I'm trying to solve the exercises from Real Analysis – Moderns Techniques and Their applications by Folland and I have questios about the exercise 2.25 b). The statement:
Let $f(x) = x^{-1/2}$ if $0<x<1$ and f(x)=0 otherwise. Let ${r_n}$ be and enumeration of the rationals, and set $g(x) = \sum_{n=1}^\infty 2^{-n} f(x-r_n)$. Then g is discontinuous at every point and unbounded on every interval, and it remains so after any modification on a Lebesgue null set.
It's proved before (item a) that $g$ is finite a.e.. If $g(x)<+\infty$, it's not so difficult to prove that $g$ is discontinuous at $x$ (there is a solution here: http://www.math.mcgill.ca/spicard/folland2.pdf) but I can't prove the same when $g(x)=+\infty$. I'm assuming that I have to see $g$ as a function $g:\mathbb R \longrightarrow [0,+\infty]$ and use the usual topology that Folland put in this set.
So, if I'm interpreting the question correctly, how can I prove that $g$ is discontinuous in a point $x$ such that $g(x)=+\infty$? The prove made in previously case doesn't help, because it shows that $g(x)$ can assume high values.
Best Answer
I claim that the function $g$ is continuous at every point where $g(x) = \infty$. Here is one way to see this. Recall that a function $h:\mathbb{R}\to {\mathbb R}\cup\{-\infty,\infty\}$ is lower semicontinuous provided $$ h(x) \le \liminf_{y \to x} h(y) $$ for every $x\in \mathbb R$. Put $h_n(x) = 2^{-n}f(x-r_n)$. Then it is easy to check that $h_n$ is lower semicontinuous.
Now it can be shown that any sum of non-negative lower semicontinuous functions is lower semicontinuous. Thus $g$ is lower semicontinuous and so $$ g(x) \le \liminf_{y\to x} g(y) $$ for all $x\in \mathbb R$. Now suppose that $g(x) = \infty$. Then we automatically get $$ \limsup_{y\to x}g(y) \le g(x). $$ Hence we have $$ \limsup_{y\to x} g(y) \le g(x) \le \liminf_{y\to x} g(y) $$ and so $g$ is continuous at $x$.