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Let $\{r_n\}_{n\in\mathbb{N}}$ be a enumeration over the rationals

Let $$g(x)=\sum_1^\infty \frac{1}{2^n} \frac{1}{\sqrt{x-r_n}} \chi_{(0,1]}$$ where $$\chi_{(0,1]} = \left\{\begin{array}{ll} 1&\mbox{if $x-r_n \in (0,1]$,}\\ 0&\mbox{otherwise.} \end{array}\right.$$

Show that $g$ is not continuous except possibly on a set of measure $0$.

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What have you tried? Where are you stuck? –  Aryabhata Feb 6 '11 at 18:53
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Hint: If $g$ were continuous on a set of positive measure, then it would be continuous on a compact set with positive measure. Think about what compact sets with positive measures in $\mathbb{R}^1$ need to look like, and show that this can't be true. –  user1736 Feb 6 '11 at 19:42
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It is very rude to post questions in the imperative mode (giving orders/assignments). You are not giving orders or assigning homework, you are asking questions. So please phrase your questions as questions, specifying what you have tried and/or where you are stuck. If this happens to be homework, consider tagging it with [homework]. –  Arturo Magidin Feb 6 '11 at 19:58
    
Hint: Show the function is unbounded on any interval. –  Zarrax Feb 6 '11 at 20:49
    
@Zarrax You can post your comment as an answer. –  Davide Giraudo Mar 14 '13 at 9:25

1 Answer 1

As Zarrax noted, the function $$g(x)=\sum_1^\infty \frac{1}{2^n} \frac{1}{\sqrt{|x-r_n|}} \chi_{(0,1]}$$ is unbounded on every open interval, because any such interval contains some $r_n$.

On the other hand, $g$ is integrable, because the integral of the $n$th term is bounded by a multiple of $2^{-n}$. Therefore, $g$ is finite almost everywhere.

Conclusion: $g$ is discontinuous at every point where it's finite, which is almost everywhere. (It is continuous in the extended sense at each $r_n$, since both the limit and the value are $+\infty$.)

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