Absolute Convergence and Fubini's Theorem

[Oxymoron]

My defintion of an absolutely convergent series is one in which you can rearrange the series and it converges to the same value, i.e.

$$\sum |x_n| < \infty$$

My question is: If one has a double series $\sum x_{m,\,n}$ which is absolutely convergent $\sum |x_{m,\,n}| < \infty$ then can I apply Fubini's Theorem to conclude that

$$\sum_n(\sum_m x_{m,\,n}) = \sum_m(\sum_n x_{m,\,n})$$

I ask this because when we covered Fubini's Theorem in class we spoke of being able to change the order of integration. So I thought, since integration is similar to sums, why can't we change the order of a sum using Fubini's Theorem?

 

[Oxymoron]

The problem I have is this: Fubini's Theorem (and Tonelli's Theorem - which might come in handy) only apply if you are working in a measure space. So I would have to have a set $X$ and consider some sigma-algebra $\mathcal{A}$ consisting of all subsets of $X$. To this I can define the counting measure on this sigma-algebra by setting

$$\mu(E) = |E|$$

if E is a finite subset of the sigma-algebra and

$$\mu(E) = \infty$$

if E is an infinite subset. Then call $(X,\mathcal{A},\mu)$ a measure space. This is the counting measure as I understand it. I can now define the measure space on the natural numbers as $(\mathbb{N},\mathcal{A},\mu)$, where $\mu$ is the counting measure on the naturals. Then $L^p(S)$, (where S is the measure space with the natural counting measure) consists of those sequences $x = \{x_n\}$ for which

$$\|x\|_p = \left(\sum_{i=1}^{\infty}|x_i|^p\right)^{1/p}$$

Now I believe I am in a position to start using Fubini's and Tonelli's theorems, right?

 

[StatusX]

Yes it's true. You can get it from the integral version by considering the counting measure on a countably infinite set.

 

[Oxymoron]

I don't exactly see how I could prove this. I mean, Fubini and Tonelli apply to integrals and I want to use these theorems to prove a result for sums. Does this mean I have to turn the sums into integrals and then back again?

 

[StatusX]

I don't know what version of the theorem you have, but there is a version that applies to the Lebesgue integral for any sigma finite measure. If your sets are both, say, the natural numbers, and the measures are the counting measure (where the size of a set is the number of elements it has), then you have:

$$\int d\mu f(n) = \sum_{n=1}^\infty f(n)$$

The product of two counting measures is the counting measure on the product space, so you can transform all the integrals that appear in the theorem into sums.

 

[Oxymoron]

This is my version:

TONELLI'S THEOREM

Suppose $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$ are $\sigma$-finite measure spaces, and $f\,:\,X \times Y \rightarrow [0,\infty]$ are $\mathcal{A}\times\mathcal{B}$-measurable. Then there are measurable functions $g\,:\,X\rightarrow[0,\infty]$ and $h\,:\,Y\rightarrow[0,\infty]$ defined by

$$g(x):=\int_Y f(x,y)\mbox{d}\nu(y)$$

$$h(y):=\int_X f(x,y)\mbox{d}\mu(x)$$

then we have

$$\int_X f\mbox{d}\mu = \int_{X \times Y}f\mbox{d}(\mu \times \nu) = \int_Y h \mbox{d}\nu$$

FUBINI'S THEOREM

Suppose $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$ are $\sigma$-finite measure spaces, and $f\,:\,X \times Y \rightarrow [0,\infty]$ are $(\mu \times \nu)$-integrable. Then there are integrable functions $g\,:\,X\rightarrow[0,\infty]$ and $h\,:\,Y\rightarrow[0,\infty]$ defined by$$g(x):=\int_Y f(x,y)\mbox{d}\nu(y) \quad \mu-a.e.$$

$$h(y):=\int_X f(x,y)\mbox{d}\mu(x) \quad \nu-a.e.$$

then we have$$\int_X f\mbox{d}\mu = \int_{X \times Y}f\mbox{d}(\mu \times \nu) = \int_Y h \mbox{d}\nu$$

 

[StatusX]

Right, so in particular, it applies with the space being the natural numbers and the measure being the counting measure. Do you see how this gives you the result for sums?

 

[Oxymoron]

Unfortunately, no. I want to turn the integrals in Fubini's theorem into sums but I am not sure I can do it.

 

[StatusX]

Ok. Then, to review, to define the Lebesque integral, you first define the integral of the characteristic function of a set to be the measure of the set. You then extend this by linearity to simple functions. Then you take as the integral of a non-negative measurable function f(x) the sup of the integrals of non-negative simple functions that are everywhere less than f. Finally, you look separately at the positive and negative, and then real and imaginary, parts of a general complex function.

So how does this go when your set is the natural numbers N and your measure is the counting measure? Well, a set with one element (eg, {5}) has measure 1, so the integral of the characteristic function of such a set is 1, and so the integral of a sequence with just one non-zero term is that term. Extend by linearity, you see that the integral of a finite sequence (which is a simple function) is the sum of the terms in the sequence. It's not hard to show that taking the sup as above extends this to infinite series.

