Exercise 1.4.33 (Basic properties of the simple integral)

Question

Let $(X,\mathcal{B},\mu)$ be a measure space, and let $f,g:X	o[0,+\infty]$ be simple unsigned functions.
\begin{enumerate}[label=(oman*)]
\item (Monotonicity) If $f \leq g$ then $\mathrm{Simp}\int_{X} f \,\mathrm{d}\mu\leq \mathrm{Simp}\int_{X} g \,\mathrm{d}\mu$.
\item (Compatibility with measure) For every $\mathcal{B}$-measurable set $E$ we have $\mathrm{Simp}\int_{X} 1_E \,\mathrm{d}\mu = \mu(E)$.
\item (Homogeneity) For every $c\in[0,+\infty]$ one has $\mathrm{Simp}\int_{X} cf \,\mathrm{d}\mu = c\mathrm{Simp}\int_{X} f \,\mathrm{d}\mu$.
\item (Finite additivity) We have $\mathrm{Simp}\int_{X} f+g \,\mathrm{d}\mu = \mathrm{Simp}\int_{X} f \,\mathrm{d}\mu+\mathrm{Simp}\int_{X} g \,\mathrm{d}\mu$.
\item (Insensitivity to refinement) Let $(X,\mathcal{B},\mu)$ be a measure space, and let $(X,\mathcal{B}',\mu')$ be its refinement. Then $\mathrm{Simp}\int_{X} f \,\mathrm{d}\mu = \mathrm{Simp}\int_{X} f \,\mathrm{d}\mu'$.
\item (Almost everywhere equivalence) If $\mu$-almost everywhere $f=g$, then $\mathrm{Simp}\int_{X} f \,\mathrm{d}\mu=\mathrm{Simp}\int_{X} g \,\mathrm{d}\mu$
\item (Finiteness) $\mathrm{Simp}\int_{X} f \,\mathrm{d}\mu < \infty$ if and only if $f$ is finite $\mu$-almost everywhere and is supported on a set of finite measure.\item (Vanishing) $\mathrm{Simp}\int_{X} f \,\mathrm{d}\mu = 0$ if and only if $f=0$ $\mu$-almost everywhere.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(oman*)]
	\item 	extit{Monotonicity}
ewline
		Let $a_1,\ldots,a_k$ be the finite values that $f$ takes on, and let $b_1,\ldots,b_{k'}$ be the finite values that $g$ takes on. (Using a rather non-rigorous notation identifying elements and singleton sets) Consider the common refinement
		$$\left\{ f^{-1}(a_i)\cap f^{-1}(b_j): 1\leq i \leq k, 1 \leq j \leq k'ight\}$$
		Since the unions $[0,+\infty]=\bigcup_{i=1}^{k}f^{-1}(a_i) = \bigcup_{j=1}^{k'}f^{-1}(b_j)$ are disjoint, we have
		\begin{align*}
\mathrm{Simp}\int_{X} f \,\mathrm{d}\mu
      &= \sum_{i=1}^{k} a_i \mu\left(f^{-1}(a_i)ight)\
      &= \sum_{i=1}^{k} a_i \mu\left(\bigcup_{j=1}^{k'}f^{-1}(a_i) \cap f^{-1}(b_j) ight)\
      &= \sum_{i=1}^{k} \sum_{j=1}^{k'}a_i \mu\left(f^{-1}(a_i) \cap f^{-1}(b_j) ight)
\end{align*}
      Similarly, we have
      \begin{align*}
\mathrm{Simp}\int_{X} g \,\mathrm{d}\mu
      &= \sum_{j=1}^{k'} b_j \mu\left(f^{-1}(b_j)ight)\
      &= \sum_{j=1}^{k'} b_j \mu\left(\bigcup_{i=1}^{k}f^{-1}(b_j) \cap f^{-1}(a_i) ight)\
      &= \sum_{j=1}^{k'} \sum_{i=1}^{k}b_j \mu\left(f^{-1}(b_j) \cap f^{-1}(a_i) ight)
\end{align*}
      Both sums are finite, and we now can easily compare them by the monotonicity of series.
\begin{align*}
\mathrm{Simp}\int_{X} f \,\mathrm{d}\mu
&= \sum_{i=1}^{k} \sum_{j=1}^{k'}a_i \mu\left(f^{-1}(a_i) \cap f^{-1}(b_j) ight)\
&\leq \sum_{i=1}^{k} \sum_{j=1}^{k'}b_j \mu\left(f^{-1}(a_i) \cap f^{-1}(b_j) ight)\
&= \mathrm{Simp}\int_{X} g \,\mathrm{d}\mu
\end{align*}

\item 	extit{Compatibility with measure}
ewline
		The indicator function is supported on $E$, and equals 1 on it. By \href{./exercise_1-4-32}{Exercise 1.4.32} we have $\mathrm{Simp}\int_{X} 1_E \,\mathrm{d}\mu = 1\cdot \mu(E)$.

