# Show that function with compact support are integrable.

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 ?

• Your proof works for bounded measurable functions with compact support (in particular for continuous functions with compact support).
– saz
Sep 18 '16 at 11:12
• 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. Sep 18 '16 at 11:12
• @Surb Mark's function is definitely not continuous. Sep 18 '16 at 11:16
• 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. Sep 18 '16 at 22:15

$$f(x) = 1/|x|,$$ for $0 < |x| < 1$ and 0 otherwise is not $L^1$.
• 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. Sep 18 '16 at 11:03