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I am working on some exercises in Folland's real analysis. In number 2.48, they ask you to prove the following question: Let $X = Y = \mathbb N$, $M = N = P(\mathbb N)$ and $\mu = \nu$ be counting measure on $\mathbb N$. Define $f(m,n) = 1$ if $m=n$, $f(m,n) = -1$ if $m = n+1$, and $f(m,n) = 0$ otherwise. Then, $\int\int f d\mu d\nu$ and $\int\int f d\nu d\mu$ exist and are unequal.

It seems to me that $\int\int f d\mu d\nu = \sum_n\sum_m f(m,n) = \sum_n f(n,n) + f(n+1,n) = \sum_n 1-1 = \sum_n 0 = 0$ and $\int\int f d\nu d\mu = \sum_m\sum_n f(m,n) = \sum_m f(m,m) + f(m,m-1) = \sum_m 1-1 = \sum_m 0 =0$. So the two integrals are equal.

What am I doing wrong? Thanks!

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Presumably, you don't mean $\sum_n\sum_m f(n,n)+f(n+1,n)$ - where does the $m$ come in? I think you mean $\sum_n\sum_m f(m.n) = \sum_n f(n,n)+f(n+1,n)$... –  Thomas Andrews Oct 29 '12 at 17:54
    
Yes exactly thank you! –  Ferenc Oct 29 '12 at 17:57
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up vote 5 down vote accepted

Represent the array as $$\begin{array}{rr} m/n&&&&\\ &1&0&0&\dots\\ &-1&1&0&\dots\\ &0&-1&1&\ddots\\ &0&0&-1&\ddots\\ &\vdots&&\ddots&\ddots \end{array}$$ When you first sum with respect to $m$, we get $0$, bu when we begin by $n$, the sum of the first row is $1$, and $0$ for the other.

So, careful with the order of integration when the function is not non-negative and not integrable.

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Thank you! This is really helpful! –  Ferenc Oct 29 '12 at 18:02
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