An Interesting Exercise in Measure Theory

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This article is the solution (sketch) to Exercise 2A, Problem 8 and 9, in Sheldon Axler’s textbook Measure, Integration & Real Analysis. Both problems look quite intuitive, but they are not that easy to prove. The solutions below are sometimes sketchy, in order to emphasize the general idea behind.

Notations

In this article, we adopt the following notations.

• $\ell(\cdot)$ denotes the length of an open interval. (The interval must be open!)

• $\abs A$ denotes the (Lebesgue) outer measure of $A\subseteq\R$. It is defined as

\[\abs A=\inf\cbra{\sum_{k=1}^\infty\ell(I_k):\{I_k\}\text{ is a sequence of open intervals with }A\subseteq\bigcup_{k=1}^\infty I_k}.\]

Exercise 2A, Problem 8

Lemma (Exercise 2A, Problem 8).

For any $A\subseteq\R$ and any bounded open interval $I$,

\[\abs A=\abs{A\cap I}+\abs{A\cap I^c}.\]

Proof.

$A=(A\cap I)\cup(A\cap I^c)\implies\text{LHS}\le\text{RHS}$. Assume that $\text{LHS}<\text{RHS}$, i.e., $\exists\{I_k\}$ s.t.

\[A\subseteq\bigcup_{k=1}^\infty I_k,\quad \sum_{k=1}^\infty\ell(I_k)<\abs{A\cap I}+\abs{A\cap I^c}.\]

Split each $I_k$ into one open interval inside $I$, collected in a new $\{I_k\}$, and two open intervals outside $I$, collected in $\{J_k\}$. Then \begin{equation}\label{eq1} A\backslash\{a,b\}\subseteq\bra{\bigcup_{k=1}^\infty I_k}\cup\bra{\bigcup_{k=1}^\infty J_k},\quad I_k\subseteq I,J_k\subseteq I^c. \end{equation} \begin{equation}\label{eq2} \sum_{k=1}^\infty\ell(I_k)+\sum_{k=1}^\infty\ell(J_k)<\abs{A\cap I}+\abs{A\cap I^c}. \end{equation} Add $(a-\varepsilon,a+\varepsilon)$ and $(b-\varepsilon,b+\varepsilon)$ to $\{J_k\}$. (\ref{eq1}) becomes \begin{equation}\label{eq3} A\subseteq\bra{\bigcup_{k=1}^\infty I_k}\cup\bra{\bigcup_{k=1}^\infty J_k}, \end{equation} and $\varepsilon$ is small enough s.t. (\ref{eq2}) still holds.

Analyze (\ref{eq1}) and (\ref{eq3}). We have

\[A\cap I\subseteq\bigcup_{k=1}^\infty I_k,\quad A\cap I^c\subseteq\bigcup_{k=1}^\infty J_k,\]

contradicting (\ref{eq2}).

$\square$

Exercise 2A, Problem 9

Problem (Exercise 2A, Problem 9).

Prove that for all $A\subseteq\R$,

\[\abs A=\lim_{t\to\infty}\abs{A\cap(-t,t)}.\]

Proof.

The case where $\abs A=0$ is trivial.

Now suppose $\abs A\in\R^+$. Let $\{I_k\}$ be s.t. \begin{equation}\label{eq4} A\subseteq\bigcup_{k=1}^\infty I_k, \end{equation} where each $I_k$ is bounded and

\[\abs A\le\sum_{k=1}^\infty\ell(I_k)<\abs A+\varepsilon.\]

Then $\exists n$ s.t.

\[\sum_{k=1}^n\ell(I_k)>\abs A-\varepsilon.\]

$\exists t$ s.t. $\bigcup_{k=1}^nI_k\subseteq(-t,t)$. Now by (\ref{eq4}), we have

\[A\cap(-t,t)^c\subseteq\bigcup_{k=n+1}^\infty I_k.\] \[\abs{A\cap(-t,t)^c}\le\sum_{k=n+1}^\infty\ell(I_k)<(\abs A+\varepsilon)-(\abs A-\varepsilon)=2\varepsilon.\]

Now suppose $\abs A=\infty$. WLOG, suppose $A\cap\Z=\emptyset$; otherwise consider $A\backslash\Z$ instead of $A$. By repeatedly using the lemma, we can prove that

\[\abs A=\sum_{n\in\Z}\abs{A\cap(n,n+1)}.\]

Because $\abs A=\infty$, $\exists m$ s.t.

\[\sum_{n=-m}^m\abs{A\cap(n,n+1)}>C.\]

Hence

\[\abs{A\cap(-m,m)}=\sum_{n=-m}^{m-1}\abs{A\cap(n,n+1)}>C.\]

Here the first equality follows from repeatedly using the lemma.

$\square$