Exercise 1.2.24 (Lebesgue measure is a metric completion of elementary measure)

(i)

Let $E,F,G$ be arbitrary sets in $2^A$. Then

• (reflexive property) We have $m(E△E)=m(\varnothing )=0$ by Lemma 1.2.15.
• (symmetric property) By the symmetric property of the symmetric set difference $E△F=F△E$ we have $m^{\text{∗}}(E△F)=0⟺m^{\text{∗}}(F△E)=0$.

• (transitive property) Suppose that $m(E△F)=0$ and $m(F△G)=0$. We then have by the subadditivity property

  $$\begin{array}{llll}\hfill m^{\text{∗}}\left(E△G\right)& =m^{\text{∗}}\left((E∕G)\cup (G∕E)\right)& \hfill & \end{array}$$

  By monotonicity of the Lebesgue outer measure

  $$\begin{array}{llll}\hfill & \le m^{\text{∗}}\left(E∕G\right)+m^{\text{∗}}\left(G∕E\right)& \hfill & \end{array}$$

  Notice that $E∕G=[(E\cap F)\cup (E∕F)]∕G=[(E\cap F)∕G]\cup [(E∕F)∕G]$; by subadditivity we again have

  $$\begin{array}{llll}\hfill & \le m^{\text{∗}}\left((E\cap F)∕G\right)+m^{\text{∗}}\left((E∕F)∕G\right)+m^{\text{∗}}\left((G\cap F)∕E\right)+m^{\text{∗}}\left((G∕F)∕E\right)& \hfill & \end{array}$$

  Finally, by monotonicity we obtain

  $$\begin{array}{llll}\hfill & \le m^{\text{∗}}\left(F∕G\right)+m^{\text{∗}}\left(E∕F\right)+m^{\text{∗}}\left(F∕E\right)+m^{\text{∗}}\left(G∕F\right)=0& \hfill & \end{array}$$

(ii)

It is easy to verify that $d$ is indeed a metric.

• (well defined) We have $d([E],[E])=0$ since $m^{\text{∗}}(E△E)=m^{\text{∗}}(\varnothing )=0$
• (positivity) Let $[E],[E^{\prime }]\in 2_{\text{∼}}^A$ be such that $[E]\ne [F]$. In other words, $E△F$ is not a null set; thus, by non-negativity of the Lebesgue outer measure its outer measure is positive.
• (symmetry) For $[E],[E^{\prime }]\in 2_{\text{∼}}^A$. We then have $d\left([E],[E^{\prime }]\right)=m^{\text{∗}}\left(E△E^{\prime }\right)=m^{\text{∗}}(F△F)=d\left([E],[E^{\prime }]\right)$.

• (triangle inequality) Let $[E],[F],[G]\in 2_{\text{∼}}^A$. Notice that

  $$\begin{array}{llll}\hfill E△F& =[E∕F]\cup [F∕E]=[E\cap G∕F]\cup [E∕G∕F]\cup [F\cap G∕E]\cup [F∕G∕G]& \hfill & \\ \hfill & \subseteq [G∕F]\cup [E∕G]\cup [G∕E]\cup [F∕G]=[E△G]\cup [G△F].& \hfill & \end{array}$$

  The assertion then follows by subadditivity and monotonicity of the Lebesgue outer measure.

• (completeness) Let ${[E_n]}_{n\in \mathbb{N}}$ be a Cauchy sequence of sets in $2^A$, i.e., for any $𝜖>0$ we have a $N\in \mathbb{N}$ such that for all $n,m>N$ we have $d([E_n],[E_m])=m^{\text{∗}}(E_n△E_m)\le 𝜖$. We argue that with respect to $d$ the sequence ${[E_n]}_{n\in \mathbb{N}}$ converges against its limit superior or inferior ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k=1}^{\infty }{\bigcap }_{m=k}^{\infty }{E}_m={\bigcap }_{k=1}^{\infty }{\bigcup }_{m=k}^{\infty }{E}_m$. We verify this. Pick an $N\in \mathbb{N}$ such that for all $n\ge N$ we have $m(E_N△E_n)\le 𝜖∕2$. By Exercise 1.2.11 (Monotone convergence theorem for measurable sets) we have

  $$\begin{array}{llll}\hfill d\left(E_N,\underset{k=1}{\overset{\infty }{\bigcup }}\underset{m=k}{\overset{\infty }{\bigcap }}{E}_m\right)& \le m^{\text{∗}}\left(\underset{k=1}{\overset{\infty }{\bigcup }}\underset{m=k}{\overset{\infty }{\bigcap }}{E}_m∕E_N\right)+m^{\text{∗}}\left(E_N∕\underset{k=1}{\overset{\infty }{\bigcup }}\underset{m=k}{\overset{\infty }{\bigcap }}{E}_m\right)& \hfill & \\ \hfill & =m^{\text{∗}}\left(\underset{k=1}{\overset{\infty }{\bigcup }}\underset{m=k}{\overset{\infty }{\bigcap }}(E_m∕E_N)\right)+m^{\text{∗}}\left(\underset{k=1}{\overset{\infty }{\bigcap }}\underset{m=k}{\overset{\infty }{\bigcup }}(E_N∕E_m)\right)& \hfill & \\ \hfill & =\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{m}^{\text{∗}}\left(\underset{m=k}{\overset{\infty }{\bigcap }}(E_m∕E_N)\right)+\underset{k\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{m}^{\text{∗}}\left(\underset{m=k}{\overset{\infty }{\bigcup }}(E_N∕E_m)\right)& \hfill & \end{array}$$

