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I am preparing for a quiz over Lebesgue Measure and the problem set I have in class is really tricky for me.

The problem I have is the following.

Let $X$ be a set and $A$ a collection of subsets of $X$ that form an algebra of sets. Suppose $l$ is a measure on $A$ such that $l(X) < \infty$. Define $\mu^\star $ as $$ \mu^\star(E) = inf \; \{\Sigma C_{i} \;\; | \;\; C_{i} \in A \; , E \subset \cup_{i=1}^{\infty}C_{i} \}. $$ Prove that a set $B$ is $\mu^\star$ measurable if and only if $$ \mu^\star(B) = l(X) - \mu^\star(B^{c}) $$

What I know so far is that $l$ is a premeasure, so that $\mu(B) = l(B)$ $\;\;$ if $B \in A$. And $\Rightarrow$ direction is quite easy because it just follows by the definition of $\mu^\star$ measurable.

Is there anybody who can help me to show the converse way?

Thanks in advance for your suggestion(s).

Block Jeong
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  • How do you define $\mu^*$ measurable in the first place? – copper.hat Sep 26 '13 at 17:57
  • As written above using infimum – Block Jeong Sep 26 '13 at 18:25
  • No, that defines $\mu^$, not what makes a set $\mu^$ measurable. So, what condition must a set satisfy to be $\mu^*$ measurable? – copper.hat Sep 26 '13 at 18:30
  • Ah! Sorry. $\mu^$ measurable is for any set $E \subset X$, $\mu^(E) = \mu(E \cap D) + \mu(E \cap D^c$) for $D \in A$ – Block Jeong Sep 27 '13 at 04:01
  • I'm confused, did you meant to have $B$ (rather than $A$) in the 'if and only if' formula above? – copper.hat Sep 27 '13 at 06:07
  • I corrected. Really sorry to bother you copper.hat. To sum up, a point I am stuck with is to prove that $\mu^(B) = \mu^(X) - \mu^(A^c)$ implies a set $B$ is $\mu^ measurable$. Thank you again for being care about this problem, copper.hat – Block Jeong Sep 27 '13 at 06:47

1 Answers1

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I think the question is this: if $(X,\mathcal M,l)$ is a finite measure space and an outer measure $\mu^*$ is defined by $$\mu^*(E) = \inf \left\{ \sum_k \ell(C_k) : E \subset \bigcup_k C_k,\ C_k \in \mathcal M \right\},$$ then a set $E$ is $\mu^*$-measurable if and only if $$\mu^*(E) + \mu^*(X \setminus E) = \mu^*(X).$$

Here is a sketch of the proof.

  1. Although $\mu^*(E)$ is defined for all $E \subset X$, it satisfies $\mu^*(E) = \ell(E)$ whenever $E \in \mathcal M$.
  2. If $E \subset X$ then there exists $C \in \mathcal M$ satisfying $E \subset C$ and $\mu^*(E) = \mu^*(C)$.
  3. If $E$ satisfies the stated assumption, then there exists $D \in \mathcal M$ satisfying $D \subset E$ and $\mu^*(D) = \mu^*(E)$. This is because, for $C \in \mathcal M$ with $X \setminus E \subset C$ and $\mu^*(X \setminus E) = \mu^*(C)$ we have $$\mu^*(E) + \mu^*(X\setminus E) = \mu^*(X) = \mu^*(C) + \mu^*(X \setminus C)$$ so that $X \setminus C \subset E$ and $\mu^*(X\setminus C) = \mu^*(E)$.
  4. With $E$ and $D$ as above, we have $\mu^*(E) = \mu^*(E \cap D) + \mu^*(E \setminus D) = \mu^*(D) + \mu^*(E \setminus D)$ because $D \in \mathcal M$. Thus $\mu^*(E \setminus D) = 0$.
  5. Again with $E$ and $D$ as above we have $\mu^*(A \cap D) = \mu^*(A \cap E)$ for every set $A \subset X$ because $$\mu^*(A \cap E) \le \mu^*((A \cap E) \cap D) + \mu^*((A \cap E) \setminus D) \le \mu^*(A \cap D) + \mu^*(E \setminus D) = \mu^*(A \cap D).$$
  6. Finally, if $A \subset X$ then $$\mu^*(A \cap E) + \mu^*(A \setminus E) \le \mu^*(A \cap D) + \mu^*(A \setminus D) \le \mu^*(A)$$ because $D \in \mathcal M$.

It follows that $E$ is $\mu^*$-measurable.

Umberto P.
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  • why is #2 true? – Akaichan Nov 21 '14 at 05:09
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    @Akaichan monotonicity is built into the definition so that if $E \subset C$ and $C \in \mathcal M$, then $\mu^(E) \le \mu^(C)$. On the other hand, for any $\epsilon > 0$ there exists a countable family ${C_k} \subset \mathcal M$ so that $$\sum_k \ell(C_k) \le \mu^(E) + \epsilon.$$ If you take $C = \cup_k C_k$ it follows that $E \subset C$, $C \in \mathcal M$ and $\mu^(C) \le \mu^*(E) + \epsilon$. Finally take a countable intersection of such sets. – Umberto P. Nov 21 '14 at 12:53