Exercise 1.7 (Linearity of the unsigned integral)

Question

Let $\left(\Omega,\mathcal{F},\mu\right)$ be a measure space.
\begin{enumerate}
	\item Let ${f: \Omega \rightarrow [0,+\infty]}$ be an unsigned measurable function which is both bounded (i.e., there is a finite $M$ such that ${|f(\omega)| \leq M}$ for all ${\omega \in \Omega}$) and has finite measure support (i.e., there is a measurable set ${E}$ with ${\mu(E) < \infty}$ such that ${f(\omega)=0}$ for all ${\omega \in \Omega \backslash E}$). Show that
	$$\displaystyle \int_\Omega f\ d\mu {=} \inf_{g \geq f} \hbox{Simp} \int_\Omega g\ d\mu$$
	holds for this function ${f}$.
		\item Establish the additivity property
    $$\displaystyle \int_\Omega f+g\ d\mu =\int_\Omega f\ d\mu + \int_\Omega g\ d\mu$$
    whenever $f, g: \Omega \rightarrow [0,+\infty]$ are unsigned measurable functions that are bounded with finite measure support.
    \item (Horizontal truncation) Show that
    $$\displaystyle \int_\Omega \min(f,n)\ d\mu \rightarrow \int_\Omega f\ d\mu$$
    as ${n \rightarrow \infty}$ whenever ${f: \Omega \rightarrow [0,+\infty]}$ is unsigned measurable.
    \item Using (iii), extend (ii) to the case where {f,g} are unsigned measurable functions with finite measure support, but are not necessarily bounded.
    \item (Vertical truncation) Show that
    $$\displaystyle \int_\Omega f 1_{f \geq 1/n}\ d\mu \rightarrow \int_\Omega f\ d\mu$$
    as ${n \rightarrow \infty}$ whenever ${f: \Omega \rightarrow [0,+\infty]}$ is unsigned measurable.
    \item Using (iii) and (v), show that (ii) holds for any unsigned measurable ${f,g}$ (which are not necessarily bounded or of finite measure support).
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}
	\item We have to demonstrate that
	$$\sup_{h \leq f} \left\{\hbox{Simp} \int_\Omega h\ d\muight\} {=} \inf_{g \geq f} \left\{\hbox{Simp} \int_\Omega g\ d\muight\}$$
		We already know that
		$$\sup_{h \leq f} \left\{\hbox{Simp} \int_\Omega h\ d\muight\} \leq \inf_{g \geq f} \left\{\hbox{Simp} \int_\Omega g\ d\muight\}$$
		So it is left to demonstrate that
		$$\sup_{h \leq f} \left\{\hbox{Simp} \int_\Omega h\ d\muight\} 
ot > \inf_{g \geq f} \left\{\hbox{Simp} \int_\Omega g\ d\muight\}$$
	Pick an $\epsilon > 0$, and let $f_\epsilon$ be $f$ rounded down to the nearest integer multiple of $\epsilon$, and let $f^\epsilon$ be $f$ rounded up to the nearest integer multiple.
	$$f_\epsilon := \max\{k\epsilon: k\in\mathbb{N}	ext{ and }k\epsilon < f\} \qquad 
		f^\epsilon := \min\{k\epsilon: k\in\mathbb{N}	ext{ and }k\epsilon \geq f\}$$
	Clearly, for all $\omega \in \Omega$ we have the pointwise bounds
	$$f_\epsilon(\omega) \leq f(\omega) \leq f^\epsilon(\omega)$$
	and
	$$f^\epsilon(\omega) - f_\epsilon(\omega) \leq \epsilon.$$
	Finally, $f$ is bounded, which means that $f_\epsilon,f^\epsilon$ can only take on finitely many values in the range of $0$,$\epsilon$, $2\epsilon$,$\ldots$, $K\epsilon \leq M \leq (K+1)\epsilon$. To sum up, for any $\epsilon>0$ we can find an $\hbox{Simp}\int_\Omega f_\epsilon$ from the left-hand side set and an $\hbox{Simp}\int_\Omega f^\epsilon$ from the right-hand side set such that the difference between them is less than an arbitrarily small quantity $\epsilon \cdot \mu(E)$. Thus, the right-hand side cannot be strictly greater than the left-hand side, and so they must be equal.
	
