Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Perhaps this has a simple answer, but I don't know (I wouldn't be asking if I did). Every set of outer measure zero is Lebesgue measurable with Lebesgue measure zero. Is the converse true? That is, if a set has Lebesgue measure zero, does it necessarily have outer measure zero?

And I suppose I should specify that I'm thinking in $\mathbb{R}^n$.

share|improve this question
2  
What definition(s) of measure are you using? –  Aryabhata Oct 20 '10 at 16:24

1 Answer 1

up vote 8 down vote accepted

If $\mu^{*}$ is an outer measure, then we define the measurable sets in terms of $\mu^{*}$; the measure is then defined to be the restriction of the outer measure to the measurable sets.

To be more explicit: if you have a $\sigma$-algebra and an outer measure $\mu^{*}$ on the algebra, then we say that a set $E$ is $\mu^{*}$-measurable if and only if for every $A$ in the $\sigma$-algebra, $$\mu^{*}(A) = \mu^{*}(A\cap E) + \mu^{*}(A\cap E').$$

As Halmos says in his book Measure Theory,

It is rather difficutl to get an intuitive understanding of the meaning of $\mu^{*}$-measurability except through familiarity with its implications.

Once you have the definiiton of $\mu^{*}$-measurable, then let $S$ be the set of all measurable sets, and you define the measure $\mu$ on $S$ by $\mu(E) = \mu^{*}(E)$ for all $E\in S$.

In particular, this holds for the Lebesgue measure: if $E$ is Lebesgue measurable, then the Lebesgue measure of $E$ is equal to the outer measure of $E$, because the Lebesgue measure of $E$ is defined to be the outer measure of $E$. This holds for any Lebesgue measure, not just for measure $0$.

The point of the theorem you state earlier is that having outer measure zero implies that the set is measurable; that's the nontrivial part of the statement (not that the Lebesgue measure of the set will then be zero).

share|improve this answer
    
+1: But, is it possible that some 'strange' definition of measure is being used? –  Aryabhata Oct 20 '10 at 16:24
    
@Moron: Not if he is talking about the "Lebesgue measure"... –  Arturo Magidin Oct 20 '10 at 16:27
    
Makes perfect sense, thank you –  Bey Oct 20 '10 at 17:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.