# Question About Function Integrability - Proposition 2.3.10 from Measury Theory by Donald Cohn

I am self-studying measure theory using Measure Theory by Donald Cohn. The book makes the following definition:

Definition$$\quad$$ Suppose that $$f:X\to[-\infty,+\infty]$$ is $$\mathscr{A}$$-measurable and that $$A\in\mathscr{A}$$. Then $$f$$ is integrable over $$A$$ if the function $$f\chi_A$$ is integrable, and in this case $$\int_Afd\mu$$, the integral of $$f$$ over $$A$$, is defined to be $$\int f\chi_Ad\mu$$.

I am confused by the proof the following proposition:

Proposition$$\quad$$2.3.10$$\quad$$ Let $$(X,\mathscr{A},\mu)$$ be a measure space, and let $$f$$ be a $$[0,+\infty]$$-valued $$\mathscr{A}$$-measurable function on $$X$$. If $$t$$ is a positive real number and if $$A_t$$ is defined by $$A_t=\{x\in X:f(x)\geq t\}$$, then \begin{align*} \mu(A_t)\leq\frac{1}{t}\int_{A_t}fd\mu\leq\frac{1}{t}\int fd\mu. \end{align*}

Proof$$\quad$$ The relation $$0\leq t\chi_{A_t}\leq f\chi_{A_t}\leq f$$ and part (c) of Proposition 2.3.4 imply that \begin{align*} \int t\chi_{A_t}d\mu \leq \int_{A_t}fd\mu\leq\int fd\mu. \end{align*} Since $$\int t\chi_{A_t} = t\mu(A_t)$$, the proposition follows.

My question with this proof is what if $$f\chi_{A_t}$$ is not integrable? According to the definition, $$\int f\chi_{A_t}d\mu = \int_{A_t}fd\mu$$ is only defined when $$f\chi_{A_t}$$ integrable. However, in this proposition, $$f$$ is a nonnegative extended real-valued function, which means it can have value $$+\infty$$ at some point in $$X$$. Then \begin{align*} \begin{split} &\int(f\chi_{A_t})^+d\mu\\ =\ &\int(f\chi_{A_t})d\mu\\ =\ &\sup\left\{\int gd\mu:g\ \text{is a nonnegative simple real-valued \mathscr{A}-measurable function on X}\\ \text{and g\leq f\chi_{A_t}}\right\}\\ =\ &\sup\left\{\sum_{i=1}^mb_i\mu(B_i):\text{g is \mathscr{A}-measurable; g=\sum_{i=1}^mb_i\chi_{B_i}\leq f\chi_{A_t};}\\ \text{and for all i b_i\in[0,+\infty), B_i's are disjoint subsets of X that belong to \mathscr{A}}\right\} \end{split} \end{align*} may have value $$+\infty$$, and so that $$f\chi_{A_t}$$ is not integrable, and so $$\int_{A_t}fd\mu$$ is not defined.

Am I missing anything? Or is the book wrong? I really appreciate any help!

Background Information: Construction of the Integral

Stage 1$$\quad$$ We begin with the simple function. Let $$(X,\mathscr{A})$$ be a measurable space. We will denote by $$\mathscr{S}$$ the collection of all simple real-valued $$\mathscr{A}$$-measurable functions on $$X$$ and by $$\mathscr{S}_+$$ the collection of nonnegative functions in $$\mathscr{S}$$.

Let $$\mu$$ be a measure on $$(X,\mathscr{A})$$. If $$f$$ belongs to $$\mathscr{S}_+$$ and is given by $$f = \sum_{i=1}^ma_i\chi_{A_i}$$ where $$a_1,\dots,a_m$$ are nonnegative real numbers and $$A_1,\dots,A_m$$ are disjoint subsets of $$X$$ that belong to $$\mathscr{A}$$, then $$\int f d\mu$$, the integral of $$f$$ with respect to $$\mu$$, is defined to be $$\sum_{i=1}^ma_i\mu(A_i)$$ (note that this sum is either a nonnegative real number or $$+\infty$$).

Stage 2$$\quad$$ As our next step, we define the integral of an arbitrary $$[0,+\infty]$$-valued $$\mathscr{A}$$-measurable function on $$X$$. For such a function $$f$$, let \begin{align*} \int fd\mu = \sup\left\{\int gd\mu:g\in\mathscr{S}_+\ \text{and}\ g\leq f\right\}. \end{align*}

Stage 3$$\quad$$ Finally, let $$f$$ be an arbitrary $$[-\infty,+\infty]$$-valued $$\mathscr{A}$$-measurable function on $$X$$. If $$\int f^+d\mu$$ and $$\int f^-d\mu$$ are both finite, then $$f$$ is called integrable (or $$\mu$$-integrable or summable), and its integral $$\int fd\mu$$ is defined by \begin{align*} \int fd\mu = \int f^+d\mu - \int f^-d\mu. \end{align*} The integral of $$f$$ is said to exist if at least one of $$\int f^+d\mu$$ and $$\int f^-d\mu$$ is finite, and again in this case, $$\int fd\mu$$ is defined to be $$\int f^+d\mu - \int f^-d\mu$$. In either case one sometimes writes $$\int f(x)\mu(dx)$$ or $$\int f(x)d\mu(x)$$ in place of $$\int fd\mu$$.

If $$f\chi_{A_t}$$ is not integrable, then $$\mu(A_t) \le \frac{1}{t}\int f\chi_{A_t}$$ holds for the reason that the right-hand side is $$+\infty$$.
• So shall I say the proposition is not stated rigorously? Because it should have stated as follows: If $f\chi_{A_t}$ is integrable, we have the relation in the original proposition. If $f\chi_{A_t}$ is not integrable, we will have instead $\mu(A_t) \leq \frac{1}{t} \int (f\chi_{A_t})d\mu \leq \int fd\mu$. Apr 20 at 3:16
• @Beerus, The proposition is stated rigorously. We just have to keep in mind that if $f\chi_{A_t}$ is not integrable, then we assign the value $+\infty$ to the integral $\int f\chi_{A_t}$. In particular, the integral is well defined as an extended real number. This is analogous to $\sup\mathbb R = +\infty$. Apr 20 at 13:26