# Integration/measure theory "paradox"?

I have encountered the following "paradox." Consider a dense countable subset of $$\mathbb{R}$$, e.g. $$\mathbb{Q}$$. Because the set is countable we may parametrise it by $$\mathbb{Q} = \{ a_n \}_{n=1}^\infty$$. Then consider the function (for some $$\epsilon >0$$) $$\sum_{n=1}^\infty \chi_{[a_n, a_n + \epsilon/2^n)}$$ where $$\chi$$ is the indicator function. Because the set $$\mathbb{Q}$$ is dense, this function converges to infinity everywhere. But its integral according to Lebesgue measure is $$\int_{\mathbb{R}} \sum_{n=1}^\infty \chi_{[a_n, a_n + \epsilon/2^n)} d\mu = \sum_{n=1}^\infty \int_{\mathbb{R}} \chi_{[a_n, a_n + \epsilon/2^n)} d\mu = \sum_{n=1}^\infty \frac{\epsilon}{2^n} = \epsilon$$ where we have commuted summation and integral using B. Levi's theorem on monotone convergence. Where is my mistake?

EDIT: Soon after posting this I realised that my intuition the function converges to infinity everywhere is wrong.

• "Because $\mathbb{Q}$ is dense, this function converges to infinity everywhere" is the flaw. For a similar reason you could say that $\bigcup_{n=1}^\infty (-\epsilon/2^n+r_n, r_n+\epsilon/2^n)$ has infinite measure. But it doesn't. In other words, it isn't true that $$\bigcup_{n=1}^\infty (-\epsilon/2^n+r_n,r_n+\epsilon/2^n)=\mathbb{R}.$$ In fact, most points in $\mathbb{R}$ are not in this union. Good luck naming a single $x$ that isn't in this union though. You could construct it so that all computable numbers are in the union.
– user123641
Nov 7, 2018 at 19:29

## 1 Answer

As a funny other example, you might consider a function like : $$x\mapsto \sum_{n\in \mathbb N} \frac{\varepsilon_n}{\sqrt{|x-a_n|}}$$ you can easily find a sequence $$(\varepsilon_n)_{n\in\mathbb N}$$ such that this sum does converges...in $$L^1$$.