5
$\begingroup$

I'm trying to show that measurable function with compact support are integrable. Can I do as following : Let $f$ with compact support. Then there is $K$ compact s.t. $supp(f)\subset K$. Then, there is $M$ s.t. $|f|<M$ (because the support is compact) and thus $$\int |f|<M\int_K=Mm(K)<\infty $$ where $m$ is Lebesgue measure. Does it work ?

$\endgroup$
4
  • 2
    $\begingroup$ Your proof works for bounded measurable functions with compact support (in particular for continuous functions with compact support). $\endgroup$
    – saz
    Sep 18 '16 at 11:12
  • 2
    $\begingroup$ If you replace “measurable” with “continuous” then, assuming you're talking about Lebesgue measure on $\mathbb{R}^n$, the bound $|f|<M$ can be stated. Just “measurable” is not sufficient. $\endgroup$
    – egreg
    Sep 18 '16 at 11:12
  • 1
    $\begingroup$ @Surb Mark's function is definitely not continuous. $\endgroup$
    – egreg
    Sep 18 '16 at 11:16
  • $\begingroup$ Bounded measurable functions with compact support are integrable, and the proof is as you wrote. On the other hand, unbounded measurable functions may not be integrable. $\endgroup$
    – Ramiro
    Sep 18 '16 at 22:15
10
$\begingroup$

It does not need to integrable eg.

$$ f(x) = 1/|x|, $$ for $0 < |x| < 1 $ and 0 otherwise is not $L^1$.

$\endgroup$
1
  • $\begingroup$ its support is $[-1,1]$. For a point not to be in the support, the function has to be zero near it. So the support is always closed. $\endgroup$
    – Mark Joshi
    Sep 18 '16 at 11:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.