Exercise 1.1.6 (Basic properties of Jordan measure)

Question

Let $E,F\subset \mathbb{R}^d$ be Jordan measurable sets. Show that
\begin{enumerate}[label=(oman*)]
\item (Boolean closure) The sets $E \cup F$, $E \cap F$, $E/F$ and $E 	riangle F$ are Jordan measurable.
\item (Non-negativity) $m(E)\geq 0$.
\item (Fininte additivity) If $E,F$ are disjoint, then $m(E \cup F) = m(E) + m(F)$.
\item (Monotonicity) If $E\subset F$, then $m(E) \leq m(F)$.
\item (Finite subadditivity) $m(E \cup F) \leq m(E) + m(F)$.
\item (Translation invariance) For any $x\in \mathbb{R}^d$, $E+x$ is Jordan measurable, and $m(E+x) = m(E)$.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(\roman*)]
\item First, we demonstrate that $E \cup F$ is Jordan-measurable, i.e.,
  \begin{equation*}\sup \left\{ m(A) : 
      \begin{array}{l} A\subset (E \cup F) \ \text{$A$ is elementary}\end{array}
    \right\} = \inf \left\{ m(B) :
      \begin{array}{l} (E \cup F) \subset B \ \text{$B$ is elementary}\end{array}
    \right\}.
  \end{equation*}
  As always, we show that we can always find an outer cover $\overline{E \cup F}$ and an inner cover $\underline{E \cup F}$ such that the difference between their simple measures does not exceed $\epsilon>0$. Thus, fix an arbitrary $\epsilon$. Since $E,F$ are Jordan-measurable, we can find outer covers $\overline{E},\overline{F}$ and inner covers $\underline{E},\underline{F}$ such that $m\left(\overline{E}\right) - m\left(\underline{E}\right) \leq \epsilon/2$ and $m\left(\overline{F}\right) - m\left(\underline{F}\right) \leq \epsilon/2$ by \href{./exercise_1-1-5}{Exercise 1.1.5}. Define
  \begin{equation*}
    \overline{E \cup F}:=\overline{E} \cup \overline{F} \hspace{1cm}
    \underline{E \cup F}:= \underline{E} \cup \underline{F}.
  \end{equation*}
  The properties of elementary measure ensure that the $\epsilon/2$ covering we have chosen give us the desired result:
  \begin{equation*}
    \begin{array}{ll}
      m\left(\overline{E \cup F}\right) - m\left(\underline{E \cup F}\right) &= \left( m\left(\overline{E}\right) + m\left(\overline{F}/\overline{E}\right) \right) - \left( m\left(\underline{E}\right) + m\left(\underline{F}/\underline{E}\right) \right)\[5pt]
                                                                             &= \left( m\left(\overline{E}\right) -  m\left(\underline{E}\right) \right) + \left(m\left(\overline{F}/\overline{E}\right) - m\left(\underline{F}/\underline{E}\right) \right)\[5pt]
                                                                             &\leq \left( m\left(\overline{E}\right) -  m\left(\underline{E}\right) \right) + \left(m\left(\overline{F}\right) - m\left(\underline{F}\right) \right)\[5pt]
                                                                             &\leq \epsilon/2 + \epsilon/2 = \epsilon.
    \end{array}
  \end{equation*}
  For $E \cap F$ we similarly define
  \begin{equation*}
    \overline{E \cap F}:=\overline{E} \cap \overline{F} \hspace{1cm}
    \underline{E \cap F}:= \underline{E} \cap \underline{F}.
  \end{equation*}
  Using some set-theoretic manipulation, we obtain
  \begin{equation*}
    \begin{array}{ll}
      m\left(\overline{E \cap F}\right) - m\left(\underline{E \cap F}\right) &=  m\left(\overline{E} \cap \overline{F}\right) - m\left(\underline{E} \cap \underline{F}\right)\[5pt]
                                                                             &= m\left( \overline{E\cap F} / (\underline{E}\cap\underline{F})\right)\[5pt]
                                                                             &= m\left((\overline{E\cap F} / \underline{E}) \cup (\overline{E\cap F}/\underline{F})\right) \[5pt]
