In the book what I've read, there is one point where the author suggest to begin the proof of the Fubini's Theorem for infinite sum in the case when is non-negative after this try to generalize.

But I really have a bad time trying to understand the entire problem for various reason: number one I'm not absolutely sure if what I've done is correct and second I have no idea of how to pass in the general case. Any suggestion, advice whatever to deal with the general case it would be great. Thanks.

Fubini's Theorem for Infinite series: Let $f: \mathbb{N} \times\mathbb{N} \rightarrow \mathbb{R} $ be a function such that $\sum _{(n,m)\in \mathbb{N} \times\mathbb{N}} f(n,m) $ is absolutely convergent. Then we have:

$\sum _{n=0}^{\infty}\sum _{m=0}^{\infty} f(n,m) =\sum _{(n,m)\in \mathbb{N} \times\mathbb{N}} f(n,m)=\sum _{(m,n)\in \mathbb{N} \times\mathbb{N}} f(n,m)= \sum _{m=0}^{\infty}\sum _{n=0}^{\infty} f(n,m)$

Proof: We may assume that for each $n,m$, that the function is non-negative, i.e., $\forall n,m \in\mathbb{N}.f(n,m)\ge0 $.

We set $L:= \sum _{(n,m)\in \mathbb{N} \times\mathbb{N}} f(n,m) $, and we want to show that $\sum _{n=0}^{\infty}\sum _{m=0}^{\infty} f(n,m)$ converges to $L$. Let $X\subset \mathbb{N^2}$ and suppose that $X$ is finite.

We claim that $\sum _{(n,m)\in X} f(n,m) \le L$. Let $g: \mathbb{N^2}\rightarrow \mathbb{N}$ be a bijective map. If we restrict the map to $X$, clearly $g[X]$ is finite, hence bounded. Since $L$ is the sup of the sequences of partial sums $\sum _{(n,m)\in \{0...N\} \times\{0...M\}} f(n,m)$ which is monotonic increasing. Thus is at most $L$.

First we have to show that $\sum _{n=0}^{\infty}\sum _{m=0}^{\infty} f(n,m)$ converges. It will suffice to to show that $\sum _{n=0}^{N}\sum _{m=0}^{\infty} f(n,m)\le L$ for any $N$. Since $\sum _{m=0}^{M} f(n,m) \rightarrow \sum _{m=0}^{\infty} f(n,m)$, $\sum _{n=0}^{N}\sum _{m=0}^{\infty} f(n,m)$ is the limit of $\sum _{n=0}^{N}\sum _{m=0}^{M} f(n,m)$ as $M \rightarrow \infty$. Then, by the monotonic convergence theorem it will suffice to show that $\sum _{n=0}^{N}\sum _{m=0}^{M} f(n,m)\le L$ for any $M$.

But, $\sum _{n=0}^{N}\sum _{m=0}^{M} f(n,m) = \sum _{(n,m)\in \{0...N\} \times\{0...M\}} f(n,m)$ by the Fubini's Theorem for finite series and since $\{0...N\} \times\{0...M\} \subset \mathbb{N^2}$ is a finite subset. the result follows for what we said above.

Now let $\varepsilon >0$ be given. Then, $L -\varepsilon$ cannot be the sup of the sequence of partial sum and we can then find a finite set $ X \subset \mathbb{N^2}$ such that $\sum _{(n,m)\in X} f(n,m) > L-\varepsilon$. Since $X$ is assumed to be finite then there is contained in some set $ \{0...N\} \times\{0...M\}$. Thus for any $N'\ge N,M\ge M'$ we have $\sum _{n=0}^{N'}\sum _{m=0}^{M'} f(n,m) = \sum _{(n,m)\in \{0...N'\} \times\{0...M'\}} f(n,m)\ge L-\varepsilon$, and hence since we already know that the sum exists

$\sum _{n=0}^{\infty}\sum _{m=0}^{\infty} f(n,m)\ge\sum _{n=0}^{N'}\sum _{m=0}^{\infty} f(n,m)\ge L-\varepsilon$.

Note: I already proved the Fubini's Theorem for finite series using induction.

Also I think we can argue as follows.

Let $g: \mathbb{N^2}\rightarrow \mathbb{N}$ be any bijective map. Since $ \sum _{(n,m)\in \mathbb{N} \times\mathbb{N}} f(n,m) $ converges absolutely then any rearrangement give us the same result. Thus $\sum _{i=0}^{\infty} f(g^{-1}(i)) = L$. Also the function is non-negative, so the sequence of partial sums converges to its supremum, i.e., $L=\text{sup}_{N\ge 0}\sum _{i=0}^{N} f(g^{-1}(i))$.

