I am studying measure theory using the lectures by Claudio Landim on youtube (https://www.youtube.com/watch?v=FOXhbYPT2tI&list=PLo4jXE-LdDTQq8ZyA8F8reSQHej3F6RFX&index=9).

Now in lecture 9 he states that if A is a Lebesgue measurable subset of $\Omega$ with finite Lebesgue outer measure, i.e. $A \in M$ s.t $\pi^*(A)\leq \infty $,

$\exists$ a set $F \in \varphi(\alpha)$ s.t. $A \subseteq F$ and $\pi^*(A)=\pi^*(F)$.

The definition of the Lebesgue outer measure I am using is

$\pi^*(A)=inf \sum \limits_{i \geq 1} \nu(E_{i})$ where $\nu$ is a $\sigma$-additive set function defined on an algebra $\alpha$ where the infimum is taken over all covers $\{E_{i}\}$ of $A$, i.e. $E_{i} \in \alpha$ and $A \subseteq \bigcup \limits_{i \geq 1} E_{i}$.


If $\pi^*(A)\leq \infty $, then by definition of the infimum $\exists$ a cover $\{E_{i}\}$ s.t.

$\pi^*(A)\leq \sum \limits_{i \geq 1} \nu(E_{i}) \leq \pi^*(A) + \epsilon$ for arbitrary $\epsilon > 0$.

My proof goes as follows:

By definition of the Lebesgue outer measure and using the observation given above

$\exists E_{i} \in \alpha$ s.t. $A \subseteq \bigcup \limits_{i \geq 1} E_{i}$ and $\sum \limits_{i \geq 1} \nu(E_{i}) \leq \pi^*(A) + \epsilon$ $\forall \epsilon > 0$.

Since $\alpha \subseteq \varphi(\alpha)$, $\pi^*(E_{i})=\nu^*(E_{i})$ and thus

$\sum \limits_{i \geq 1} \pi(E_{i}) \leq \pi^*(A) + \epsilon$ $\forall \epsilon > 0$

Now set $E = \bigcup \limits_{i \geq 1} E_{i}$ and note that $E \in \varphi(\alpha)$ by definition.

Additionally by $\sigma$-subadditivity, $\pi^*(E) \leq \sum \limits_{i \geq 1} \pi^*(E_{i})$ and so

$\pi^*(E) \leq \pi^*(A) + \epsilon$.

Since $\epsilon$ was arbitrary this proves that $\pi^*(E) \leq \pi^*(A)$.

This implies that $\pi^*(E) = \pi^*(A)$ because

$\pi^*(E) \geq \pi^*(A)$ also holds by monotonicity of $\pi^*$

This proves the claim.

Now I've got 2 questions on this.

1) Is my proof correct? 2) I did not use the condition that A is a measurable set because I'm only working with the Lebesgue outer measure. If my proof is correct we do not need this condition and the claim should hold for arbitrary subsets of a set $\Omega$. Is my conclusion correct?

  • 1
    $\begingroup$ I think what you have done is correct. I don't see why they wanted to assume measurability. $\endgroup$ – Kavi Rama Murthy Jul 12 at 8:29
  • $\begingroup$ Thanks for the quick reply @KaviRamaMurthy. $\endgroup$ – DerivativesGuy Jul 12 at 8:48

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