Exercise 1.2.16 (Equivalent formulations of finite Lebesgue measurability)

We use the chain of implications to demonstrate $(i)⟺(\mathit{ii})⟺(\mathit{iii})$ first.

• $(i)⟹(\mathit{ii})$
  By Lebesgue measurability we can find an open set $U$ such that $m(U∕E)\le 𝜖$. By countable additivity $m(U)\le m(E)+𝜖$. Since $m(E)$ and $𝜖$ are both finite, $m(U)$ must be finite too. In other words, our usual choice of the open set $U$ from the definition is already sufficient.

• $(\mathit{ii})⟹(\mathit{iii})$
  Pick an arbitrary $𝜖>0$. By (ii) we can find an open set $U\in O({ℝ}^d):E\subseteq U$ such that $m^{\text{∗}}(U∕E)=m^{\text{∗}}(U△E)\le 𝜖∕3$. We do not, however, know whether $U$ is bounded or not. This is easily fixable by cutting out its "small" (in measure theoretic sense) unbounded part as follows. For all $r\in \mathbb{N}$ let $U_r:=U\cap {(-r,r)}^d$ where each $U_r$ is open and bounded. Let us look closely at $E△U_r$

  • item We obviously have $U_1∕E\subseteq U_2∕E\subseteq \dots$ and ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{r=1}^{\infty }{U}_r∕E=U∕E$. Thus, by Exercise 1.2.11 we have

    $$\underset{r\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{m}^{\text{∗}}(U_r∕E)=m^{\text{∗}}(U∕E)$$

    Pick an $r_1$ large enough such that $-𝜖∕3\le m^{\text{∗}}(U∕E)-m^{\text{∗}}(U_r∕E)$; in other words, $m^{\text{∗}}(U_r∕E)\le m^{\text{∗}}(U∕E)+𝜖∕3$.

  • Similarly notice that $E∕U_1\supseteq E∕U_2\supseteq \dots$ and ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{r=1}^{\infty }E∕U_r=E∕U=\varnothing$. Again, by Exercise 1.2.11 we have

    $$\underset{r\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{m}^{\text{∗}}(E∕U_r)=m^{\text{∗}}(E∕U)=0$$

    In other words, we can pick $r_2\in \mathbb{N}$ large enough so that $m^{\text{∗}}(E∕U_r)\le 𝜖∕3$.

  Set $r:=\max <!--\; FUNCTION\; APPLICATION\; -->\{r_1,r_2\}$; then

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

  as desired.

• $(\mathit{iii})⟹(i)$
  Pick an arbitrary $𝜖>0$ and choose $U\in O({ℝ}^d)$ such that $m^{\text{∗}}(U△E)=m^{\text{∗}}(U∕E)+m^{\text{∗}}(E∕U)\le 𝜖∕3$. Notice that $U$ is already almost sufficient for us - the only problem is a possible part of $E$ not covered by $U$, i.e., $E∕U$. Since it is not very big ($m^{\text{∗}}(E∕U)\le 𝜖$) we can choose a small open set $U_1$ to cover $E∕U$, and then simply take $U^{\prime }:=U\cup U_1$. This is possible by outer regularity of the Lebesgue outer measure (Lemma 1.2.12): we can always find an open outer cover $U_1$ of $E∕U$ such that $m(U_1)-m(E∕U)\le 𝜖∕3$. Then we obviously have $E\subseteq U\cup U_1$, the latter being open. Thus,

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

  and we also have $m(U^{\prime })\le m(U)+m(U_1)<\infty$.

Next we verify $(i)⟺(\mathit{iv})⟺(v)$

• $(i)⟹(\mathit{iv})$
  Since $E$ is Lebesgue measurable, we can find a closed set $F\subseteq E$ such that $m(E∕F)=m(E)-m(F)\le 𝜖∕2$. If $E$ is bounded, then $F$ is bounded also, and we are done. If $E$ is not bounded, notice that we can make $F$ bounded by approximating it using the sequence $F_r:=F\cap B(0,r)$ for any $r\in \mathbb{N}$. By Exercise 1.2.11 by $F_1\subseteq F_2\subseteq \dots$ and ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{r=1}^{\infty }{F}_r=F$ we have $\underset{r\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}m(F_r)=m(F)$. Thus, pick $r\in ℝ$ large enough so that $m(F)-m(F_r)\le 𝜖∕2$, i.e., $-m(F_r)\le -m(F)+𝜖∕2$. We then obtain

  $$m(E∕F_r)=m(E)-m(F_r)\le m(E)-m(F)+𝜖∕2\le 𝜖$$

• $(\mathit{iv})⟹(v)$
  Follows directly by taking the same $E$ for (v) as the $E$ for (iv).
• $(v)⟹(\mathit{vi})$
  Follows trivially from (v) since any compact set is automatically a bounded Lebesgue measurable set.
• $(\mathit{vi})⟹(\mathit{vii})$
  Follows trivially, since any bounded Lebesgue measurable set can be contained in a ball $B(0,r)$ for some $r\in ℝ$ large enough; by $m(F)\le m(B(0,r)<\infty$ it must have a finite Lebesgue measure.

