## IMT Exercise 0.0.2

(Tonelli’s theorem for series over arbitrary sets). Let ${A, B}$ be sets (possibly infinite or uncountable), and ${(x_{n,m})_{n\in A, m\in B}}$ be a doubly infinite sequence of extended non-negative reals ${x_{n, m}\in [0,+\infty]}$ indexed by ${A}$ and ${B}$. Show that

$\displaystyle \sum_{(n,m)\in A\times B} x_{n,m} =\sum_{n\in A}\sum_{m\in B} x_{n,m} =\sum_{m\in B}\sum_{n\in A} x_{n,m}$

We know that

$\displaystyle \sum_{(n,m)\in A\times B} x_{n,m} =\underset{F\subset A\times B, F\mbox{\footnotesize{ finite}}}{\sup}\sum_{(n,m)\in F}x_{n,m}$

Let ${F=U\times V}$ with ${U}$ and ${V}$ finite and ${U\subset A}$ and ${V\subset B}$. As the ${x_{n,m}}$ are positive, or zero, we can write

$\displaystyle \sum_{(n,m)\in U\times V} x_{n,m} =\sum _{n\in U}\sum_{m\in V} x_{n,m} =\sum_{m\in V}\sum_{n\in U} x_{n,m}$

and the result follows. ${\Box}$

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### 3 Responses to IMT Exercise 0.0.2

Hi, I have just started Terry Tao’s book on Measure theory. I found your blog to be quite interesting, as I an engineering student and am self studying mathematics. For this exercise, I have some comments.
I think we need to work a little bit more to establish the result. The definition you are using will show that $\sum_{(m,n)\in A\times B}x_{(m,n)}\le\sum_{m\in A}\sum_{n\in B}x_{(m,n)}$. For the reverse direction, I think, we need to show, along the lines of exercise 0.0.1, that $\sum_{m\in A}\sum_{n\in B}x_{(m,n)}\le\sum_{(m,n)\in A\times B}x_{(m,n)}$.

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• cjpn says:

Hi, thank you for your comment.

The part of the proof that I skipped over was that

$\displaystyle\underset{U\subset A, V\subset B, U\mbox{\footnotesize{ and }}V\mbox{\footnotesize{ finite}}}{\sup}\sum_{n\in U}\sum_{m\in V} x_{n,m}=\sum_{n\in A}\sum_{m\in B} x_{n,m}$,

giving the result. However, perhaps that is a step too far.

In view of the hint that was given I did consider dividing the proof into two parts depending on whether the sum was finite or not, but I hoped to avoid this. I’ll think about this some more, meanwhile if you want to post your own proof, that will be great.

I did edit your comment to get the latex to display correctly, please see the latex help on the wordpress site.

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2. Troy Woo says:

I agree with Samrat Mukhopadhyay that you need to prove the other way round:
$$\sum_A\sum_Bx_{n,m}\leq\sum_{A\times B}x_{n,m}$$
This can be proved for LHS being $+\infty$ or not separately (the hint was actually meant for this part). For the finite case, you first reduce $\sum_A\sum_Bx_{n,m}$ to $\sum_{A_\omega}\sum_Bx_{n,m}$ with $A_\omega\subset A$ at most countable.

Of course for each $n\in A_\omega$, $\sum_Bx_{n,m}$ is necessarily finite. This in turn leads to
$$\sum_{B}x_{n,m}=\sum_{B_{\omega,n}}x_{n,m}$$
where $B_{\omega,n}$ is at most countable. Then $B_\omega:=\bigcup_{n\in A_\omega} B_{\omega,n}$ is also at most countable. And we have:
$$\sum_A\sum_Bx_{n,m}=\sum_{A_\omega}\sum_{B_\omega}x_{n,m}\leq \sum_{A_\omega\times B_\omega}x_{n,m}\leq \sum_{A\times B}x_{n,m}$$

The more difficult part is when $\sum_A\sum_Bx_{n,m}=+\infty$. In this case, we have for any positive number $M>0$, there exists $A_F\subset A$, $A_F$ finite, such that $\sum_{A_F}\sum_Bx_{n,m}>M$. Denote the cardinality of $A_F$ by $N$, then we claim that exists $n\in A_F$ and a finite subset $B_F\subset B$ such that $\sum_{B_F}x_{n,m}>M/N$. For otherwise $\sum_{B}x_{n,m}\leq M/N$ for any $n\in A_F$ and then $\sum_{A_F}\sum_Bx_{n,m}\leq M$ which is a contradiction. Since $M$ is arbitrary, we see that there exists $A_F,B_F$ such that $\sum_{A_F}\sum_{B_F}x_{n,m}$ can be arbitrarily large.

Finally,
$$\sum_{A_F}\sum_{B_F}x_{n,m}=\sum_{A_F\times B_F}x_{n,m}\leq \sum_{A\times B}x_{n,m}$$
This makes the RHS also blow up to infinity: $\sum_{A\times B}x_{n,m}=+\infty$. And therefore
$\sum_A\sum_Bx_{n,m}\leq \sum_{A\times B}x_{n,m}$ in both cases.

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