Exercise 1.1.5 (Equivalent formulation of Jordan measurability)

Question

Let $E \subset \mathbb{R}^d$ be bounded. Show that the following are equivalent:
\begin{enumerate}[label=(oman*)]
	\item $E$ is Jordan measurable
	\item For every $\epsilon>0$ there exist elementary sets $A\subset E \subset B$ such that $m(B/A) \leq \epsilon$.
	\item For every $\epsilon>0$ there exists an elementary set $A$ such that $m^{*,(J)}(A	riangle E) \leq \epsilon$.
\end{enumerate}
In particular, we see that elementary measure coincides with Jordan measure on its domain.

Elu Thingol · Accepted Answer

Notice that Jordan outer and inner measures are monotone and that a cover of a larger set is also a cover of any of its subsets. Thus, a smaller subset has in a sense ``more'' covers than its superset, and thus a possibly larger supremum or a smaller infimum.

\begin{proof}
  $(1)\implies (2)$ By definition, (1) means
  \begin{equation*}\sup \left\{ m(A) : 
      \begin{array}{l} A\subset E \ \text{$A$ is elementary}\end{array}
    \right\} = \inf \left\{ m(B) :
      \begin{array}{l} B\supset E \ \text{$B$ is elementary}\end{array}
    \right\}
  \end{equation*}
  By definition of supremum and infimum, we can find always find a set $A \subset E$ and $B \supset E$ such that $m(B) - m(E) \leq \epsilon/2$ and $m(E) - m(A) \leq \epsilon/2$. Since $A \supset B$, from $m(B) = m(B/A) + m(A)$ we deduce that
  $$m(B/A) = m(B) - m(A) \leq \epsilon/2 + m(E) - (m(E) - \epsilon/2) = \epsilon.$$\par

$(2) \implies (3)$ Unwrapping the definitions one by one, we have to demonstrate that
  \begin{equation*}
    \forall \epsilon > 0 \quad \exists A \in \mathcal{E}(\mathbb{R}^d): \quad \inf \left\{ m(C) :
      \begin{array}{l} C\supset (A/E \cup E/A) \ \text{$C$ is elementary}\end{array}
    \right\} \leq \epsilon
  \end{equation*}
  Fix $\epsilon>0$. Then we can find sets $A,B: A \subset E \subset B$ such that $m(B/A)\leq \epsilon$. Set $A_\epsilon := A$. Then $A \triangle E = A/E \cup E/A = E/A$. Since $E\subset B$ we have $E/A \subset B/A$. From this deduce that
  \begin{align*}
    \left\{ m(C) :
    \begin{array}{l} C\supset E/A \ \text{$C$ is elementary}\end{array}
    \right\} \subset \left\{ m(C') :
    \begin{array}{l} C'\supset B/A \ \text{$C'$ is elementary}\end{array}
    \right\}.
  \end{align*}
  Since the latter set is the larger one, by monotonicity we conclude that
  $m^{*,J}(A\triangle E) \leq m(B/A) \leq \epsilon$
  as desired.\par

