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Let $\{q_k:\ k \in \mathbb N^+\}$ be an enumeration of the rational numbers in $[0,\ 1)$ and let $(a_k)$ be a sequence of strictly positive real numbers such that

$$ \sum_{k=1}^\infty a_k = 1 $$

Denote $S(x) = \{k \in \mathbb N^+:\ q_k \in [0,\ x)\}$ i.e. indices of rational numbers in $[0,\ x)$ and define $f:[0,\ 1] \rightarrow \mathbb R$ by

$$ f(x) = \begin{cases} \sum_{k \in S(x)} a_k & (x > 0) \\ 0 & (x = 0) \end{cases} $$

I know that $f(x)$ is discontinuous at every positive rational number. I speculate that $f(x)$ is continuous at irrational numbers, but I have no idea how to prove or disprove it.

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You can find an answer in this thread. Although the situation is more special, because there $a_k = 2^{-k}$ is chosen, the proof doesn't change. Let $x \in \mathbb{R}$ be irrational, and let $N \in \mathbb{N}$ with $\sum_{k=N1}^\infty a_n < \varepsilon/2$. Now define $$\delta:= \min_{i=1,\ldots,N}|q_i-x|.$$ For any $|y-x| < \delta$ we have $q_k \in [0,x)$ if and only if $q_k \in [0,y)$ for all $k=1,\ldots,N$. Thus $$S(x) \cap [1,N] = S(y) \cap [1,N]$$ and therefore $$|f(x)-f(y)| \le 2 \sum_{k=N+1}^\infty a_k<\varepsilon.$$

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