A conjecture on writing a function as a sum of uncountably many points 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: 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 :-)
