# Measure Theory Question

1. Sep 25, 2008

### Thorn

I was told that you can find a disjoint sequence of sets...say {Ei} such that

m*(U Ei) < Σ m*(Ei).. That is the measure of the union of all these sets is less than the sum of the individual measure of each set.... This is obvious if the sets aren't disjoint...But can someone give me an example of this? Thanks.

2. Sep 25, 2008

### morphism

Is m* Lebesgue outer measure? If so you will need to use nonmeasurable sets, because equality always holds for measurable sets.

3. Sep 25, 2008

### Thorn

yeah... m* is the Lebesgue outer measure. So, you can say things like m*(E) when E isn't even measurable!? I didn't think m would be defined from a non measurable set...

4. Sep 25, 2008

### morphism

That's the point of using outer measure: it's defined for every subset of R.

5. Sep 25, 2008

Yeah, it doesn't make much sense to talk about the measure of a non-measureable set. You could come up with ways to interpret the inequality in that case, but the result would be pretty vacuous, unless I'm missing something.

I'd always understood sigma-additivity (i.e., sum of a the measures of a countable set of disjoint subsets = the measure of the union of the subsets) to be part of the definition of any measure.

6. Sep 25, 2008

### Thorn

So I take it, then there isn't an example of this...even for non-measurable sets..?

7. Sep 25, 2008

### morphism

The Lebesgue outer measure is not a measure for this very reason.

8. Sep 25, 2008

### Thorn

Ha....well then it seems that EVERY non measurable set would be an example of

m*(U Ei) < Σ m*(Ei)..

9. Sep 26, 2008

### morphism

What do you mean? What are you taking as your disjoint sets?

10. Oct 4, 2008

### WHOAguitarninja

Be careful, as morphism is saying, what you are talking about is the outer measure, NOT the measure. A non measurable set still has outer measure.

In any case, have you looked at the Vitali non measurable set? You may be able to construct an example if you consider that.

11. Dec 2, 2008

### vigvig

Like someone mentioned above, the set E has to be non lebesgue measurable. Proof by contradiction. Remember that m* is the outer measure, while m is the Lebesgue measure (by assumption). Assume that U= disjoint union of (E_i) such that m*(U)<sum_i(m*E_i). Assuming U is measurable, m(U)=m*(U). Since the set of all measurable sets of R is a sigma algebra, you can easily prove that each E_i must belong to the sigma algebra and are therefore measurable. For each i, then m*(E_i)=m(E_i).Thus you have m(U)< sum_i(m(E_i)), which violates one of the major property of measure (not outer measure though). So U must not be measurable.

Now, back to your question, it's hard to visualize non measurable subset of R. In fact "m(U)< sum_i(m(E_i))" is the property that was used to construct a non measurable subset of R by using R/Q.

Vignon S. Oussa

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