Understanding Lebesgue Measure and Rudin's Principles of Mathematical Analysis

[r4nd0m]

Hi,
I'm just reading Rudin's Principles of mathematical analysis - the last chapter on Lebesgue integration and I am having a bit trouble understanding the motivation of the definition of Lebesgue measure.

This is how I understand it:

We want to measure sets in \mathds{R}^n so what we have to do is to find some \sigma-algebra on \mathds{R}^n and to define measure on the \sigma-algebra. Now the sets, which we want to have in the \sigma-algebra are mainly intervals and their countable unions.
So we seek, and find out, that there exists such a \sigma-algebra (denoted by- \mathfrak{M} (\mu)) consisting of so called \mu-measurable sets and there also exists a regular, countably additive, nonnegative (did I forget something?) set function \mu.

Now my questions are:
1. Is there some "larger" \sigma-algebra containing \mathfrak{M} (\mu) or is \mathfrak{M} (\mu) the largest?
2. Does \mu have to be regular on \mathfrak{M} (\mu)?

 

[StatusX]

I don't know what you mean by \mathfrak{M} (\mu). You started with a sigma-algebra that you defined the measure \mu on. Is \mathfrak{M} (\mu) some different sigma algebra, like maybe the completion of the original sigma algebra with respect to the measure?

 

[mathwonk]

your specific questions seem to have nothing to do with motivation, so i do not know if you want an answer to the question of how lebesgue measure is motivated, or just how it is defined.

 

[r4nd0m]

I don't know what you mean by \mathfrak{M} (\mu). You started with a sigma-algebra that you defined the measure \mu on. Is \mathfrak{M} (\mu) some different sigma algebra, like maybe the completion of the original sigma algebra with respect to the measure?

yeah I'm sorry, it is the same sigma-algebra.

your specific questions seem to have nothing to do with motivation, so i do not know if you want an answer to the question of how lebesgue measure is motivated, or just how it is defined.

Well I just want to know, why is the Lebesgue measure so important - because if there was a larger class of measurable sets containing Lebesgue measurable sets we could use that one instead.

 

[mathwonk]

well it depends wht you are using it for. if you want to be able to assert that soemthing is true for all sets of certain kind, you want thre to be a lot of them.

but if yo have to check something is true for all of them yiou want there to be as few as possible.

so you sort of want two sets, one smaller than the ither, such that anything true for the smaller set is also true for the larger collection. so there is the concept of the completion of a sigma algebra which entails enlarging it in a trivial way.

 

[StatusX]

Start with the open intervals. The sigma algebra generated by these intervals (which also includes all open and closed sets) is called the Borel sigma algebra. We define a measure on this by assigning \mu((0,1))=1, and then extending so that the measure is invariant under translation. This is called the lebesgue measure. It can be shown that any measure that is invariant under translation is equal to some constant multiple of this measure. Finally, the sigma algebra is extended slightly to what is called the completion of the original borel sigma albegra with respect to the lebesgue measure, which just means any set which is a subset of a set with measure zero is designated to be measurable itself and is assigned a measure of zero. It can be shown that no more sets can be added to this sigma algebra and assigned a measure without invalidating one of the desired properties (eg, coutable additivity, translation invariance, etc).

 

[r4nd0m]

It can be shown that no more sets can be added to this sigma algebra and assigned a measure without invalidating one of the desired properties (eg, coutable additivity, translation invariance, etc).

Thanks, that claryfies many things to me.