Measure theory in R^n and in abstract spaces

[WWGD]

Hi:

I am trying to review the way L^p spaces are treated differently

in Royden. In Ch.6, he treats them under "Classical Banach Spaces",

and then again, in his Ch.11 , under "Abstract Spaces".

This is what I understand: (Please comment/correct)

In the case of abstract spaces, we do not deal with R^n,

or subsets of R^n, where the idea of measure is "natural" , and

where we start by defining the measure of an open "box" ( a

subset bounded by open intervals (a1,b1),(a2,b2),..,(an,bn))

which we define to have measure (a1-b1)(a2-b2)...(an-bn)

and build our measure around this, by defining the measure of a set

using the infimum of coverings by these boxes.



In abstract spaces, instead of R^n, we choose any topological space,

then define a sigma algebra , like, e.g., the Borel algebra generated by

open sets, or one generated by the compact subsets, etc. , which we

define to be the collection of measurable sets.

Then we define measurability in terms of being a member of

the sigma algebra, and , given abstract spaces X= (X,s_X,mu_X), Y=(Y,s_Y,mu_Y) , with

s_X,s_Y sigma algebras and mu_X,mu_Y measures (satisfying the basic

axioms ;subadditivity, monotonicity eyc.) , we then define f:X-->Y ; X,Y as above

is measurable, if for U_y in s_Y (the sigma -algebra in Y ), f^-1(U_Y) is in s_X (the

sigma-algebra of X).



What I am not clear on, is the issue of integrability: in abstract

spaces, we integrate against a measure. How is this different from

Lebesgue integration, where we do a sum :Sumf(y)( mu_X (x):f(x)=y)

i.e., multiply any value f(y) in the range by the measure of its preimage?.

It seems to be the same?.

Question:

Is any of these abstract spaces , or measure triples (X, s_X,mu_X) a Hilbert space,

or are they all Banach spaces?. Are there also non-measurable subsets in these

abstract spaces?.

Thanks for Comments/Corrections/Examples/Refs

 

[yyat]

In abstract spaces, instead of R^n, we choose any topological space,

A measure space (a set with sigma algebra and measure) does not need to be a topological space (elementary measure theory requires no topology).

What I am not clear on, is the issue of integrability: in abstract

spaces, we integrate against a measure. How is this different from

Lebesgue integration, where we do a sum :Sumf(y)( mu_X (x):f(x)=y)

i.e., multiply any value f(y) in the range by the measure of its preimage?.

It seems to be the same?.

The integral of real or complex valued functions can be defined for any measure space, in the case of the Borel sigma algebra on R^n it is Lebesgue's integral. For this reason the general version of this integral for arbitrary measure spaces is also often called the Lebesgue integral.

Is any of these abstract spaces , or measure triples (X, s_X,mu_X) a Hilbert space,
or are they all Banach spaces?

I think you are confusing measure spaces with L^p spaces. For any measure space X one can define L^p(X), these are Banach spaces and L^2(X) is also a Hilbert space.

Are there also non-measurable subsets in these

abstract spaces?.

Yes, all subsets not in the sigma-algebra (by definition).

 

[WWGD]

Thanks For Your Comments, YYAT:

A measure space (a set with sigma algebra and measure) does not need to be a topological space (elementary measure theory requires no topology).

Thanks. What I meant is that we can define a collection of measurable sets
in any set , without an "intuitive" construction, like that of starting with an
interval (a,b) , whose measure is intuitively clearly b-a, since it is what we
commonly understand as distance .
What I meant by having a top. space is that , unlike the case of R^n, (and many other spaces), we do not need a metric to replicate the idea of m(a,b)=b-a ; we can say that b-a
is the distance between a and b, which does not exist in a non-metric (metrizable)
space. What I mean is that the notion of measure in non-metrizable spaces does
not ( or cannot) agree with some intrinsic property of the space, as m(a,b)=b-a
agrees with the standard Euc. metric in R^n.




The integral of real or complex valued functions can be defined for any measure space, in the case of the Borel sigma algebra on R^n it is Lebesgue's integral. For this reason the general version of this integral for arbitrary measure spaces is also often called the Lebesgue integral.


Thanks. It just seemed the case that in all these cases, the integral came down
to doing a sum (e.g., like a Lebesgue sum , where the integral of f is given by
Sum ( y ( m(x): f(x)=y ) ) , so that the value of the integral in all these cases
was of this type. But this point was never explicitly made .


So a question was:

Is it true that integration in abstract measure spaces is always done by a sum of this type, i.e.:

Sum ( y=f(x)( m(x): f(x)=y))

i.e., as a weighted sum of the images y=f(x) by the measure of the preimage of f(x)?


Sorry, this part below got mixed up:

I think you are confusing measure spaces with L^p spaces. For any measure space X one can define L^p(X), these are Banach spaces and L^2(X) is also a Hilbert space.


Yes, sorry, this is what I meant, that given a measure triple , we turn it into a measure
space this way, by using the L^p -norm Int_S ( f^p )^1/p , and considering those
that are finite.
.

"are there non-measurable sets"

Yes, all subsets not in the sigma-algebra (by definition).

Do we ever work with P(X) as the sigma-algebra on X, to have all subsets be measurable?

I know we cannot do this with the Lebesgue measure.


Thanks again.

 

[yyat]

Is it true that integration in abstract measure spaces is always done by a sum of this type, i.e.:

Sum ( y=f(x)( m(x): f(x)=y))

i.e., as a weighted sum of the images y=f(x) by the measure of the preimage of f(x)?

This formula only works for "simple functions" (linear combinations of indicator functions of measurable sets). For general measurable functions one uses approximation by simple functions to define the integral, as explained in http://en.wikipedia.org/wiki/Lebesgue_integral#Integration".

Do we ever work with P(X) as the sigma-algebra on X, to have all subsets be measurable?

I know we cannot do this with the Lebesgue measure.

Take, for example, the counting measure on Z. Note that the measure is really a set theoretic concept in this case (cardinality).