Upper bound for the approximation of Lebesgue integral

Question

Let $u:[-T/2,T/2]\to\mathbb{R}$ be a nonnegative measurable function $u(t)\ge0$. For each integer $n \ge 1$, define the $n$th order approximation to the Lebesgue integral as follows: $$I_n=\sum_{m}m2^{-n}\mu_{m,n} \ \ \ \ \text{where}\ \ \mu_{m,n}=\mu(t:m2^{-n}\le u(t)\lt (m+1)2^{-n})$$Assume that $I_1$ is finite. I want to show that the change in going from the approximation of order $n$ to $n+1$ is nonnegative and upper bounded by $T2^{−n−1}$. According to the MIT lecture notes, this fact is obvious from the figure: I don't see how the figure shows that $I_{n+1} - I_{n}\le T2^{−n−1}$. Also I don't know how to prove it mathematically. We should have: $$I_{n+1} - I_{n} = \sum_{m}m(2^{-(n+1)}\mu_{m,n+1} - 2^{-n}\mu_{m,n}) = \sum_{m}m2^{-n}(2^{-1}\mu_{m,n+1}-\mu_{m,n})\le T2^{−n−1} \implies \sum_{m}m(2^{-1}\mu_{m,n+1}-\mu_{m,n})\le T2^{−1}$$ So we need to find a relation between $\mu_{m,n+1}$ and $\mu_{m,n}$ in general.

Answer 1

An idea to prove this formally is to write $$ \mu_{m,n}=\mu_{2m,n+1}+\mu_{2m+1,n+1}, $$ which follows from the fact that $$ m2^{-n}\leq u(t)<(m+1)2^{-n} \quad \Leftrightarrow \quad 2m2^{-n-1}\leq u(t)<2(m+1)2^{-n-1}. $$ This gives $$ I_n= \sum_m m2^{-n} \mu_{m,n}= \sum_m [m2^{-n}\mu_{2m,n+1} +m2^{-n}\mu_{2m+1,n+1}]\\ = \sum_m [2m2^{-n-1}\mu_{2m,n+1} + (2m+1)2^{-n-1}\mu_{2m+1,n+1}] - \sum_m 2^{-n-1}\mu_{2m+1,n+1}\\ = I_{n+1}- \sum_m2^{-n-1}\mu_{2m+1,n+1}. $$ Finally, we just note that, owing to countably additivity of Lebesgue measure, $$ \sum_m 2^{-n-1} \mu_{2m+1,n+1} \leq T2^{-n-1}. $$