We claim that if $X$ is a finite subset $\mathbb{N^2}$. Then $\sum _{(n,m)\in X} f(n,m) \le L$.

The sequence $\big(g(n,m)\big)_{(n,m)\in X} $ is finite and hence bounded. Let $N$ be an upper bound, i.e., $g(n,m)\le N$ for any $(n,m)\in X$. We set $\{ i\in \mathbb{N}: i\le N \}$. Then $g [X] \subset \{ i\in \mathbb{N}: i\le N \}$.

We have:

$\sum _{(n,m)\in X} f(n,m)=\sum _{n\in g[X]} f(g^{-1}(n)) \le \sum _{n\in \{ i\in \mathbb{N}: i\le N \}} f(g^{-1}(n))=\sum _{n=0}^{N} f(g^{-1}(n))\le L$. Thus $\sum _{(n,m)\in X} f(n,m) \le L$ as desired.

Let $n, M\in\mathbb{N} $ be arbitrary. Then $\sum _{(i,j)\in \{n\} \times \{0....M\}} f(i,j)=\sum _{m=0}^{M} f(n,m) \le L$ for each $M\ge 0$. Thus $\sum _{m=0}^{M} f(n,m)$ converges and is at most $L$, i.e., $\sum _{m=0}^{\infty} f(n,m)\le L$ for a fixed $n$.

We claim that $\sum _{n=0}^{N}\sum _{m=0}^{\infty} f(n,m)\le L$ for any $N\in \mathbb{N}$. We argue by contradiction suppose that there is some $N\ge 0$ such that $\sum _{n=0}^{N}\sum _{m=0}^{\infty} f(n,m)> L$. For the sake of simplicity let $S_{N,\infty}:=\sum _{n=0}^{N}\sum _{m=0}^{\infty} f(n,m)$, and $S_{N,M}:=\sum _{n=0}^{N}\sum _{m=0}^{M} f(n,m)$.

We may assume that $S_{N,\infty}> L$. Let us fix some $\varepsilon>0$ such that $\varepsilon< S_{N,\infty}-L$. Since $S_{N,M}\rightarrow S_{N,\infty}$. Let some sufficient large $M$ such that $S_{N,\infty}-S_{N,M}\le \varepsilon<S_{N,\infty}-L$.

Then $L<S_{N,M} = \sum _{n=0}^{N}\sum _{m=0}^{M} f(n,m) =\sum _{(n,m)\in \{0...N\} \times\{0...M\}} f(n,m)\le L$ [since $\{0...N\} \times\{0...M\}$ is a finite subset of $\mathbb{N^2}$] a contradiction. This means that $S_{N,\infty}\le L$ for each $N\ge 0$ and hence $S_{\infty,\infty}\le L$.

Let $X\subset \mathbb{N^2}$ which is finite. Then $X\subset \{0...N\}\times \{0...N\}$. So $\sum _{(n,m)\in X} f(n,m)\le \sum _{(n,m)\in \{0...N\} \times\{0...N\}} f(n,m)=\sum _{n=0}^{N}\sum _{m=0}^{N} f(n,m)\le S_{\infty,\infty}$. Since this hold for any finite set of $\mathbb{N^2}$, then $L\le S_{\infty,\infty}$. Hence $L = S_{\infty,\infty}$

Is this correct? Any suggestion for the general case (the sign case), please. What about if we show that $f(n,m)$ can be express as a part which is just positive and other which is negative, with this almost done, right? What do you think about it?


  • $\begingroup$ Isn't it true in general that $\sum_x\sum_y f(x,y)=\sum_y\sum_xf(x,y)$? (even for infinite sums) $\endgroup$ – Ragnar Dec 22 '13 at 1:14
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    $\begingroup$ @Ragnar No. ${}{}{}{}$ $\endgroup$ – Pedro Tamaroff Dec 22 '13 at 1:21
  • $\begingroup$ @PedroTamaroff, that's interesting, could you give an easy example for which equality does not hold? $\endgroup$ – Ragnar Dec 22 '13 at 1:25
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    $\begingroup$ @Ragnar: $f(n,m)=1$ when $n=m$, and $f(n,m)=-1$ when $n=m+1$ and $0$ otherwise $\endgroup$ – Jose Antonio Dec 22 '13 at 1:30
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    $\begingroup$ Here you can find a proof I think is great.- $\endgroup$ – Pedro Tamaroff Dec 22 '13 at 1:33

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