(Tonelli’s theorem for series over arbitrary sets). Let be sets (possibly infinite or uncountable), and be a doubly infinite sequence of extended non-negative reals indexed by and . Show that

We know that

Let with and finite and and . As the are positive, or zero, we can write

and the result follows.

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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 . For the reverse direction, I think, we need to show, along the lines of exercise 0.0.1, that .

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Hi, thank you for your comment.

The part of the proof that I skipped over was that

,

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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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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