1

I was reading over this post and its responses, but I see that the assumption is always that $f$ itself is integrable or $\mu$-integrable. So I was thinking suppose we have some Borel measure $\mu$ on $[0, \infty)$ and a continuous function $f:[0, \infty) \to \mathbb{R} $. When is the map $$x \mapsto \int_{(0, x)} f(s) d \mu$$ continuous?

I think the question is easier when we are integrating over a closed interval since that would imply $f$ is also bounded. Without it I cannot be sure that $f$ is $\mu$-integrable.

It obviously also depends on $\mu$. For example if we take the dirac delta measure $\delta_a$, the map is not continuous at $a$, unless $f$ happens to be $0$ at $a$. So maybe a continuous $\mu$ is probably also required?

CharlieCornell
  • 482
  • 1
    If $f$ is continuous on $[0,\infty)$, then isn't it continuous on $[0,x]$ for any $x$? – Ted Shifrin Jan 25 '22 at 19:58
  • Oh yes, that is totally my mistake for missing something so obvious. My question remains on the measure part though, there ought to be restrictions on the measures (especially discrete ones) that would make it not continuous? – CharlieCornell Jan 25 '22 at 20:04
  • 1
    My days of measure theory are way behind me, but I think you are basically right. It doesn't make sense to talk about continuity of a measure, but I think that you want $\mu$ to be absolutely continuous with respect to standard Lebesgue measure on $\Bbb R$. – Ted Shifrin Jan 25 '22 at 20:06
  • 1
    Hint 1. Show that $F(x)=\int_{(0,x)} f,d\mu$ is continuous on the left. Hint 2. Consider $\int_{{x_0}} f,d\mu$ for a few moments. Hint 3. You don't need continuity for anything other than integrability. Hint 4. Naah. That's enough. – B. S. Thomson Jan 25 '22 at 20:25
  • Thank you for the hints! It took me a while to write everything out rigorously but I followed them and arrived at the conclusion that we just need $\mu$ to assign measure $0$ to countably large sets. Is that correct? – CharlieCornell Jan 25 '22 at 23:20
  • 1
    @CharlieCornell My guess (without checking details) is that (i) assume $\mu$ is a finite Borel measure on $(0,\infty)$. (ii) let $S={ x: \mu({x})>0}$. (iii) conclude that a necessary and sufficient condition for $F(x)=\int_{(o,x)} f,d\mu$ to be everywhere continuous is that $f$ is zero at each point of the countable set $S$. – B. S. Thomson Jan 26 '22 at 00:44
  • 1
    P.S. Your version ("we just need μ to assign measure 0 to countable sets") would be correct but a weaker answer. – B. S. Thomson Jan 26 '22 at 01:14
  • "It doesn't make sense to talk about continuity of a measure" it is classical to identify (or even sometimes define ... see e.g. Bourbaki) measures with distributions of order $1$. It is then quite meaningful to talk about a measure being continuous ... but then it indeed means that it is absolutely continuous with respect to the Lebesgue measures and its Radon-Nikodim "derivative" is continuous. – LL 3.14 Jan 26 '22 at 06:00

0 Answers0