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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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    @Zarrax You can post your comment as an answer.2013-03-14

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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 $+\infty$.)