• $(\mathit{vii})⟹(i)$
  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

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

  Where it is easy to verify that $m(U)<\infty$.

Now we demonstrate the equivalence of the "pixelation" criteria $(i)⟺(\mathit{viii})⟺(\mathit{ix})$.

• $(i)⟹(\mathit{viii})$
  By the definition of Lebesgue measure we have

  $$m(E)=\left\{\sum _{n=1}^{\infty }\left|B_n\right|:\begin{array}{c}{B}_n\text{ are boxes}\hfill \\ E\subset \bigcup _{n=1}^{\infty }{B}_n\hfill \end{array}\right\}<\infty$$

  Pick an arbitrary ${(B_n)}_{n\in \mathbb{N}}$ such that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }\left|B_n\right|-m(E)\le 𝜖∕2$. For each $N\in \mathbb{N}$ define $F_N:={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^N{B}_n$. Notice that

  • em The set $E∕F_1\supseteq E∕F_2\supseteq \dots$, and we have ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{N=1}^{\infty }E∕F_n=\varnothing$; thus $\underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}m(E∕F_N)=0$ by Exercise 1.2.11. Pick an $N_1\in \mathbb{N}$ large enough such that $m(E∕F_{N_1})\le 𝜖∕2$.
  • The set $F_1∕E\subseteq F_2∕E\subseteq \dots$, and we have ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{N=1}^{\infty }{F}_N∕E={\bigcup }_{n=1}^{\infty }{B}_n∕E$; thus, $\underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}m(F_N∕E)=m({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n∕E)=m({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n)-m(E)\le 𝜖∕3$. Pick an $N_2\in \mathbb{N}$ large enough such that $m(F_{N_2}∕E)\le 𝜖∕2$.

  Now set $N=\max <!--\; FUNCTION\; APPLICATION\; -->\{N_1,N_2\}$, and we have

  $$m(E△F_N)\le m(E∕F_N)+m(F_N∕E)\le 𝜖∕2+𝜖∕2=𝜖$$

  as desired.

• $(\mathit{viii})⟹(\mathit{ix})$
  We use the following lemma:

  Lemma 1. Let $A,B,C\in {ℝ}^d$ be sets such that $A$ approximates $B$ in measure $m^{\text{∗}}(A△B)\le 𝜖$, and $B$ approximates $C$ in measure $m^{\text{∗}}(B△C)\le 𝜖$. Then $A$ approximates $C$ in measure $m^{\text{∗}}(A△C)\le 𝜖$.

  Proof. Since we can represent both sets disjointly as $A=(A\cap B)\cup (A∕B)$ and $B=(B\cap C)\cup (B∕C)$ by monotonicity of the Lebesgue outer measure it follows that

  $$\begin{array}{llll}\hfill m^{\text{∗}}(A△C)& \le m^{\text{∗}}(A∕C)+m^{\text{∗}}(C∕A)& \hfill & \\ \hfill & \le m^{\text{∗}}((A\cap B)∕C)+m^{\text{∗}}((A∕B)∕C)+m^{\text{∗}}((B\cap C)∕A)+m^{\text{∗}}((B∕C)∕A)& \hfill & \\ \hfill & \le m^{\text{∗}}((A\cap B)∕C)+m^{\text{∗}}(A∕B)+m^{\text{∗}}((B\cap C)∕A)+m^{\text{∗}}(B∕C)& \hfill & \\ \hfill & \le m^{\text{∗}}((A\cap B)∕C)+𝜖+m^{\text{∗}}((B\cap C)∕A)+𝜖& \hfill & \\ \hfill & =m^{\text{∗}}(A∕C\cap B∕C)+𝜖+m^{\text{∗}}(B∕A\cap C∕A)+𝜖& \hfill & \\ \hfill & \le m^{\text{∗}}(B∕C)+𝜖+m^{\text{∗}}(B∕A)+𝜖& \hfill & \\ \hfill & \le 4𝜖.& \hfill & \end{array}$$ □

  Now we have by theorem assumption an elementary set $F$ that approximates $E$, i.e., $m^{\text{∗}}(F△E)\le 𝜖$. But $F$ is an elementary (Jordan measurable) set. Thus, by Exercise 1.1.14 we can find a $n\in \mathbb{N}$ large enough so that the measure of all dyadic cubes $Q:={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{E(F,2^{-n})}{Q}_n$ of sidelength $2^{-n}$ contained in $F$ satisfies $m(F)-m(Q)=m(F∕Q)=m(F△Q)\le 𝜖$. By the previous lemma, from $m^{\text{∗}}(E△F)\le 𝜖$ and $m(F△Q)\le 𝜖)$ we deduce $m^{\text{∗}}(E△Q)\le 𝜖$.

• $(\mathit{ix})⟹(i)$
  Since the union of dyadic cubes $Q$ is a finitely measurable set, and since $E$ differs from $Q$ by an arbitrarily small Lebesgue outer measure, by part (vii) $E$ must be measurable too, with finite measure.