Approximating characteristic functions by cutting the real axis into smaller and smaller pieces

Question

Let $\Lambda_r^*=\frac{1}{2\pi r} \mathbb{Z} \subset\mathbb{R} (r>0)$, let $E\subset\mathbb{R}$ be a Lebesgue measurable set with finite measure $|E|$, define $J_r=(-\frac{1}{4\pi r}, \frac{1}{4\pi r}), J_r(\alpha)=\alpha+J_r$ for $\alpha\in\Lambda_r^*$, $\{r_n\}$ any positive sequence with $r_n\to\infty$ as $n\to \infty$. Define $$f_n =\sum_{\alpha\in\Lambda_{r_n}^*} 2\pi r_n |E\cap J_{r_n}(\alpha)|\chi_{J_{r_n}(\alpha)}.$$

Then is it true that $f_n \to \chi_E$ (a.e. pointwise or in $L^2$ sense) as $n\to \infty$?

I think it's easy when $E$ is an interval, but I can't prove it for general $E$. Can any one give some help?

Answer 1

You just asked this in response to my answer to this question, but you didn't give me a chance to answer it there!

By expressing the problem in this very concrete way, you've made it harder. It is much easier to understand if you look at it abstractly. Let $X_r$ be the subspace of $L^2(\mathbb{R})$ consisting of the functions which are constant on each of these intervals $\alpha + J_r$, and let $P_r$ be the orthogonal projection from $L^2(\mathbb{R})$ onto $X_r$. Then $f_n = P_{r_n}(\chi_E)$, so you are asking whether $P_{r_n} \chi_E \to \chi_E$.

But as you notice (and as I said in my previous answer), this is easy when $E$ is an interval. Taking linear combinations, we get $P_{r_n}f \to f$ whenever $f$ is a finite linear combination of characteristic functions of intervals. But these functions are dense in $L^2(\mathbb{R})$, so we finally get $P_{r_n}f \to f$ for any $f \in L^2(\mathbb{R})$ by an $\frac{\epsilon}{3}$ argument.