An exercise on weak convergence from Henrot's book

Question

I am working on the following exercise from Hernot and Pierre's "Shape Variation and Oprimization" (EMS edition): in $\mathbb{R}$ let us consider the following sequence of open sets: $$\Omega_n:=\bigcup_{k=0}^{2^{n-1}-1}\left(\frac{2k}{2^n},\frac{2k+1}{2^n}\right).$$ Prove that $\chi_{\Omega_n}$ converges weakly in $L^p(0,1)$, for any $1\leq p\leq\infty$ to the constant function equal to $1/2$. $\chi_E$ stands for the charactersitic function of $E$.

Say that we start with $1<p<\infty$, we want to see that for any $u\in L^{p'}(0,1)$ with $\frac{1}{p}+\frac{1}{p'}=1$ we have $$\lim_{n\rightarrow\infty}\int_{(0,1)}\chi_{\Omega_n}u\:dx=\frac{1}{2}\int_{(0,1)}u\:dx.$$

A natural place to start is to prove first the case $u=\chi_{(a,b)}$ with $0\leq a<b\leq 1$; the general result follows rather directly by approximation. Now in this case we want to see that $$\lim_{n\rightarrow\infty}\int_{(a,b)}\chi_{\Omega_n}\:dx=\lim_{n\rightarrow\infty}\sum_{k=0}^{2^{n-1}-1}\left|\left(\frac{2k}{2^n},\frac{2k+1}{2^n}\right)\cap(a,b)\right|=\frac{b-a}{2}.$$ ($|\cdot|$ denotes the Lebesgue measure.)

I have a good intuition about why this should be true. Basically, the sets $\Omega_n$ occupy half of the interval $(0,1)$ at each level and they become thinner and thinner as $n$ grows. Hence, in the limit they should cover half of the interval $(a,b)$. I am trying to formalize that reasoning.

For $n\in\mathbb{N}$ we know that there exist some $k_0,k_n\geq0$ such that $$\left(\frac{2k}{2^n},\frac{2k+1}{2^n}\right)\cap(a,b)=\emptyset$$ if either $k\leq k_0$ or $k\geq k_n$. Then, we have $$\sum_{k=0}^{2^{n-1}-1}\left|\left(\frac{2k}{2^n},\frac{2k+1}{2^n}\right)\cap(a,b)\right|=\frac{2k_0+1}{2^n}-a+\frac{1}{2^n}+\ldots+\frac{1}{2^n}+b-\frac{2k_n}{2^n}.$$ In the (very restrictive) case that $a$ and $b$ coincide with extremes of the $\Omega_n$s the terms in the middle are exactly $2^{n-1}(b-a)$ and there are $2^n$ of them so we would be done. I believe in general we have roughly that situation "plus an error'' that should vanish as $n$ goes to $\infty$ but I am not sure how to translate that hand waving into a rigorous proof.

Any ideas would be appreciated.

PS: while writing this post I found this entry which deals with the same problem, but there is only a hint which suggest to start where I started, so that is not much of a help.

Answer 1

The special case you showed is when $a$ and $b$ are themselves dyadic numbers. In the general case, find dyadic numbers of the form $a_n=2p/2^n<a$ and $b_n=(2q+1)/2^n>b$ such that (say) $0<a-a_n,b_n-b < 10/2^n$. Then $|\Omega_n\cap(a,b)| \le |\Omega_n\cap(a_n,b_n)|\le|\Omega_n\cap(a,b)|+ 20/2^n$ (very crude). In the limit, you'll get $$ \limsup_{n\to\infty}|\Omega_n\cap(a,b)|\le \lim_{n\to\infty}\frac{b_n-a_n}{2}\le\liminf_{n\to\infty}(|\Omega_n\cap(a,b)|+20/2^n). $$ The middle term comes by recycling the calculation you made in your original post, and it is a limit that can be evaluated directly.