Define the sum of the non-negative numbers $\{r\_s \mid s \in S\}$ $S$ uncountable to be


$$\sup \_{D \subseteq S} \sum \_{d \in D} r\_d$$


($D$ being finite), which exists if this supremum is finite.


Define a point function to be a function from $[0, 1]$ to $\mathbb R$ that is $0$ everywhere except for a single point, where it takes a positive value.


Suppose we have an uncountable family of point functions $f\_r: [0, 1] \to \mathbb R$ indexed by $r \in [0, 1]$. Define the pointwise sum function $S[a, b]: [0, 1] \to \mathbb R$ as


$$S[a, b] (x) = \sum \_{r \in [a, b]} f\_r (x) \ .$$


It can be shown that if $S[0, 1]$ is well defined, then so is $S[0, a]$ for any $a$ such that $0 \le a < 1$.


Assume that if $S[0, 1]$ is well defined. Does it follow that for Lebesgue almost every $a \in [0, 1]$ the function $S[0, a]$ is discontinuous at at least one point?



A conjecture on writing a function as a sum of uncountably many points
OK, here goes as promised. 


Fix $\delta>0$ and consider the set $A\_\delta$ of all points $a$ such that $S[0,a]$ is continuous and $\max f\_a>\delta$ (this maximum is just that exceptional positive point value in your case but we can do arbitrary non-negative not identically $0$ functions). If we could find $a\_n,a\in A\_\delta$ such that $a\_1>a\_2>\dots\to a$, then $S[0,a\_n]$ would be a decreasing sequence of continuous functions converging to the continuous limit $S[0,a]$ pointwise but not uniformly (because $\max(S[0,a\_n]-S[0,a])\ge\max f\_{a\_n}>\delta$). Thus, for every $a\in A\_\delta$, there exists an open interval $(a,b\_a)$ free from points in $A\_\delta$. Since we can place at most countably many disjoint open intervals on $\mathbb R$, $A\_\delta$ is countable for each $\delta>0$. Hence the set of $a$ for which $S[0,a]$ is continuous (which is, say $\cup\_{k\ge 1}A\_{1/k}$) is also countable and, thereby, of Lebesgue measure $0$. 


Just made it community wiki not to collect reputation from homeworks :-)


