Let $f geq 0$ be a function. Show that

$$lim_n frac{1}{2^n}sum_{j=1}^{2^{2n}} chi_{{f > j/2^n}} = f$$ pointwise. Here $chi_A$ is the indicator function on the set $A$ and ${f > j/2^n}:= {x : f(x) > j/2^n}$

Attempt:

I’m a bit lost here. The sum looks really complicated and it’s hard to see for me what’s going on.

I tried to estimate $$|f(x)- frac{1}{2^n}sum_{j=1}^{2^{2n}} chi_{{f > j/2^n}}(x)|$$

but could not get anything that becomes small.