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}

This entry was posted in IMT Exercises, Measure Theory. Bookmark the permalink.

3 Responses to IMT Exercise 0.0.2

  1. Samrat Mukhopadhyay says:

    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)} .


    • 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.


  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
    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.

    \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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