\item 	extit{Homogeneity}
ewline
		Suppose that $f$ takes on finitely many values $a_1,\ldots,a_k$, with the corresponding sets $f^{-1}(a_1),\ldots,f^{-1}(a_k)$. Then it is easy to verify that $cf$ also takes on finitely many values $b_1,\ldots,b_k$, and these are exactly $ca_1,\ldots,ca_k$ on the sets $f^{-1}(a_1),\ldots,f^{-1}(a_k)$. (If $a_i$ is a value in the image of $f$, then $f(x)=a_i$, and $(cf)(x)=cf(x)=ca_i$. Conversely, if $b_i$ is some value in the image of $f$, then $b_i = (cf)(x)=cf(x)$, i.e, there is some $a_j$ such that $a_j=b_i/c$.) In other words, the image of $cf$ is $\{ca_1,\ldots,ca_k\}$, and for any $1\leq i \leq k$ we have $(cf)^{-1}(ca_i) = f^{-1}(a_i)$. This implies
		$$\mathrm{Simp}\int_{X} cf \,\mathrm{d}\mu = \sum_{i=1}^{k}ca_i\cdot\mu\left((cf)^{-1}(ca_i)ight) = c\sum_{i=1}^{k}a_i\cdot\mu\left(f^{-1}(a_i)ight) = c\cdot \mathrm{Simp}\int_{X} f \,\mathrm{d}\mu$$

\item 	extit{Finite additivity}
ewline
		Again, we use the common refinement for both $f$ and $g$ as in part (i):
\begin{align*}
\mathrm{Simp}\int_{X} f \,\mathrm{d}\mu &= \sum_{i=1}^{k} \sum_{j=1}^{k'}a_i \mu\left(f^{-1}(a_i) \cap f^{-1}(b_j) ight)\
\mathrm{Simp}\int_{X} g \,\mathrm{d}\mu &= \sum_{i=1}^{k} \sum_{j=1}^{k'}b_j \mu\left(f^{-1}(a_i) \cap f^{-1}(b_j)ight)
\end{align*}
		We then have
		\begin{align*}
        \mathrm{Simp}\int_{X} f \,\mathrm{d}\mu+\mathrm{Simp}\int_{X} g \,\mathrm{d}\mu &= \sum_{i=1}^{k} \sum_{j=1}^{k'}a_i \mu\left(f^{-1}(a_i) \cap f^{-1}(b_j) ight) + \sum_{i=1}^{k} \sum_{j=1}^{k'}b_j \mu\left(f^{-1}(a_i) \cap f^{-1}(b_j) ight) \
                                  &= \sum_{i=1}^{k} \sum_{j=1}^{k'}(a_i+b_j) \mu\left(f^{-1}(a_i) \cap f^{-1}(b_j) ight)
		\end{align*}
		We argue that the last term is exactly $\mathrm{Simp}\int_{\mathbb{R}^{d}}x{(f+g)}$. First of all $f+g$ is simple, since $\{(f+g)(x):x\in X\} = \{a_i+b_j:1\leq i \leq k, 1 \leq j \leq k'\}$, and the latter has cardinality $k''$ of at most $k''\leq k\cdot k'$. Denote the elements of $\{a_i+b_j:1\leq i \leq k, 1 \leq j \leq k'\}$ by $(d_l)_{l=1}^{k''}$. We then have
      \begin{align*}
        \mathrm{Simp}\int_{\mathbb{R}^{d}}x{(f+g)} &= \sum_{l=1}^{k''} d_l\cdot \mu\left((f+g)^{-1}(d_l)ight) = \sum_{l=1}^{k''} d_l\cdot \mu\left(\{x\in X: f(x)+g(x)=d_l\}ight)\
                         &=  \sum_{l=1}^{k''} d_l\cdot \sum_{(i,j):a_i+b_j=d_l}\mu\left(\{x\in X: f(x)=a_i,g(x)=b_j\}ight)\
                         &=  \sum_{l=1}^{k''} \sum_{(i,j):a_i+b_j=d_l}(a_i+b_j)\mu\left(f^{-1}(a_i)\cap g^{-1}(b_j)ight) = \sum_{(i,j)}(a_i+b_j)\mu\left(f^{-1}(a_i)\cap g^{-1}(b_j)ight)\
                         &= \sum_{i=1}^{k}\sum_{j=1}^{k'}(a_i+b_j)\mu\left(f^{-1}(a_i+b_j)ight)
      \end{align*}

