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This is a follow-up question to a previous post.

For any open interval $I = (a,b)$, let $\ell(I) = b-a$ denote the length of $I$. For any set of real numbers $A$, define the following set functions: \begin{align} m^*(A) &= \inf\left\{\sum_{i=1}^\infty \ell(I_n): \bigcup_{n=1}^\infty I_n \supseteq A, I_n \text{ is an open, bounded interval for all $n$}\right\} \\ m^{**}(A)&= \inf\{m^*(\mathcal{O}): \mathcal{O} \supseteq A, \mathcal{O} \text{ open}\} \\ m^{***}(A)&= \sup\{m^*(F): F \subseteq A, F \text{ closed}\} \end{align}

Caratheodory defined a set $A$ to be measurable if for any set of real numbers $E$,

$$ m^*(E) = m^*(E \cap A) + m^*(E \cap A^C) \qquad (1) $$

Other authors define a set $A$ to be measurable if

$$ m^{**}(A) = m^{***}(A) \qquad (2) $$

I'm trying to confirm these two definitions are equivalent. I have been able to show that if $m^{***}(A) < \infty$, then (1) $\iff$ (2). However I am having trouble showing that if a set satisfies $m^{**}(A) = m^{***}(A) = \infty$, then (1) must hold. Any ideas on how to proceed?

Thanks!

David
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1 Answers1

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$m^{**}(E)=m^{***}(E)=\infty$ does not imply that $E$ is measurable (in the usual sense of Carathéodory).

Proof.

  • First some terminologies and notations. $m^{**}$ is the Lebesgue outer measure, and $ m^{***}$ is the Lebesgue inner measure. Both are defined on $\mathfrak{P}(\mathbb{R})$, the power set of $\mathbb{R}$. Let us denote the $\sigma$-algebra of all $m^{**}$-measurable set by $\mathfrak{M}_L$.

  • We need two small lemmas, which are pretty standard:

Lemma 1. There exists $E_0\subset[0,1)$ such that $E_0\notin\mathfrak{M}_L$.

Lemma 2. For every $E\in\mathfrak{P}(\mathbb{R})$, we have $m^{***}(E)\leq m^{**}(E)$.

  • Proof of Lemma 2. Let $E\in\mathfrak{P}(\mathbb{R})$. Let $C_0$ be an arbitrary closed set in $\mathbb{R}$ such that $C_0\subset E$ and let $O_0$ be an arbitrary open set in $\mathbb{R}$ such that $O_0\supset E$. Then $C_0\subset O_0$ so that $m^{*}(C_0)\leq m^{*}(O_0)$ (this is the monotonicity of the Lebesgue measure $m^{*}$). Since this hold for an arbitrary open set $O_0$ such that $O_0\supset E$, we have

    $m^{*}(C_0)\leq\inf\{m^{*}(O) : O\supset E, O \text{ open}\}=m^{**}(E)$.

    Since this holds for an arbitrary closed set $C_0$ such that $C_0\subset E$, we have $\sup\{m^{*}(C) : C\subset E, C \text{ closed}\}\leq m^{**}(E)$, that is, $m^{***}(E)\leq m^{**}(E).$

  • Consider the set $E=E_0\cup[1,\infty)$. Since $m^{***}([1,\infty))=\infty$, we have $m^{***}(E)=\infty$ by the monotonicity of $m^{***}$. Then by lemma $2$, we have $m^{**}(E)=m^{***}(E)=\infty$. To show $E\notin\mathfrak{M}_L$, first note that since $E_0$ and $[1,\infty)$ are disjoint we have $E_0=E\setminus[1,\infty)$. Since $[1,\infty)\in\mathfrak{M}_L$, if $E\in\mathfrak{M}_L$, then we would have $E_0\in\mathfrak{M}_L$, which is a contradiction. This shows that $E\notin\mathfrak{M}_L$.

Conclusion. If $m^{**}(E)=m^{***}(E)$, then $E\in\mathfrak{M}_L$ provided that $m^{**}(E)<\infty$.

user0221
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