Exercise 1.2.7 (Equivalent formulations of Lebesgue measurablity)

We prove the equivalence by $(i)⟺(\mathit{ii})⟺(\mathit{iii})$ and $(i)⟺(\mathit{iv})⟺(v)$.

$(i)⟺(\mathit{ii})⟺(\mathit{iii})$

• (i) $⟺$ (ii) follows directly from the definition
• (ii) $⟹$ (iii). Fix $𝜖$ and by (ii) find a set $U$ containing $E$ such that $m^{\text{∗}}(U∕E)\le 𝜖$. Since $E\subseteq U$ we have $U△E=(U∕E)\cup (E∕U)=U∕E$; thus, $m^{\text{∗}}(U△E)=m^{\text{∗}}(U∕E)\le 𝜖$.

• (iii) $⟹$ (ii). We find an open set $U$ such that $m^{\text{∗}}(U△E)\le 𝜖$. The trick is to cover the difference between two sets $U△E$ by another open set $V$; by Lemma 1.2.12 we find an open cover $V$ of $E∕U$ with $𝜖\le m^{\text{∗}}(V)\le 2𝜖$. Then, $E\subseteq U\cup V$ (and $U\cup V$ is open) with

  $$m^{\text{∗}}((U\cup V)∕E)=m^{\text{∗}}(U∕E\cup V∕E)\le m^{\text{∗}}(U∕E)+m^{\text{∗}}(V∕E)\le m^{\text{∗}}(U△E)+m^{\text{∗}}(V)\le 3𝜖$$

$(i)⟺(\mathit{iv})⟺(v)⟺(\mathit{vi})$

• (i) $⟹$ (iv). Then $E$ is Lebesgue measurable, and by Lemma 1.2.13 its complement $E^c$ is Lebesgue measurable too. By definition we can find an open cover $U$ of $E^c$ such that $m^{\text{∗}}(U∕E^c)\le 𝜖$. Notice that $U^c\subseteq {(E^c)}^c=E$; thus, $F:=U^c$ is a closed inner cover of $E$. Furthermore, we have $U∕E^c=U\cap E=E∕U^c$; thus, we obtain

  $$m^{\text{∗}}(E∕U^c)=m^{\text{∗}}(U∕E^c)\le 𝜖$$

• (iv) $⟹$ (v). Trivial.
• (v) $⟹$ (vi). Since every closed set is Lebesgue measurable (Lemma 1.2.13), the assertion follows directly.

• (vi) $⟹$ (i). Pick an arbitrary $𝜖>0$, and take a measurable set $E_{𝜖}$ such that $m^{\text{∗}}(E△E_{𝜖})\le 𝜖∕3$ as in (vii). On one hand, notice that we can replace the measurable set $E_{𝜖}$ by an open set $U_1\supseteq E_{𝜖}$ such that $m^{\text{∗}}(U_1)-m^{\text{∗}}(E_{𝜖})\le 𝜖∕3$. On the other hand, by Lemma 1.2.12 we can replace $E∕E_{𝜖}$ by an open set $U_2$ such that $m^{\text{∗}}(U_2)-m(E∕E_{𝜖})\le 𝜖∕3$. Then we have an open set $U:=U_1\cup U_2$ such that $E\subseteq U$ and

  $$\begin{array}{llll}\hfill m^{\text{∗}}(U∕E)& \le m^{\text{∗}}(U_1∕E)+m^{\text{∗}}(U_2∕E)& \hfill & \\ \hfill & =m^{\text{∗}}(U_1∕E)+\frac{𝜖}{3}& \hfill & \\ \hfill & \le m^{\text{∗}}((U_1∕E_{𝜖}\cup E_{𝜖})∕E)+\frac{𝜖}{3}& \hfill & \\ \hfill & \le m^{\text{∗}}(U_1∕E_{𝜖}∕E)+m^{\text{∗}}(E_{𝜖}∕E)+\frac{𝜖}{3}& \hfill & \\ \hfill & \le m^{\text{∗}}(U_1∕E_{𝜖})+m^{\text{∗}}(E_{𝜖}△E)+\frac{𝜖}{3}& \hfill & \\ \hfill & \le 𝜖& \hfill & \end{array}$$

$(1)⟺(2)$ is the definition.
$(2)\Rightarrow (3)$ follows by taking the same open set $U\supset A$. $(3)\Rightarrow (2):$ Fix $𝜀>0$ and let $U$ be such that $m^{\text{∗}}(U△A)<𝜀$. Let $V\supset U△A$ be an open set such that $m^{\text{∗}}(V)<2𝜀$, by outer regularity. Note that $A\subset U\cup (A\setminus U)\subset U\cup V$, which is an open set, and $m^{\text{∗}}((U\cup V)\setminus A)\le m^{\text{∗}}(U\setminus A)+m^{\text{∗}}(V\setminus A)\le m^{\text{∗}}(U△A)+m^{\text{∗}}(V)<3𝜀$. So $(1)⟺(2)⟺(3)$ $(1)\Rightarrow (4):A$ is Lebesgue measurable, so also $A^c$ is. Fix $𝜀>0$ and let $U$ be an open set such that $m^{\text{∗}}\left(U△A^c\right)<𝜀$. Let $F=U^c$. So $m^{\text{∗}}(F△A)=m^{\text{∗}}\left(U△A^c\right)<𝜀$, and $F$ is closed. (4) $\Rightarrow (5):$ Fix $𝜀>0$ and let $F$ be a closed set such that $m^{\text{∗}}(A△F)<𝜀$. By outer regularity let $U\supset A△F$ be an open set such that $m^{\text{∗}}(U)<2𝜀$. Then $F\cap U^c$ is a closed 36 set and $F\cap U^c\subset F\cap A\subset A$, and

$(5)\Rightarrow (6):$ Fix $𝜀>0$ and let $F\subset A$ be a closed set such that $m^{\text{∗}}(A\setminus F)<𝜀$. The $C=F$ is Lebesgue measurable, and $m^{\text{∗}}(A△C)=m^{\text{∗}}(A\setminus F)<𝜀$. $(6)\Rightarrow (3):$ Fix $𝜀>0$ and let $C$ be Lebesgue measurable such that $m^{\text{∗}}(A△C)<𝜀$. Let $U\supset C$ be an open set such that $m^{\text{∗}}(U△C)=m^{\text{∗}}(U\setminus C)<𝜀$. Note that $A△U\subset$ $A△C\cup C△U$, so