# Did Spivak leave out Jordan-measurability too in his definition of partition of unity?

This is a continuation of these two questions that are asking the same thing as each other:

An application of partitions of unity: integrating over open sets.

Is this definition missing some assumptions?

In the answers to these two questions it was concluded that in item (4) of Spivak's definition of a partition of unity, he should have required that the supports of the functions $\varphi \in \Phi$ be compact, not merely closed. This is so the integral $\int \varphi\cdot |f|$ would be defined for locally bounded and almost everywhere continuous $f$. However, it seems that the support of $\varphi$ must also be Jordan-measurable for this integral to be defined. So am I right in thinking that Spivak should have required in item (4) that the supports of the functions $\varphi \in \Phi$ be both compact and Jordan-measurable, or is Jordan-measurability unnecessary or implied by other conditions?

• Why does it seem "that the support of $\phi$ ... to be defined"? $\;$ – user57159 Apr 27 '15 at 3:17
• It seemed that way because there is a theorem that the integral of a bounded almost everywhere continuous function on a compact Jordan-measurable set is defined. It doesn't seem that way anymore though, I see how I was thinking wrong. – Joshua Meyers Apr 27 '15 at 3:30
• I just answered my own question. – Joshua Meyers Apr 27 '15 at 3:38

No, stipulating that the supports of the functions $\varphi\in\Phi$ be Jordan-measurable is not needed. It is sufficient for the existence of the integral $\int \phi\cdot |f|$ that $f$ be locally bounded and almost everywhere continuous and the support of $\varphi$ be compact. This is because $\varphi(x)=0$ at the boundary of $\textrm{Supp}\,\varphi$, so there would be no discontinuities there.
Here is the argument in more detail: Let $S=\textrm{Supp}\,\varphi$. Then for all $x\in S$, there exists a neighborhood $N_x$ of $x$ such that $f|_{N_x}$ is bounded. Since $\{N_x | x\in S \}$ covers $S$, there exists a finite subcover with bounded union $U\supset S$. Then $f$ is bounded on U. The function $\varphi$ is everywhere continuous, so if $\varphi\cdot |f|$ is discontinuous at $x\in U$, then $f$ is discontinuous at $x$. Thus $\varphi\cdot |f|$ is almost everywhere continuous. As $\varphi\cdot |f|$ vanishes outside $S$, the integral $\int_R \varphi\cdot |f|$over a rectangle $R\supset U$ exists.