Exercise 1.2.17 (Carathéodory criterion of Lebesgue measurability)

We use the chain of implications.

• $(i)⟹(\mathit{ii})$
  Since both $E$ and $B$ are Lebesgue measurable, and since intersection and set difference of two Lebesgue measurable sets is again Lebesgue measurable by Lemma 1.2.13, we can apply the additivity property of Lebesgue measure

  $$m(B)=m((B\cap E)\cup (B∕E))=m(B\cap E)+m(B∕E)$$

• $(\mathit{ii})⟹(\mathit{iii})$
  Let $A$ be an elementary set, i.e., $A={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^M{C}_n$ for some boxes $C_1,\dots ,C_M$. By subadditivity of the outer Lebesgue measure we have

  $$m(A)\le m^{\text{∗}}(A\cap E)+m^{\text{∗}}(A∕E)$$

  By Lemma 1.1.2 we can represent it as a union of disjoint boxes $A={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^N{B}_n$. We then have

  $$\begin{array}{llll}\hfill m^{\text{∗}}(A\cap E)+m^{\text{∗}}(A∕E)& \le \underset{n=1}{\overset{M}{\sum }}{m}^{\text{∗}}(B_n\cap E)+\underset{n=1}{\overset{N}{\sum }}{m}^{\text{∗}}(B_n∕E)& \hfill & \\ \hfill & =\underset{n=1}{\overset{M}{\sum }}\left(m^{\text{∗}}(B_n\cap E)+m^{\text{∗}}(B_n∕E)\right)& \hfill & \\ \hfill & =\underset{n=1}{\overset{M}{\sum }}m(B_n)=m(A).& \hfill & \end{array}$$

  Thus, both quantities have to be equal.

• $(\mathit{iii})⟹(i)$
  We have two cases.

  • Case I: $m^{\text{∗}}(E)$is finite.
    By definition of $m^{\text{∗}}(E)$ we have boxes ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n\supset E$ such that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }|B_n|\le m^{\text{∗}}(E)+𝜖∕2$ for an arbitrary $𝜖>0$. Notice that all of these boxes can be enlargened just a little bit to make them open, i.e., we consider the open box cover $A:={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n^{\prime }\supset E$, such that $m(B_n^{\prime })\le m(B_n)+𝜖∕2\cdot 2^n$. From this we deduce:

    $$\begin{array}{llll}\hfill m^{\text{∗}}(E)+m^{\text{∗}}(B∕E)& =m^{\text{∗}}\left(\underset{n=1}{\overset{\infty }{\bigcup }}E\cap B_n^{\prime }\right)+m^{\text{∗}}\left(\underset{n=1}{\overset{\infty }{\bigcup }}{B}_n^{\prime }∕E\right)& \hfill & \\ \hfill & \le \underset{n=1}{\overset{\infty }{\sum }}{m}^{\text{∗}}(E\cap B_n^{\prime })+\underset{n=1}{\overset{\infty }{\sum }}{m}^{\text{∗}}(B_n^{\prime }∕E)& \hfill & \\ \hfill & \le \underset{n=1}{\overset{\infty }{\sum }}\left(m^{\text{∗}}(E\cap B_n^{\prime })+m^{\text{∗}}(B_n^{\prime }∕E)\right)& \hfill & \\ \hfill & \stackrel{(\mathit{ii})}{\text{=}}\underset{n=1}{\overset{\infty }{\sum }}m(B_n^{\prime })\le \underset{n=1}{\overset{\infty }{\sum }}m(B_n)+𝜖& \hfill & \\ \hfill & \le m^{\text{∗}}(E)+𝜖& \hfill & \end{array}$$

    Thus, $m(B∕E)\le 𝜖$ for an arbitrary $𝜖>0$ and we are done.

  • Case II: $m^{\text{∗}}(E)$is infinite.
    Let $A$ be an elementary set, and let $B$ be a box. Since $A∕(E\cap B)\subseteq (A∕B)\cup ((A\cap B)∕E)$

    $$\begin{array}{llll}\hfill m^{\text{∗}}(A\cap E\cap B)+m^{\text{∗}}(A∕(E\cap B))& \le m^{\text{∗}}(A\cap B\cap E)+m^{\text{∗}}((A\cap B)∕E)+m(A∕B)& \hfill & \\ \hfill & =m(A\cap B)+m(A∕B)=m(A)& \hfill & \end{array}$$

    Thus, for any elementary $A$ we have the Carathéodory criterion for $A\cap B$:

    $$m(A)=m^{\text{∗}}(A\cap (E\cap B))+m^{\text{∗}}(A∕(E\cap B))$$

    Thus, $E\cap B$ is Lebesgue measurable by the finite version of this assertion. But $E$ can be represented as a countable union of $E={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }E\cap B(0,n)$ where $B(0,n)$ are boxes of side length $n$ around 0. Thus $E$ is Lebesgue measurable.