$(3)\implies (1)$ To prove that $E$ is Jordan measurable we show that we can always find a set $B$ on the outside of $E$ and the set $A$ inside of $E$ such that the difference between them is less than any $\epsilon>0$ in elementary measure. Thus, fix an arbitrary $\epsilon$. By (2), we can find an elementary set $A$ such that $m^{*}(A/E \cup E/A) \leq \epsilon/4$. By monotonicity of Jordan outer measure, we also have $m^{*}(A/E) \leq \epsilon/4$ and $m^{*}(E/A)\leq \epsilon/4$ (in other words, we have outer covers $\overline{A/E}$ and $\overline{E/A}$ both with measure less than, say, $\epsilon/2$). We construct an inner and outer cover using these both sets:
  \begin{equation*}
    \begin{array}{ll}
      \overline{E}&:=A\cup(\overline{E/A})\[5pt]
      \underline{E}&:= \underline{E\cap A} := \underline{A/(\overline{A/E})}.
    \end{array}
  \end{equation*}
  We now hope that the above covers do not differ in measure by more than an $\epsilon$:
  \begin{equation*}
    \begin{array}{l}
      m\left(\overline{E}\right) =m\left(A\right)+m\left(\overline{E/A}/A\right) \leq m\left(A\right)+m\left(\overline{E/A}\right) \leq m\left(A\right)+\epsilon/2\[5pt]
      m\left(\underline{E}\right) =m\left(A\right)-m\left(\overline{A/E}/A\right) \leq m\left(A\right)-m\left(\overline{A/E}\right) \leq m\left(A\right)-\epsilon/2\[5pt]
    \end{array}
  \end{equation*}
  Thus,
  \begin{equation*}
    m\left(\overline{E}\right) - m\left(\underline{E}\right) \leq \epsilon
  \end{equation*}
  as desired.
\end{proof}

Prakash Anand · Answer

Before we start, we verify a simple lemma.

Lemma 1. $J^{}$ and $J_{}$ are monotone; $J^{}$ is subadditive.

Proof. Let $A \subset C$ . Let ${(E_{n})}_{n}, {(F_{n})}_{n}$ be sequences of elementary sets such that $E_{n} \subset A$ , $C \subset F_{n}$ and $m (E_{n}) \to J_{∗} (A), m (F_{n}) \to J^{∗} (C)$ . Note that for every $n$ ,

J_{∗} (C) \geq m (E_{n}) \to J_{∗} (A) and J^{∗} (A) \leq m (F_{n}) \to J^{∗} (C)

Also, if ${(G_{n})}_{n}$ is a sequence of elementary sets such that $A \subset G_{n}$ and $m (G_{n}) \to J^{∗} (A)$ , then $A \cup C \subset G_{n} \cup F_{n}$ so

J^{∗} (A \cup C) \leq m (G_{n} \cup F_{n}) \leq m (G_{n}) + m (F_{n}) \to J^{∗} (A) + J^{∗} (C)

□

Using this lemma, we proceed with the proof.

Proof. For any $𝜀 > 0$ there exist elementary sets $E \subset A \subset F$ such that $m (E) > J_{∗} (A) - 𝜀$ and $m (F) < J^{∗} (A) + 𝜀$ . So if $A$ is Jordan measurable then $m (F) < J^{∗} (A) + 𝜀 = J_{∗} (A) + 𝜀 <$ $m (E) + 2 𝜀 .$ Since $E \subset F$ we have that $m (F) = m (E) + m (F ∖ E)$ , so $m (F ∖ E) < 2 𝜀 .$

Now assume that there exist $E, F \in E_{0}, E \subset A \subset F$ such that $m (F ∖ E) < 𝜀$ . Then, $A ∖ E \subset F ∖ E$ , so $J^{∗} (A ∖ E) \leq m (F ∖ E) < 𝜀 .$

Finally assume that for every $𝜀 > 0$ there exists an elementary set $E_{𝜀} \subset A$ such that $J^{∗} (A ∖ E_{𝜀}) < 𝜀 .$ Since $E_{𝜀} \subset A$ we have that $m (E_{𝜀}) \leq J_{∗} (A) .$ Using the subadditivity of $J^{∗}$ we get

J^{∗} (A) \leq J^{∗} (A ∖ E_{𝜀}) + m (E_{𝜀}) < 𝜀 + J_{∗} (A)

Taking $𝜀 \to 0$ and recalling that $J_{∗} (A) \leq J^{∗} (A)$ by definition, we have that $J^{∗} (A) = J_{∗} (A)$ and so $A$ is Jordan measurable. □

Exercise 1.1.5 (Equivalent formulation of Jordan measurability)

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Exercise 1.1.5 (Equivalent formulation of Jordan measurability)

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