 

[Oxymoron]

So if you have a 1 element set (these sets consist of elements from $\mathbb{N}$ right?) then you are saying that

$$\int_{\{a\,:\,a\in\mathbb{N}\}}\chi\mbox{d}\mu_C = 1$$

that is, the Lebesgue integral of the characteristic function over a 1-element set is one. And this is because the counting measure, $\mu_C$, says that the measure of any 1-element set is one.

But I don't see how the following quote follows from this:

Posted by StatusX:

...and so the integral of a sequence with just one non-zero term is that term.

A sequence of what exactly? A sequence of characteristic functions? A sequence of natural numbers?

If $f$ is simple, then

$$f = \sum_i a_i\chi_{A_i}$$

right? Where $A_i$ is some collection of natural numbers (in fact, I think it is a measurable subset of natural numbers!). Then, if we use the counting measure on the naturals we can define

$$\int f\mbox{d}\mu_C = \sum_i^n a_i \mu(A_i)$$

which will be infinite if $A_i$ is infinite, or, if the subset is finite, then the integral will be simply the sum of the number of elements in $A_i$.

So then

$$\int_{E_n}f\chi_{E_n}\mbox{d}\mu_C = \int_{E_n}\sum_i a_i\chi_{A_i \cap E_n}\,\mbox{d}\mu = \sum_i a_i\mu(A_i \cap E_n)$$

 

[StatusX]

Sorry, by "sequence" I mean a function from the natural numbers to the real or complex numbers.

 

[Oxymoron]

The way you define a simple function is like this

$$f = \sum_{i=1}^n a_i\chi_{A_i}$$

where $a_i \in \mathbb{N}$ and $A_i$ is a measurable subset of $\mathbb{N}$, right?

Are you treating $\int f$, where f is simple, like a sum of a sequence of natural numbers? I mean, the integral of a simple function is defined as

$$\int f\mbox{d}\mu_C = \sum_{i=1}^n a_i\mu_C(A_i)$$

which is a sum of a sequence of natural numbers. And so when you said

and so the integral of a sequence with just one non-zero term is that term.

you meant that the integral of a simple function is the sum of a sequence of natural numbers.

 

[StatusX]

The sequences consist of real (or complex) numbers, not natural numbers. They are functions from the natural numbers to the real numbers, which just means they are indexed by N (ie, (a_1, a_2,...), where the a_n are real).

Now, every subset of N is measurable with the counting measure. So the following sequences are simple functions on this measure space:

1) (0,3,0,0,0...)
2) (1,2,3,2,1,0,0,0,...)
3) (1,0.5,0,5,1,1,1,...).

The integral of the first is 3 times the measure of {2}, which is 3*1=3. The integral of the second is 1 times the measure of {1,5} plus 2 times the measure of {2,4} plus 3 times the measure of {3}, or 1*2+2*2+3*1=9. It should be obvious that any finite sequence is a simple function, and its integral is its (finite) sum.

The integral of the third is 1 times the measure of {1,4,5,6,...} plus 0.5 times the measure of {2,3}, which is infinite because of the first term, and so this function is not integrable.

The sequence (1,1/2,1/4,1/8,...) is not simple, precisely because its range is not a finite set, but it can be approximated from below by the sequences (1,0,0,0,...),(1,1/2,0,0,...),(1,1/2,1/4,0,...), which are simple and whose integrals converge to 2, the sum of the infinite series.

 

[Oxymoron]

I appreciate your help StatusX but I still don't know how to start the justification. I have to prove that

$$\sum_n(\sum_m x_{m,\,n}) = \sum_m(\sum_n x_{m,\,n})$$

by applying Tonelli's and Fubini's theorems.

The first thing I would do is define my measure space. This is easy because it is simply the set of natural numbers, together with the counting measure which says that the measure of a subset is the number of elements within.

Now, for suitable functions, we have the following analogy:

$$\int \left( \int f(x,y)\,\mbox{d}\mu(x)\right)\mbox{d}\nu(y) = \int \left( \int f(x,y)\,\mbox{d}\nu(y)\right)\mbox{d}\mu(x)$$

which looks just like what I have to prove except that there are integral signs instead of summation signs, and functions instead of series.

If I could show that the Lebesgue Integral is equivalent to a summation and that the function $f(x,y)$ defined as $f\,:\,X\times Y \rightarrow \mathbb{R}$ is equivalent to a series of real (or complex) numbers $x_{m\,n}$. Is this basically what I have to do? Because once I do that then Fubini's Theorem gives the desired result instantly. I just haven't caught onto the concept of being able to prove the equivalencies yet. :(

 

[StatusX]

I'm not sure which part you're not seeing. What I showed above was that if a_n is a sequence, then there is a set (the natural numbers), a measure (the counting measure) and a measurable function in this measure space (f(n)=a_n) such that:

$$\int f d\mu = \sum_{n=1}^\infty a_n$$.

So, by working in this measure space, applying the theorem, and then using the above formula to rewrite the integrals in terms of sums, you get the result for sums. I'm really just repeating myself at this point. If you still don't understand, maybe you should go back and read what I wrote above again.

 

[Oxymoron]

Actually it all makes perfect sense now! Admittedly I did go back an re-read everything you said, but I can now rewrite the integrals as sums and everything. Apply Fubini and Tonelli and get my result which is what I wanted. Thankyou so much for your help and guidance.