defining integrals with partition of unity

Question

I have a question about the following bit of text: They say that $\int_A\phi\cdot\vert f\vert$ exists for each $\phi\in\Phi$. I don't see why this is the case. Since $\Phi$ is a subordinate cover, we know that for each $\phi$, there is a $U$ in $\mathcal O$, such that $\phi$ is zero outside some closed set $V$ contained in $U$. This means that our integral is reduced to $$ \int_A\phi\cdot\vert f\vert=\int_V\phi\cdot\vert f\vert. $$ However, they only say that $f$ is bounded in some open set, say $B$. Let's assume $B\subset V$. In that case, our integral still might not have a bounded integrand, so we might still have a problem. Could someone clarify why the integral exists?

Answer 1

But as a part of partition of unity, $\phi\in C^{0}$ on the compact set $V$, so $\phi$ is bounded, then so is $\phi\cdot|f|$.

The whole story is like this: First choose compact set such that $\phi$ is supported in $V$. Now choose a finite subcover $\{G_{i}\}$ of $V$ such that $G_{i}\subseteq A$. Now by assumption $f$ is bounded on each $G_{i}$, and all such $G_{i}$ are finitely many, then a common upper bound is then obtained.