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[My question concerns part of Exercise 2.25 in Folland's Real Analysis text.]

I'm looking at the function $g(x)=\sum_{n=1}^\infty 2^{-n}f(x-r_n)$, where $f(x)=x^{-1/2}$ for $x\in (0,1)$ and $f(x)=0$ elsewhere, and $\{r_n\}$ is some enumeration of the rational numbers. I am trying to prove that $g$ is discontinuous at every point. It is easy enough to see that $g$ is discontinuous wherever it is finite (since it is unbounded on every interval), and that it is finite almost everywhere (since it is in particular in $L^1$). However, it seems to me that to show that $g$ is discontinuous at a point $x_0$ with $g(x_0)=+\infty$, one would have to exhibit a sequence $\{x_n\}$ converging to $x_0$ such that $\{g(x_n)\}$ is bounded. It seems intuitively obvious to me that such a sequence exists, but I have not been able to construct one. Any suggestions? (Hints preferred to answers.)

1234
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1 Answers1

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Hint: if a function is continuous at a point, it is bounded in some neighborhood of that point.

PA6OTA
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  • Hmm... even for extended real-valued functions? As I understand, such a function (for the sake of definiteness, the function $g$ above) may be continuous even at a point $x_0$ where it attains the value $\infty$, provided that $\lim_{x\to x_0} g(x)=\infty$. Or am I misinformed? – 1234 May 19 '14 at 22:11
  • I don't have the book, so I'm not sure what the "$\infty$" point means there. You can notice that the left and right limits at a rational point (if they exist) are $-\infty$ and $+\infty$, and that can't be the same thing. May be, consider $g(x)=\sum_{n=1}^\infty 2^{-n}|x-r_n|^{-1/2}$? – PA6OTA May 19 '14 at 22:20
  • Ah--you've brought to my attention a careless mistake I originally made when defining the function. It has now been corrected-- my apologies. Given the corrected definition of the function $g$, is it possible to show that, given a point $x_0$ where $g(x_0)=+\infty$, there is a point $y$ arbitrarily close to $x_0$ such that $g(y)\le M$ for some fixed $M$? – 1234 May 19 '14 at 22:34