Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know this is sort of a nonissue, but in one of the exercises the author asks us to prove

If $0\leq c<+\infty$, then $\int cf=c\int f$ where $f:\mathbb{R}^{d}\to[0,+\infty]$ (note $f$ is not assumed measurable, just unsigned).

The integral is the "lower unsigned Lebesgue integral" as is usually defined, except that as stated $f$ is any unsigned function.

This is easy enough to show, however, he adds a parenthetical note stating: "(The case $c=+\infty$ unfortunately is not true, but this is somewhat tricky to show.)

I can't think of a counter-example when the convention $\infty\cdot0=0$ is used. Anyone have an idea?

share|cite|improve this question
Just some thoughts: The equality holds when $c=\infty$ and $f$ is measurable, which follows since in this case both sides are zero iff f is essentially zero. – Nikolas Kuhn Apr 23 at 16:16
1) By definition of the integral, the left hand side is zero iff the support of $f$ does not contain any measurable set of positive measure, and $\infty$ else. In particular, if the left hand side vanishes, any simple function smaller than $f$ is essentially zero, so the right hand side vanishes, too. 2) Hence, the only way for $f$ to violate the equality is that the support of $f$ contains a measurable set $A$ of positive measure, so LHS $=\infty$ but for any $\lambda>0$, the set $f^{-1}([\lambda,\infty])$ does not contain any measurable set of positive measure. – Nikolas Kuhn Apr 23 at 16:30
Elaborating on my last comment, I guess for a 'suitable' choice of bijection $\alpha:\mathbb{R}/\mathbb{Q}\to[0,1]$ setting $f$ as the projection $\mathbb{R}\to\mathbb{R}/\mathbb{Q}$ followed by $\alpha$ would work. – Nikolas Kuhn Apr 23 at 16:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.