A function finite almost everywhere but NOT Lebesgue integrable

lebesgue-integrallebesgue-measuremeasure-theory

I was trying to solve a problem on measure theory from the book of Folland and got stuck in one problem(Prob-25,page-59). My comments are in italics.

Let $m$ be the Lebesgue measure on $\mathbb{R}$ and $L^1(m)$ be the space of all m'ble functions $g$ such that $|g|$ is integrable.

Problem:
Let $f(x)=x^{-\frac{1}{2}}$ for $0<x<1$ and $f(x)=0$ else. Let $\{r_n\}_{n=1}^\infty$ be an enumeration of rationals.

Define $g(x)=\sum_n\frac{f(x-r_n)}{2^n}$.

Show that

(1) $g\in L^1(m)$.

(2) $g$ is discontinuous everywhere and unbounded on every interval. It remains so after a correction on a set of measure $0$.

(3) $g^2<\infty$ a.e. but $g^2$ is not integrable over any interval.

Since $f$ is non negative, by applying monotone convergence theorem I could solve 1. But I'm completely blind about 2 and 3 even about how to start. Any kind of suggestions will be appreciated.

Thank you for your help in advance.

Best Answer

Hint: On the interval $(a,b)$, pick a rational $r$ such that $a<r<b$. Note that $g(x) \ge 2^{-n} f(x-r)$, where $n$ is such that $r = r_n$.

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