Exercise 1.4.40 (Basic properties of absolutely convergent integral)

Question

$
ewcommand{\intx}[1]{\int_{X} #1 \mathop{}\!\mathrm{d}{\mu}}$ Let $(X,\mathcal{F},\mu)$ be a measure space, and let $L^1$ denote the space $L^1((X,\mathcal{F},\mu)	o\mathbb{C})$ of all measurable functions from $(X,\mathcal{F})$ to $(\mathbb{C},\mathcal{B})$.
\begin{enumerate}[label=(oman*)]
	\item Show that $L^1$ is a vector space over $\mathbb{C}$.\item Show that the integration map $$\int: L^1	o\mathbb{C}, \qquad f \mapsto \int_{X} f \,\mathrm{d}{\mu}$$ is a linear transformation.\item Establish the triangle inequality
	$$\|f+g\|_{L^1}  \leq \|f\|_{L^1} + \|g\|_{L^1}$$
	and the homogeneity property, for all $f,g\in L^1$ and $c\in\mathbb{C}$:
	$$\|c f \|_{L^1} = |c| \cdot \|f\|_{L^1} $$\item If $\mu$-almost everywhere $f=g$ then $\int_{X} f \,\mathrm{d}{\mu}=\int_{X} g \,\mathrm{d}{\mu}$.
	\item If $(X,\mathcal{F}',\mu')$ is a refinement of $(X,\mathcal{F},\mu)$ then
	$$\int_{X} f \,\mathrm{d}{\mu}' = \int_{X} f \,\mathrm{d}{\mu}$$\item $\|f\|_{L^1} = 0$ if and only if $\mu$-almost everywhere $f=0$.\item If $Y\subseteq X$ is $\mathcal{F}$ measurable, and $f\in L^1$ then $festriction_Y \in L^1$, and $\int_Y f estriction_Y \ \mathrm{d}\mu = \intx{f1_Y}$.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(oman*)]
	\item By Theorem 1.3 from \href{../linear_algebra}{Linear Algebra} it suffices to demonstrate that
	\begin{itemize}
	\item $0_f \in L^1$. A zero function is a simple function with integral zero, and is thus absolutely integrable.\item $f,g\in L^1 \implies f+g \in L^1$. If $\intx{\Re f}$ and $\intx{\Re g}$ are finite quantities, then by Theorem 1.4.37
	$$\intx{\Re(f+g)} = \intx{(\Re f + \Re g)} = \intx{\Re f} + \intx{\Re g}$$
	is finite as well. Since the same is true for the imaginary part, the assertion follows.\item $c\in\mathbb{C}, f\in L^1 \implies cf \in L^1$, and by assumption $\Re c, \Im c, \intx{\Re f}, \intx{\Im f} < \infty$. By the linearity of the integration operator (Theorem 1.4.37 and Exercise 1.4.35 (iii)) we then have        $$\intx{\Re(cf)} = \intx{\Re\left[ (\Re c + i\Im c)(\Re f + i \Im f)ight]} = \intx{\Re c \Re f - \Im c \Im f} = \Re c \intx{\Re f} - \Im c \intx{\Im f} < \infty$$
	\end{itemize}
	
