I am reading "Calculus on Manifolds" by Michael Spivak.
An open cover $\mathcal{O}$ of an open set $A\subset\mathbb{R}^n$ is admissible if each $U\in\mathcal{O}$ is contained in $A$. If $\Phi$ is subordinate to $\mathcal{O}$, $f:A\to\mathbb{R}$ is bounded in some open set around each point of $A$, $\{x:f\text{ is discontinuous at }x\}$ has measure $0$, then each $\int_A \varphi\cdot |f|$ exists.
The author wrote $\int_A \varphi\cdot |f|$ exists without a proof.
So I forced to check $\int_A \varphi\cdot |f|$ exists.
At first, it was necessary to consider why the above conditions are required.
One of the above conditions is "$f:A\to\mathbb{R}$ is bounded in some open set around each point of $A$".
I am new to such a property of functions.
So, I want you to check if I am using the condition "$f:A\to\mathbb{R}$ is bounded in some open set around each point of $A$" correctly.
I used the following theorems to check $\int_A \varphi\cdot |f|$ exists.
3-8 Theorem.
Let $A$ be a closed rectangle and $f:A\to\mathbb{R}$ a bounded function. Let $B=\{x:f\text{ is not continuous at }x\}$. Then $f$ is integrable if and only if $B$ is a set of measure $0$.
3-11 Theorem.
Let $A\subset \Bbb R^n$ and let $\mathcal{O}$ be an open cover of $A$. Then there is a collection $\Phi$ of $C^\infty$ functions $\varphi$ defined in an open set containing $A$, with the following properties:(1). For each $x \in A$ we have $0 \leq \varphi(x) \leq 1$.
(2). For each $x \in A$ there is an open set $V$ containing $x$ such that all but finitely many $\varphi \in \Phi$ are $0$ on $V$.
(3). For each $x \in A$ we have $\sum_{\varphi \in \Phi}\varphi(x)=1$ (by (2) for each $x$ their sum is finite in some open set containing $x$).
(4). For each $\varphi \in \Phi$ there is an open set $U$ in $\mathcal{O}$ such that $\varphi = 0$ outside of some closed set contained in $U$.
(A collection $\Phi$ satisfying (1) to (3) is called a $C^\infty$ partiion of unity for $A$. If $\Phi$ also satisfies (4), it is said to be subordinate to the cover $\mathcal{O}$. In this chapter we will only use continuity of the functions $\varphi$.)
Even if we change "closed" in (4) with "compact", Theorem 3-11 still holds.
About this, please see https://math.stackexchange.com/a/243671/1226161.
Let $\varphi\in\Phi$.
Then by (4), there is an open set $U\in\mathcal{O}$ such that $\varphi=0$ outside of some compact set $C$ contained in $U$.
Since $\mathcal{O}$ is admissible, $C\subset U\subset A$.
Since $\varphi$ is continuous, $\varphi(x)=0$ for $x$ on the boundary of $C$.
Let $a$ be an any point on the boundary of $C$.
For any positive $\varepsilon$, there exists a neighborhood $V_1$ of $a$ such that $|\varphi(x)|=|\varphi(x)-\varphi(a)|<\varepsilon$ for any $x\in V_1$ since $\varphi$ is continuous at $a$ and $\varphi(a)=0$.
For any positive $\varepsilon$, there exists a neighborhood $V_2$ of $a$ and $M$ such that $||f|(x)|<M$ for any $x\in V_2$ since $|f|$ is bounded in some open set around each point of $A$.
Then, $|\varphi\cdot |f|(x)-\varphi\cdot |f|(a)|=|\varphi(x) |f|(x)|=|\varphi(x)|\cdot |f(x)|<\varepsilon\cdot M$ for any $x$ in some neighborhoood of $a$.
So, $\varphi\cdot |f|$ is continuous at each $a$ on the boundary of $C$.
Since $\varphi\cdot |f|(x) = 0$ if $x\in A-C$, so $\int_A \varphi\cdot |f|=\int_C \varphi\cdot |f|$.
Let $B$ be a closed rectangle such that $C\subset B$.
Let $g:B\to\mathbb{R}$ be a function such that $g(x) = \varphi\cdot |f|(x)$ if $x\in C$ and $g(x) = 0$ if $x\in B-C$.
Then, $g$ is continuous at each point $a$ in $B-C$ and $g$ is continuous at each point $a$ on the boundary of $C$ and $\{x\in C:\varphi\cdot |f|\text{ is discontinuous at }x\}\subset \{x\in A:f\text{ is discontinuous at }x\}$ has measure $0$.
So, $\int_C \varphi\cdot |f|$ exists.
So, $\int_A \varphi\cdot |f|$ exists.
I wrote $\int_A \varphi\cdot |f|$ as if I knew what $\int_A \varphi\cdot |f|$ means.
But we don't know what $\int_A \varphi\cdot |f|$ means if $A$ is not bounded.
So, I want to defne as follows:
An open cover $\mathcal{O}$ of an open set $A\subset\mathbb{R}^n$ is admissible if each $U\in\mathcal{O}$ is contained in $A$. Let $\mathcal{O}$ be admissible. Let $\Phi$ be subordinate to $\mathcal{O}$. By (4) in Theorem 3-11, there is an open set $U\in\mathcal{O}$ such that $\varphi=0$ outside of some compact set $C_{\varphi}$ contained in $U$. Since $\mathcal{O}$ is admissible, $C_{\varphi}\subset U\subset A$. If $f:A\to\mathbb{R}$ is bounded in some open set around each point of $A$, $\{x:f\text{ is discontinuous at }x\}$ has measure $0$, then each $\int_{C_{\varphi}} \varphi\cdot |f|$ exists. The value of $\int_{C_{\varphi}} \varphi\cdot |f|$ does not depend on the choice of $C_{\varphi}$. We define $\int_A \varphi\cdot |f|:=\int_{C_{\varphi}} \varphi\cdot |f|$.