# defining integrals with partition of unity

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?

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.
• I am not familiar with $L^\infty$. Also, $\phi$ is $C^\infty$? – Sha Vuklia Jan 23 '18 at 9:07
• Yes, in particular, $\phi$ is continuous, this is what $C^{0}$ meant. $L^{\infty}$ in some sense means bounded. – user284331 Jan 23 '18 at 9:09
• So it's impossible to provide an argument without $L^\infty$? – Sha Vuklia Jan 23 '18 at 9:13
• Okay, like this: Choose $V$ as a compact set, then the continuous function $\phi$ on the compact set must be bounded, then the multiplication $\phi\cdot|f|$ is bounded too. – user284331 Jan 23 '18 at 9:14
• First, choose $V$ to be compact such that $\phi$ is supported in $V$, then find a finite subcover $\{G_{i}\}$ of $K$ and use the fact that $f$ is bounded on each $G_{i}$, this subcover is finite, the common bound can be obtained. – user284331 Jan 23 '18 at 9:19