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This is a theorem in Rudin RCA p.22

Let $(X,\mathfrak{M},\mu)$ be a measure space. If $\{f_n\}$ is a sequence of measurable functions such that $f_n:X\rightarrow [0,\infty]$ and $f(x)=\sum_{n=1}^{\infty} f_n(x)$, then $\int_X f d\mu = \sum_{n=1}^{\infty} \int_X f_n d\mu$.

And then, there is a corollary of this theorem in the book.

"If $a_{ij}≧0$ is a double sequence, then $\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} a_{ij}=\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_{ij}$" (Hint:use counting measure $\mu$ such that $\mu(E)=\infty$ for infinite set $E$ and $\mu(E)$ is finite ordinal of $E$ for finite set $E$.

I tried an hour to formulate a function suitable to $f_n$ which leads this corollary, but i'm still stuck.. Help!

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$f_i(j) = a_{ij}$? –  user27126 Jan 12 '13 at 7:31
    
@Sanchez That works.. Thank you. Since I tried to find a function on $\omega \times \omega$, it was troublesome –  Katlus Jan 12 '13 at 7:44

1 Answer 1

up vote 3 down vote accepted

$f_{i}(j) = a_{ij}$ would work.

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