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 ?

It does not need to integrable eg. 
$$
f(x) = 1/|x|,
$$
for $0 < |x| < 1 $ and 0 otherwise is not $L^1$.