	\item In view of superadditivity (href{./exercise_1-5}{Exercise 1.5 (i)}) we only have to demonstrate that
		$$\int_\Omega f+g\ d\mu \leq \int_\Omega f\ d\mu + \int_\Omega g\ d\mu$$
		Just as in the previous part, define $f_\epsilon,f^\epsilon$ and $g_\epsilon,g^\epsilon$. We give ourselves an $\epsilon$ of a room and see that
		$$f+g \leq f_\epsilon + g_\epsilon + 2\epsilon$$
		from $f^\epsilon(\omega) - f_\epsilon(\omega) \leq \epsilon$ and $g^\epsilon(\omega) - g_\epsilon(\omega) \leq \epsilon.$ By \href{./exercise_1-5}{Exercise 1.5} and properties of the simple integral,
		\begin{align*}
		\int_\Omega f+g\ d\mu &\leq \int_\Omega f_\epsilon + g_\epsilon + 2\epsilon \ \mathrm{d}\mu\
		&= \hbox{Simp}\int_\Omega f_\epsilon + g_\epsilon + 2\epsilon \ \mathrm{d}\mu\
		&= \hbox{Simp}\int_\Omega f_\epsilon \ \mathrm{d}\mu + \hbox{Simp}\int_\Omega g_\epsilon \ \mathrm{d}\mu+ 2\epsilon\mu(E)\
		&= \left[\int_\Omega f \ \mathrm{d}\mu - \epsilonight] + \left[\int_\Omega g \ \mathrm{d}\mu - \epsilonight]+ 2\epsilon\mu(E)
		\end{align*}
		where we have used the supremum definition of the unsigned integral in the last line. Letting $\epsilon 	o 0$ and using the assumption that $\mu(E)$ is finite, we obtain the claim.
	\item See the equivalent \href{../an_introduction_to_measure_theory/exercise_1-4-35}{Exercise 1.4.35 (ix)} from \href{..}{Terence Tao}'s \href{../an_introduction_to_measure_theory}{An Introduction to Measure Theory}.
	\item Now we continue to assume that $\mu$ is a finite measure, but now we do not assume that $f,g$ are bounded. Then for any natural number $n$, we can use the previous case to deduce that
		$$\int_\Omega \min(f,n) + \min(g,n) \ \mathrm{d}\mu \leq \int_\Omega \min(f,n) \ \mathrm{d}\mu + \int_\Omega\min(g,n) \ \mathrm{d}\mu.$$
		Furthermore, since $\min(f+g,n) \leq \min(f,n) + \min(g,n)$, we conclude that
		$$\int_\Omega \min(f+g,n) \ \mathrm{d}\mu \leq \int_\Omega \min(f,n) \ \mathrm{d}\mu + \int_\Omega\min(g,n) \ \mathrm{d}\mu.$$
		Taking limits as $n	o\infty$ using horizontal truncation, we obtain the claim.
	
	\item See the equivalent \href{../an_introduction_to_measure_theory/exercise_1-4-35}{Exercise 1.4.35 (x)} from \href{..}{Terence Tao}'s \href{../an_introduction_to_measure_theory}{An Introduction to Measure Theory}.
		\item Finally, we no longer assume that $\mu$ is of finite measure, and also do not require $f,g$ to be bounded. If either $\int f$ or $\int g$ is infinite, then by monotonicity, $\int f + g$ is infinite as well, and the claim follows; so we may assume that $\int f$ and $\int g$ are both finite. By \href{./exercise_1-5}{Markov's inequality (Exercise 1.5(v))}, we conclude that for each natural number $n\in\mathbb{N}$, the set $E:=\{\omega \in \Omega: f(\omega) > \frac{1}{n}\} \cup \{\omega \in \Omega: g(\omega) > \frac{1}{n}\}$ is has finite measure. These sets are increasing in $n$, and $f,g,f+g$ are supported on $\bigcup_{n\in\mathbb{N}}E_n$, and so by vertical truncation
	$$\int_\Omega (f+g)\ d\mu = \lim_{n	o\infty}\int_\Omega (f+g) \cdot 1_{E_n}\ d\mu.$$
	From the previous case, we have
	$$\int_\Omega (f+g) \cdot 1_{E_n}\ d\mu \leq \int_\Omega f 1_{E_n}\ d\mu + \int_\Omega g 1_{E_n}\ d\mu$$
	Letting $n	o\infty$ and using horizontal truncation we obtain the claim.
\end{enumerate}

Exercise 1.7 (Linearity of the unsigned integral)

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Exercise 1.7 (Linearity of the unsigned integral)

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