I have $A \subset E^n$, which has positive volume. I also have $f$ which is a positive-valued function on $A$ such that $\int_{A}f$ exists. I want to show that $\int_{A}f > 0$.
I have been stuck with this for a few hours, I feel like I need to use compactness somehow, but I do not see how it is applicable. Any help would be great!
On
let $A_n$ be the subset of $A$ where $f > 1/n.$ Now suppose So $$ \int f \geq \int f I_{A_n} $$
Now suppose $$ \int f I_{A_n} =0 $$ for all $n.$
Let $g_n = fI_{A_n}$ it follows from the monotone convergence theorem that $$ \int g_n \to \int f $$ so $\int f=0$ if and only if $\int f I_{A_n}=0$ for all $n.$
But $\int f I_{A_n} =0$ implies that $f \leq 1/n$ everywhere which cannot be true for all $n.$
Since $f$ is strictly positive there exists a $t>0$ and a subset $U \subset A$ such that $f > t$ on $U$. Since $f$ is positive and $A$ contains $U$, the integral over $A$ should be bigger than that of $f$ over $U$ (try to prove this for yourself if you haven't seen this). Hence, $$\int_A f d\mu \geq \int_U f d\mu > \int_U t d\mu = t \int_U d\mu > 0$$