	\item This is a direct concequence of additivity from Theorem 1.4.37 and homogeneity from \href{./exercise_1-4-35}{Exercise 1.4.35}.\item By the pointwise triangle inequality we have $|f+g|\leq|f|+|g|$. Since all of these functions are unsigned, we can use the properties from the \href{./exercise_1-4-35}{Exercise 1.4.35}. By the monotonicity property of the integral, we have $\int_{X} |f+g| \,\mathrm{d}{\mu} \leq \int_{X} |f| \,\mathrm{d}{\mu}+\int_{X} |g| \,\mathrm{d}{\mu}$. The homogeneity property follows from the pointwise identity $|cf| = |c|\cdot |f|$ and by the homogeneity property of the unsigned integral in general.\item Suppose that $\mu$-almost everywhere we have $f = g$. Let $E$ be the null set on which $f
eq g$. \      First, assume that $f,g$ are both real valued. By \href{./exercise_1-4-35}{Exercise 1.4.35} (xi) we split our integrals. Using the analogous property of the unsigned integral from \href{./exercise_1-4-35}{Exercise 1.4.35} (i) we obtain:
	\begin{align*}
	\int_{X} f \,\mathrm{d}{\mu} 
	&= \intx{f^+}-\intx{f^-}\
	& = \int_{X/E}f^+\ \mathrm{d}\mu + \int_{E}f^+\ \mathrm{d}\mu  - \int_{X/E}f^-\ \mathrm{d}\mu  - \int_{E}f^-\ \mathrm{d}\mu \
	& = \int_{X/E}f^+\ \mathrm{d}\mu  - \int_{X/E}f^-\ \mathrm{d}\mu\
	& = \int_{X/E}g^+\ \mathrm{d}\mu  - \int_{X/E}g^-\ \mathrm{d}\mu\
	& = \int_{X/E}g^+\ \mathrm{d}\mu + \int_{E}g^+\ \mathrm{d}\mu  - \int_{X/E}g^-\ \mathrm{d}\mu  - \int_{E}g^-\ \mathrm{d}\mu \
	&= \intx{g^+}-\intx{g^-}\
	&= \int_{X} g \,\mathrm{d}{\mu}
	\end{align*}
	Now assume $f,g$ are real valued. We similiarly obtain
	\begin{align*}
	\int_{X} f \,\mathrm{d}{\mu} 
	&= \intx{\Re f} + i \intx{\Im f}\
	&= \int_{X/E}{\Re f}\ \mathrm{d}\mu + \int_{E}{\Re f}\ \mathrm{d}\mu + i \int_{X/E}{\Im f}\ \mathrm{d}\mu + i \int_{E}{\Im f}\ \mathrm{d}\mu\
	&= \int_{X/E}{\Re f}\ \mathrm{d}\mu + i \int_{X/E}{\Im f}\ \mathrm{d}\mu \
	&= \int_{X/E}{\Re g}\ \mathrm{d}\mu + i \int_{X/E}{\Im g}\ \mathrm{d}\mu \
	&= \int_{X/E}{\Re g}\ \mathrm{d}\mu + \int_{E}{\Re g}\ \mathrm{d}\mu + i \int_{X/E}{\Im g}\ \mathrm{d}\mu + i \int_{E}{\Im g}\ \mathrm{d}\mu\
	&= \intx{\Re g} + i \intx{\Im g} = \int_{X} g \,\mathrm{d}{\mu}
	\end{align*}
	
