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 $\int fd\mu$.


1 Answer 1


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 $+\infty$.

  • $\begingroup$ 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 $\mu(A_t) \leq \frac{1}{t} \int (f\chi_{A_t})d\mu \leq \int fd\mu$. $\endgroup$
    – Beerus
    Apr 20 at 3:16
  • 1
    $\begingroup$ @Beerus The Lebesgue integral is always well-defined for nonnegative function (even it isn’t integrable). $\endgroup$
    – rfloc
    Apr 20 at 12:53
  • $\begingroup$ @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 $\sup\mathbb R = +\infty$. $\endgroup$
    – Alex Ortiz
    Apr 20 at 13:26

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