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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, $\iint f\ \mathsf d\mu \mathsf d\nu$ and $\iint f\ \mathsf d\nu\mathsf d\mu$ exist and are unequal.

It seems to me that \begin{align}\iint f\ \mathsf d\mu\mathsf 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\end{align} and \begin{align}\iint f\ \mathsf d\nu\mathsf 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.\end{align} So the two integrals are equal.

What am I doing wrong? Thanks!

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    $\begingroup$ 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 $\sum_n\sum_m f(m.n) = \sum_n f(n,n)+f(n+1,n)$... $\endgroup$ Oct 29 '12 at 17:54
  • $\begingroup$ Yes exactly thank you! $\endgroup$
    – Ferenc
    Oct 29 '12 at 17:57
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Represent the array as $$\begin{array}{r|rrrr} m\backslash n &1&2&3&\cdots\\\hline 1&1&0&0&\dots\\ 2&-1&1&0&\dots\\ 3&0&-1&1&\ddots\\ 4&0&0&-1&\ddots\\ \vdots&\vdots&\vdots&\ddots&\ddots \end{array}$$ When you first sum with respect to $m$, we get $0$, but 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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    $\begingroup$ Thank you! This is really helpful! $\endgroup$
    – Ferenc
    Oct 29 '12 at 18:02

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