Fubini's theorem states that for $\sigma$-finite measure spaces $X,Y$ and any integrable function $f(x,y)$ we have that

$$\int_X \int_Y f(x,y)dydx = \int_Y \int_X f(x,y)dxdy$$

Now suppose we're working in discrete settings, so we have sequences $a:\mathbb{N}\times\mathbb{N}\rightarrow [0,1]$. The most natural definition of an integral over one variable would be

$$\int_{\mathbb{N}} a(n,m) dn : = \lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n=1}^N a (n,m)$$ Note that this "integral" does not correspond to a measure on $\mathbb{N}$!

Can we deduce from the Fubini's theorem (or in some other way) that

$$\int_{\mathbb{N}}\int_{\mathbb{N}} a(n,m)dndm = \int_{\mathbb{N}}\int_{\mathbb{N}}a(n,m)dmdn$$


No. Let $a(n,m)=1$ if $n\geq m$ and $a(n,m)=0$ if $n<m$ for a counterexample.


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