\item 	extit{Insensitivity to refinement}
ewline
		Since taking simple integrals both w.r.t. $\mathcal{B}$ and $\mathcal{B'}$ implies that $f$ is both $\mathcal{B}$-measurable and $\mathcal{B}'$-measurable, we see that for the finite values $a_1,\ldots,a_k$ of $f$ we have $f^{-1}(a_i)\in\mathcal{B}$.
\begin{align*}
\mathrm{Simp}\int_{X} f \,\mathrm{d}\mu &= \sum_{i=1}^{n} a_i\cdot \mu'(f^{-1}(a_i))\
&= \sum_{i=1}^{n} a_i\cdot \mu'estriction_{\mathcal{B}}(f^{-1}(a_i))\
&= \sum_{i=1}^{n} a_i \cdot \mu(f^{-1}(a_i))
&= \mathrm{Simp}\int_{X} f \,\mathrm{d}\mu'
\end{align*}

\item 	extit{Almost everywhere equivalence}
ewline
		Let $E$ be the null set on which $f(x)
eq g(x)$. Let $a_1,\ldots,a_k$ be the finite values of $f$ and $g$ on $X/E$. Let $a_{k+1},\ldots,a_{k+k'}$ be the values of $f$ on $X$, and let $b_{k+1}$, $\ldots$, $b_{k+k''}$ be the values of $g$ on $X$. We then have for 
      \begin{align*}
        \mathrm{Simp}\int_{X} f \,\mathrm{d}\mu &= \sum_{i=1}^{k+k'} a_i \cdot \mu(f^{-1}(a_i)) = \sum_{i=1}^{k} a_i \cdot \mu(f^{-1}(a_i)) \
                     &= \sum_{i=1}^{k} a_i \cdot \mu(g^{-1}(a_i)) + \sum_{i=k+1}^{k+k''} b_i \cdot \mu(g^{-1}(b_i)) = \mathrm{Simp}\int_{X} g \,\mathrm{d}\mu
      \end{align*}

\item 	extit{Finiteness}
ewline
		Suppose that $\mathrm{Simp}\int_{X} f \,\mathrm{d}\mu = \sum_{i=1}^{k}a_i\mu(f^{-1}(a_i))$ is finite. Then, each element $a_i\mu(f^{-1}(a_i))$ is finite. By the laws of multiplication for measure spaces, this implies that for all $1\leq i \leq k$ we have one of the following
		\begin{itemize}
			\item $i < \infty$ and $\mu(f^{-1}(a_i)) < \infty$ or
			\item $a_i = 0$ and $\mu(f^{-1}(a_i)) = \infty$ or
			\item $a_i = \infty$ and $\mu(f^{-1}(a_i)) = 0$
		\end{itemize}
		The second case is impossible by definition since we always trivially exclude the case where one of the simple values is zero. Thus, the components of the simple sum are either finite (like in the first case) or zero (like in the last case). Thus $f$ is finite almost everywhere and has finite measure support.
ewline
      Conversely, finite almost everywhere and finite measure support assumption implies that the components of the sum are finite or zero, and thus the integral itself is finite.

\item 	extit{Vanishing}
ewline
		Suppose that $\sum_{i=1}^{k} a_i \cdot \mu(f^{-1}(a_i))$. Since every term in summation is non-negative, every term must itself be zero. This is only possible when the measures $f^{-1}(a_i), 1\leq i \leq k$ are all zero.
ewline
      Conversely, suppose that $f$ is zero almost everywhere. Then the inverse image of every non-zero simple value $a_i$ returns a null set $f^{-1}(a_i)$ resulting in a zero integral.

\end{enumerate}

Exercise 1.4.33 (Basic properties of the simple integral)

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Exercise 1.4.33 (Basic properties of the simple integral)

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