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My book states that the Lebesgue measure is complete, but does not give a proof. Is the proof difficult?

What I know about the Lebesgue Measure_

It contains

  1. $\emptyset$

  2. All finite half open intervals $(a,b]$

  3. All infinite intervals $(\infty,b]$, and $(a,\infty)$.

  4. I also know that it contains the Borel sigma algebra, that is the sigma algebra generated by all open sets.

Here is how I try to prove it but I get stuck:

I need to prove that if $\mu(A)=0$, and $B \subseteq A$. Then B is also in the sigma algebra.

I start with the set $A$ and assume it is 0, and $B$ is any subset of $A$. Since the measure of $A$ is zero, I know that it can not contain any type of interval, because then the measure would be not 0. But $A$ can consist of points. Here is where I get stuck.

If I can prove that $A$ only can consist of a countable number of points, then it is easy to see that $B$ can also be only a countable number of points. And a each point is in the sigma-algebra, and a countable union is also in the sigma algebra.

But how do I know that $A$ only can consist of a countable number of points?(Or maybe it can be larger, how do I prove it then?)

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It depends on what definition you use.

Using the Caratheodory criterion you can show that every set with outer measure zero is measurable.

Alternatively, a set $E$ is measurable, by definition, if for all $\varepsilon>0$ there exists an open set $U\supset E$ such that $m^*(U \setminus E)<\varepsilon$. Now this will easily imply completeness.

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  • $\begingroup$ Yeah, my book uses the first definition. Thank you I will try to use that. $\endgroup$ – user119615 May 12 '14 at 0:07

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