Exercise 1.2.18 (Lebesgue inner measure)

Let $A$ be an arbitrary elementary set containing $E$.

(i)

Let $A^{\prime }$ be another arbitrary elementary sets containing $E$. We demonstrate that $$m(A)-m^{\text{∗}}(A∕E)=m(A^{\prime })-m^{\text{∗}}(A^{\prime }∕E).$$

Notice, however, that we can also demonstrate that both quantities are equal to

$$m(A\cap A^{\prime })-m^{\text{∗}}((A\cap A^{\prime })∕E)$$

since $A\cap A^{\prime }$ is still elementary with $E\subseteq A\cap A^{\prime }$, but we now enjoy the nestedness property $A\cap A^{\prime }\subseteq A,A^{\prime }$. In the following we only show this for $A$, the assertion for $A^{\prime }$ follows similarly.

• Notice that we can break up $A∕E$ into three layers: the part of $A$ that does to contain $A\cap A^{\prime }$ and the part of $(A\cap A^{\prime })$ that does not contain $E$. In other words, $A=\left[A∕(A\cap A^{\prime })\right]\cup \left[(A\cap A^{\prime })∕E\right]$. By the subadditivity property of the Lebesgue outer measure we have

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

• Notice that we can rewrite $m(A)=m(A\cup A^{\prime })+m(A∕A^{\prime })$ by Exercise 1.2.17. This allows to rephrase the assertion we want to prove as

  $$\begin{array}{llll}\hfill & m(A)-m^{\text{∗}}(A∕E)\le m(A\cap A^{\prime })-m^{\text{∗}}((A\cap A^{\prime })∕E)& \hfill & \\ \hfill & ⟺m(A∕A^{\prime })+m^{\text{∗}}\left((A\cap A^{\prime })∕E\right)\le m^{\text{∗}}(A∕E).& \hfill & \end{array}$$

  Let ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n$ be a collection of boxes such that $A∕E\subseteq {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n$ and $m({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n)\le m^{\text{∗}}(A∕E)+𝜖$.

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

  Since $(A\cap A^{\prime })∕E=(A∕E)\cap (A^{\prime }∕E)\subseteq (A∕E)\cap {(A∕A^{\prime })}^C=(A∕E)∕(A∕A^{\prime })\subseteq {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n∕(A∕A^{\prime })$ we obtain by monotonicity

  $$\begin{array}{llll}\hfill & \le m\left(\underset{n=1}{\overset{\infty }{\bigcup }}{B}_n\cap (A∕A^{\prime })\right)+m\left(\underset{n=1}{\overset{\infty }{\bigcup }}{B}_n∕(A∕A^{\prime })\right)& \hfill & \end{array}$$

  We can safely use the additivity property, since we are only dealing with measurable sets

  $$\begin{array}{llll}\hfill & =\underset{n=1}{\overset{\infty }{\sum }}m\left(B_n\cap (A∕A^{\prime })\right)+m\left(B_n∕(A∕A^{\prime })\right)& \hfill & \end{array}$$

  Applying Carathéodory criterion we obtain

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

  And since this is true for any $𝜖>0$ we see that

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

• Now we demonstrate that $m(A)-m^{\text{∗}}(A∕E)\le m^{\text{∗}}(E)$. Notice that $A=(E\cap A)\cup (A∕E)$. By subadditivity of the Lebesgue outer measure $m(A)\le m^{\text{∗}}(E\cap A)+m^{\text{∗}}(A∕E)=m^{\text{∗}}(E)+m^{\text{∗}}(A∕E)$. Thus, we have $m(A)-m^{\text{∗}}(A∕E)\le m^{\text{∗}}(E)$.
• Both direction follow by the Carathéodory criterion since $m^{\text{∗}}(A\cap E)=m^{\text{∗}}(E)$.

Let $E,F$ be Lebesgue measurable sets such that $A\subset F\subset E$ and $m(E)<\infty$.

Fix $𝜀>0$. Let ${\left(B_n\right)}_n$ be a sequence of boxes such that $E\setminus A\subset {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_n{A}_n$ and ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_n\left|B_n\right|\le$ $m^{\text{∗}}(E\setminus A)+𝜀$. Note that $F\setminus A\subset (E\setminus A)\setminus (E\setminus F)\subset {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_n\left(B_n\setminus (E\setminus F)\right)$ and $E\setminus F\subset {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_n\left(B_n\cap (E\setminus F)\right)$. So,

Taking $𝜀\to 0$ and rearranging we have that

Since $E\setminus A\subset (E\setminus F)\cup (F\setminus A)$ we have

So we have show that

whenever $A\subset F\subset E$.

For general Lebesgue measurable sets $E,F$ such that $A\subset F$ and $A\subset E$ and $m(E)<$ $\infty ,m(F)<\infty$, we have that $A\subset E\cap F$, so

This shows that $m_{\text{∗}}$ is well defined.

Now, if $A$ is Lebesgue measurable then for any Lebesgue measurable set $E\supset A$ with $m(E)<\infty$ we have that $m(E\setminus A)=m(E)-m(A)$. So $m_{\text{∗}}(A)=m(E)-m(E\setminus A)=$ $m(A)=m^{\text{∗}}(A).$

On the other hand, assume that $m^{\text{∗}}(A)=m_{\text{∗}}(A)$. Fix $𝜀>0$ and let $U\supset A$ be an open set such that $A\subset U$ and $m(U)\le m^{\text{∗}}(A)+𝜀$. Note that

so $m^{\text{∗}}(U\setminus A)\le 𝜀$.