	\item It is important to make clear that $f$ is measurable with respect to the original $\sigma$-algebra $\mathcal{B}$; in other words, for every open set $U \subseteq \mathbb{C}$ the inverse $f^{-1}(U)$ is $\mathcal{B}$-measurable, even if we are looking at the space $(X,\mathcal{B}',\mu')$.
ewlie
	A trick is to recall the results of \href{./exercise_1-4-35}{Exercise 1.4.35} part (ix) and (x) which says that for an increasing sequence $E_1\subseteq E_2 \subseteq \ldots$ of measurable functions we have
	$$\lim_{n	o\infty} \intx{\min(f,n)	imes 1_{E_n}} = \int_{X} f \,\mathrm{d}{\mu}$$
	This allows us to restrict our attention to horizontally and vertically truncated functions; thus, it suffices to demonstrate the theorem for bounded functions with finite measure support $S \subseteq X$. We have to demonstrate that
	$$\sup \left\{\mathrm{Simp}\int_{X} h \,\mathrm{d}\mu: \begin{array}{l} 	ext{$h$ is simple in $\mathcal{B}$}\ 0 \leq h \leq f \end{array} ight\} = \sup \left\{\mathrm{Simp}\int_{X} g \,\mathrm{d}\mu': \begin{array}{l} 	ext{$g$ is simple in $\mathcal{B}'$}\ 0 \leq g \leq f \end{array} ight\}.$$
	\begin{itemize}
		\item[$\leq$)] Pick an arbitrary $\mathcal{B}$-simple function $h$ minorising $f$. From $\mathcal{B}\subseteq \mathcal{B}'$ we see that $h$ is automatically also a $\mathcal{B}'$-simple function, and so $\mathrm{Simp}\int_{X} h \,\mathrm{d}\mu'$ is contained on the left-hand side.\item[$\geq$)] Pick an arbitrary $\mathcal{B}'$-simple function $g = c_11_{F_1} + \ldots + c_n1_{F_n}$ minorizing $f$. Notice that $F_1\cup\ldots\cup F_n = S$. Let $E_1,\ldots,E_k$ be a partition of $S$. Define
		\begin{align*}
		\hat{c}_1 &= \max\left\{ c_j: 1\leq j \leq n, \ E_1 \cap F_j 
eq \emptysetight\}\
		& \vdots\
		\hat{c}_n &= \max\left\{ c_j: 1\leq j \leq n, \ E_n \cap F_j 
eq \emptysetight\}
		\end{align*}
		and
		$$h := \hat{c}_1 1_{E_1} + \ldots + \hat{c}_n 1_{E_n}$$
		We then have
		\begin{align*}
		\mathrm{Simp}\int_{X} g \,\mathrm{d}\mu &= c_1\mu\left({F_1}ight) + \ldots + c_n\mu\left({F_n}ight) = c_1\mu\left(\bigcup_{i=1}^{n}F_1\cap E_iight) + \ldots + c_n\mu\left(\bigcup_{i=1}^{n}F_n\cap E_iight)\
		&= \sum_{i=1}^{n}c_1m\left(F_1 \cap E_iight) + \ldots + \sum_{i=1}^{n}c_nm\left(F_n \cap E_iight)\
		&= \sum_{j=1}^{n}c_jm\left(F_j \cap E_1ight) + \ldots + \sum_{j=1}^{n}c_jm\left(F_j \cap E_night)\
		&\leq \sum_{j=1}^{n}\hat{c}_jm\left(F_j \cap E_1ight) + \ldots + \sum_{j=1}^{n}\hat{c}_jm\left(F_j \cap E_night)\
		&=\hat{c}_1\mu\left(\bigcup_{j=1}^{n}F_j\cap E_1ight) + \ldots + \hat{c}_n\mu\left(\bigcup_{j=1}^{n}F_j\cap E_night) = \hat{c}_1\mu\left(E_1ight) + \ldots + \hat{c}_n\mu\left( E_night)\
		&= \mathrm{Simp}\int_{X} h \,\mathrm{d}\mu
		\end{align*}
		and we are done.
	\end{itemize}
	
	\item We first prove $(\impliedby)$. First, suppose $f$ is real valued. Since $\int_{X} f \,\mathrm{d}{\mu} = \intx{f^+} - \intx{f^-}$, and the both unsigned terms are zero by the analogous assertion from \href{./exercise_1-4-35}{Exercise 1.4.35} (viii), we have $\int_{X} f \,\mathrm{d}{\mu}=0$. In the case when $f$ is complex valued, breaking down the integral into the real and imaginary parts and using the previous sentence yields the result.
ewline
	Now we demonstrate the converse statement $(\implies)$. First suppose that $f$ is real valued. By non-negativity of the integral, both $\intx{f^+}$ and $f^{-1}$ must be also zero. By \href{./exercise_1-4-35}{Exercise 1.4.35} (viii) both $f^+$ and $f^-$ must be zero $\mu$-almost everywhere. Sum of two $\mu$-almost everywhere zero functions is again a $\mu$-almost everywhere zero function, and we are done. The assertion for the complex valued $f$ follows similarly.
	
	\item As in the previous parts of the exercise, we first break down the real valued integral into positive and negative parts, and then apply the analogous statement for unsigned functions from the \href{./exercise_1-4-35}{Exercise 1.4.35}. Similarly, the assertion follows for complex-valued integrals.
	
\end{enumerate}

Exercise 1.4.40 (Basic properties of absolutely convergent integral)

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Exercise 1.4.40 (Basic properties of absolutely convergent integral)

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