# Lebesgues's Sigma Algebra

1. Nov 28, 2005

### Palindrom

I'm now studying Lebesgues measure on R^n, and when we finished constructing L (all measurable sets), we moved on to Borel sets and so on.

My question is, is there is no way of extending the Lebesgues measure on a larger family than L? Can it be done while still saving the nice properties of the Lebesgues measure (invariance under isometries, at the least)?

If so, how, and if not, why (as in proof)?

2. Nov 28, 2005

### matt grime

It is a theorem that the Borel measure on R^n is the unique lebesgue measure on R^n satisfying u([a,b]) = b-a, so whilst there may be other possible measures on the space they will not give the measure of the basic sets as what you want. Of course, you're free to define other measures too. The thing is that the sets (a,b] are a base for the borel sets on R (obvious generalization to R^n) and so specifying the measure on them fixes the measure.

So, the answer is (i'm almost sure), yes there are other measures, but no they're not natural and don't measure (a,b] correctly.

But i'm sure there is a different way to read your request. Perhaps you could elaborate.

Last edited: Nov 28, 2005
3. Nov 28, 2005

### fourier jr

if u:A->R+ is a measure on a ring A then hahn's extension theorem (NOT the hahn-banach theorem) allows a sigma-additive extension u' onto the sigma-ring generated by A. not only that, if u is sigma-finite on A then u' is also sigma finite & that extension is unique. not sure if that's exactly what palindrom was looking for but it sounds close.

Last edited: Nov 28, 2005
4. Nov 29, 2005

### Palindrom

While you are both saying interesting things, they're not answering my question. :)

I have a function $$\lambda :L \to \left[ {0,\infty } \right]$$ (when L is the Lebesgues sigma Algebra) called the Lebesgues measure. I'm asking wether there exists a function $$\mu :S \to \left[ {0,\infty } \right]$$, s.t.:
1. $$\mu$$ is a measure on R^n.
2. S strictly contains L.
3. $$\mu \left| {_L = \lambda } \right.$$

If the answer is yes, can I add the requirement that $$\mu$$ be invariant under isometries?

Last edited: Nov 29, 2005
5. Nov 29, 2005

### matt grime

And as we both imply, that would depend upon what L is already.

6. Nov 30, 2005

### Palindrom

What do you mean? L is the Lebesgue sigma Algebra: The set of all sets B for whom for all sets A s.t.:
1. A is of finite external measure.
2. A's inner and external measures are equal.
The set (B intersection A) keeps 1. and 2.
I thought it was widely common to make that construction before talking about Borel sets.
So now, do you have an answer?

7. Nov 30, 2005

### matt grime

The sets that are measurable are determined by the measure, so no, you can't extend L since L is the set of all measurable sets. Once you specify that u((a,b]) is b-a, that fixes the meaure and the set of all measurable sets with respect to that measure. It's like asking if you can make the discrete topology any finer.

After all, given the basic sets (a,b] in R we can take the power set of them, which is the smallest sigma algebra containing these elements. This is not the set of all lebesgue measurable sets but is a sigma algebra. You yourself required that L is the set of all measurable sets, thus it cannot be extended.

Last edited: Nov 30, 2005
8. Nov 30, 2005

### fourier jr

what about the caratheodory process? the version in royden's book (chap 12) goes like this: "let u be a measure on an algebra A & u* the outer measure induced by u. the the restriction $$\overline{u}$$ of u* to the u*-measureable sets is an extension of u to a sigma algebra containing A"

9. Nov 30, 2005

### matt grime

Hmm, but we're already talking about the largest possible sigma algebra containing the basic (a,b] so we cannot get anything else.

10. Dec 1, 2005

### Palindrom

I defined L using inner and outer measure. What you are saying, is that the largest sigma algebra containing the open sets with a measure function, specified on open sets, defined on it, must be L. It might be trivial to you, but I don't see why that's true.

In other words: Define S to be the largest sigma algebra containing special boxes, on which one can define a measure that is consistent with the volume of one said box. Can you prove that if A is in S, then it is in L- i.e., it keeps the requirements that i mentioned earlier (inner and outer measure)?

11. Dec 1, 2005

### Palindrom

That is, if S even exists.

12. Dec 1, 2005

### matt grime

You stated that L is the set of *all* measurable sets, not me. I don't think it is trivial or not trivial, I have no opinion on the subject as I know nothing about it and am merely trying to piece together the definitions consistently.

13. Dec 1, 2005

### Palindrom

Measurable not as in "one can define its measure", but as in "we have defined its measure".

I've defined L above with inner and outer measure, and I want to know if I can expand the measure on a larger family than L. Is the question now clear?

14. Dec 1, 2005

### matt grime

If L is the set of all sets for which you "have defined the measure" we'd need to know for which ones you chose to define them, wouldn't we? If you chose to define the measure for all sets for which one may define the measure then of course you cannot extend, and if you chose not to, then of course you can extend it. So, how did you chose those sets exactly? Personally, I have no knowledge of measure theory so cannot say if the set of all sets for which you "have defined" it is the same as all sets for which you "may define" it (note it took you until post 6 to clairfy how you chose to define L)

15. Dec 1, 2005

### Palindrom

Let's stop going in circles. I clarified exactly how I define L. I didn't think I had to in the beginning, because it is a well known construction in every basic measure theory class.