  We approximate these limits by taking elements close enough to them, i.e, we set $K=\max <!--\; FUNCTION\; APPLICATION\; -->\{N,K_{𝜖,1},K_{𝜖,2}\}$.

  $$\begin{array}{llll}\hfill & \le m^{\text{∗}}\left(\underset{m=K}{\overset{\infty }{\bigcap }}(E_m∕E_N)\right)+𝜖+m^{\text{∗}}\left(\underset{m=K}{\overset{\infty }{\bigcup }}(E_N∕E_m)\right)-𝜖\le 𝜖& \hfill & \end{array}$$

  Thus, we have given a constructive proof on how to find a limit of any Cauchy sequence in this metric space.

(iii)

We demonstrate both denseness and closure.

• $E_{\text{∼}}$is dense in $L_{\text{∼}}$.
  Pick an arbitrary Lebesgue measurable set $[E]\in L_{\text{∼}}$. By Exercise 1.2.16 (Equivalent formulations of finite Lebesgue measurability) we can find an elementary set $F$ which differs from $E$ by not more than $𝜖>0$ in measure. Thus, $d([E],[F])=m(E△F)\le 𝜖$ and we are done.
• $L_{\text{∼}}$is the closure of $E_{\text{∼}}$.
  By the previous exercise, we can make any Cauchy sequence ${[E_n]}_{n\in \mathbb{N}}$ of elementary sets converge to $[{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k=1}^{\infty }{\bigcap }_{m=k}^{\infty }{E}_m]$ in measure. Every elementary set is Lebesgue measurable. By Lemma 1.2.13 countable unions and intersections of Lebesgue measurable sets must be measurable; thus $[{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k=1}^{\infty }{\bigcap }_{m=k}^{\infty }{E}_m]\in L_{\text{∼}}$ as desired.

(iv)

We demonstrate that $m$ is the only continuous extension of the elementary measure.

• continuity
  We want to demonstrate that

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->𝜖>0\exists <!--\; FUNCTION\; APPLICATION\; -->{\delta }_{𝜖}>0\forall <!--\; FUNCTION\; APPLICATION\; -->[E],[F]\in L_{\text{∼}}:d([E],[F])\le {\delta }_{𝜖}⟹|m(E)-m(F)|\le 𝜖$$

  It suffices to set ${\delta }_{𝜖}:=𝜖$. We then have from $m(E△F)=m(E∕F)+m(F∕E)\le 𝜖$ that

  $$\begin{array}{llll}\hfill |m(E)-m(F)|& =|m(E)-m(E\cap F)+m(F\cap E)-m(F)|& \hfill & \\ \hfill & =|m(E∕F)-m(F∕E)|\le m(E∕F)+m(F∕E)\le 𝜖.& \hfill & \end{array}$$
• unique continuous extension of the elementary measure Let ${[E_n]}_{n\in \mathbb{N}}$ be an arbitrary convergent sequence of elementary sets with $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}[E_n]=E$ for some $[E]\in L_{\text{∼}}$. By continuity of $m$ we have $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}m([E_n])=m([E])$. Any other value would contradict the continuity.

For parts of ii)

Let $[E]=[F]$ and $[E^{\prime }]=[F^{\prime }]$, and let $𝜀$ be arbitrary, so that $m^{\text{∗}}(E\mathrm{\Delta }F)<𝜀$ and $m^{\text{∗}}(E^{\prime }\mathrm{\Delta }{F}^{\prime })<𝜀$.

Consider the set $(E\cap F)\setminus (E^{\prime }\cup F^{\prime })$ and its relation to $(E\setminus E^{\prime })$.

We can split $E\setminus E^{\prime }$ into a 4-element partition:

Then, by subadditivity:

Similarly,

By monotonicity, we also have:

and

So that

Since $𝜀$ was arbitrarily small, so is the difference, and thus:

An analogous argument gives us:

and so in total:

Does $d$ give a metric space?

We have $m^{\text{∗}}(E\mathrm{\Delta }E)=m^{\text{∗}}(\varnothing )=0$.

Then for $[E]\ne [F]$, there is $m^{\text{∗}}(E\mathrm{\Delta }F)>0$ by definition of $d$.

Since $E\mathrm{\Delta }F=F\mathrm{\Delta }E$, symmetry is given, and the last property to tackle is the triangle inequality. Let $[E],[F],[G]\in 2^A$. Notice that:

This is contained in:

Then,

Thus, we have a metric space.

Proof of completeness is lacking;