals from $\ n$-th binary division of $ \ [0,1]. \ $ It is obviously enough to check the thesis for $\ u\ge 0. \ $ In fact, I believe that I can check that $ \ (\star) \ $ is correct for simple functions. How to go from there? If $\ u_k \ $ are simple functions monotonically growing up to $\ u, \ $ we have $$\lim_{k\to \infty} \ \lim_{n\to \infty} \ \int_{A_n}u_k \ \mathrm{d}t \ \ = \ \ \lim_{k\to\infty} \ \frac{1}{2}\int_{0}^{1} u_k \ \mathrm{d}t \ \ = \ \ \frac{1}{2}\int_{0}^{1}u \ \ \mathrm{d}t.$$ That would end the proof if we could interchange the limits on the left side. Can we do that? How to solve this problem?