                                                                             &\leq m\left(\overline{E\cap F} / \underline{E}\right) + m\left(\overline{E\cap F}/\underline{F}\right) \[5pt]
                                                                             &\leq m\left(\overline{E} / \underline{E}\right) + m\left(\overline{F}/\underline{F}\right) \[5pt]
                                                                             &\leq \epsilon/2 + \epsilon/2 = \epsilon. \[5pt]
    \end{array}
  \end{equation*}
  We repeat the manipulations with the set difference $E/F$:
  \begin{equation*}
    \begin{array}{ll}
      m\left(\overline{E/F}\right) - m\left(\underline{E/F}\right) &= m\left(\overline{E}/\overline{E\cap F}\right) - m\left(\underline{E}/\underline{E\cap F}\right)\[5pt]
                                                                   &\leq m\left(\overline{E}/\underline{E\cap F}\right) - m\left(\underline{E}/\underline{E\cap F}\right)\[5pt]
                                                                   &= m\left(\overline{E}\right) - m\left(\underline{E\cap F}\right) - m\left(\underline{E}\right) + m\left(\underline{E\cap F}\right)\[5pt]
                                                                   &\leq \epsilon/2.
    \end{array}
  \end{equation*}
  Finally, the fact that the symmetric difference $E\triangle F$ is Jordan measurable follows from the fact that set differences and unions of Jordan measurable sets are Jordan measurable.
\item Non-negativity follows from the fact that the infimum/supremum of non-negative reals (which elementary measures are) is necessarily non-negative.
\item Given $m(E),m(F)$ exist and $E\cap F=\emptyset$, we show that
  $$m(E \cup F) = m(E) + m(F)$$
  or, in other words, we show that the sum of the infimums of the former and the latter set is equal to the infimum of the union set
  \begin{equation*}
    \inf\left\{ m(B') : \begin{array}{l} E \subset B' \ \text{$B'$ is elementary}\end{array} \right\} + 
    \inf\left\{ m(B'') : \begin{array}{l} E \subset B'' \ \text{$B''$ is elementary}\end{array} \right\} = 
    \inf\left\{ m(B) : \begin{array}{l} (E \cup F) \subset B \ \text{$B$ is elementary}\end{array} \right\}.
  \end{equation*}
  To show that the equation holds, we employ the usual trick of showing that the left-hand side is both greater than or equal to and less than or equal to the right-hand side.
 \begin{itemize}
\item $\leq$ Pick an arbitrary cover $\overline{E}$ and
    $\overline{F}$. Then the set $\overline{E} \cup (\overline{F}/\overline{E})$ covers $E \cup F$ with $m(\overline{F} \cup (\overline{F}/\overline{E})) \leq m(\overline{E}) + m(\overline{F})$.
\item $\geq$ Pick an arbitrary cover $\overline{E \cup F}$. Set $\epsilon := m\left(\overline{E \cup F}\right) - m(E \cup F)$. Since $E,F$ are Jordan measurable, we can find covers $\overline{E},\overline{F}$ such that $m(\overline{E}) - m(E) \leq \epsilon/3$ and $m(\overline{F}) - m(F) \leq \epsilon/3$.
\end{itemize}

\item This is a corollary of the previous result:
  $$m(F) = m\left((F/E)\cup(E)\right) \stackrel{(3)}{=} m(F/E) + m(E) \stackrel{(2)}{\geq} m(E)$$

\item Similarly,
  $$m(E\cup F) = m\left( (E/F)\cup (F/E) \right) \stackrel{(5)}{=} m(E/F) + m(F/E) \stackrel{(4)}{\leq} m(E) + m(F).$$
  This follows by the fact that any cover $\overline{E+x}$ of $E+x$ is the translated cover $\overline{E}+x$ of $E$, and vice versa. Since the underlying elementary measures are translation invariant, the set of measures of covers of both sets coincide.
\end{enumerate}

Exercise 1.1.6 (Basic properties of Jordan measure)

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Exercise 1.1.6 (Basic properties of Jordan measure)

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