7
$\begingroup$

Suppose we have an uncountable family of functions $f_r: [0, 1] \to R$ indexed by $r \in [0, 1]$ such that for each $r$, there exists a unique $x$ in $[0, 1]$ such that $f_{r}$ is positive on $x$ and $0$ elsewhere.

Define the pointwise sum function $S[a, b]: [0, 1] \to R$ as $S[a, b] (x) = \sum_{r \in [a, b]} f_r (x)$.

If $S[0, 1]$ is well defined, then so is $S[a, b]$ for any $a, b \in R$.

Suppose that $S[0, 1]$ is well defined and that for every $x \in [0, 1]$, the set $\{r \in [0, 1]: f_r (x) > 0\}$ is dense in $[0, 1]$. Is it true that for a.e. $r \in [0, 1]$, the function $S[0, r]$ is discontinuous a.e.?

$\endgroup$
8
  • $\begingroup$ What do you mean by positive on a single value? And do you ask for the discontinuity of $S[0,a]$ for almost every $a\in[0,1]$? $\endgroup$ Feb 21, 2019 at 8:24
  • $\begingroup$ Sorry, I will clarify. And yes. $\endgroup$ Feb 21, 2019 at 8:25
  • $\begingroup$ What do you mean by $S[a,b]$ is well defined? Is the sum always finite? Since each $f_r$ is not zero and not negative, the sum always exists. $\endgroup$ Feb 21, 2019 at 10:03
  • $\begingroup$ Yes the sum is always finite. $\endgroup$ Feb 21, 2019 at 10:04
  • $\begingroup$ Perhaps I'm not understanding correctly, but: if for one $x$ the set is dense, in particular it is infinite, and the sum $S[0,1](x)$ is $+\infty$, hence not defined? $\endgroup$
    – efs
    Feb 24, 2019 at 23:15

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.