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Let $S = [0,1] \times [0,1]$ and let $D = \{(x,y) \in S : x = y \} \subset S$. Let $\mu \times \nu = \lambda$ be the product of the Lebesgue measure $\mu$ on $[0,1]$ and the counting measure $\nu$ on $[0,1]$ (s.t. if $|E| \in \mathbb{N}$, then $\nu(E) = |E|$ else $\nu(E) = +\infty$). Finally, let $f$ be the characteristic function of $D$ on $S$ (i.e., $f = \chi_D$).

I would like to calculate the following three integrals:

$ (1) \int_{[0,1]} d\mu(x) \int_{[0,1]} f(x,y) d\nu(y) $

$ (2) \int_{[0,1]} d\nu(y) \int_{[0,1]} f(x,y) d\mu(x) $

$ (3) \int_D f(x,y) d\lambda(x,y) $


$$ \int_{[0,1]} d\mu(x) \int_{[0,1]} f(x,y) d\nu(y) = 1 \cdot \int_{[0,1]} f(x,y) d\nu(y) = 1 \cdot (\nu([0,1]) \cdot 1) = +\infty $$


$$ \int_{[0,1]} d\nu(y) \int_{[0,1]} f(x,y) d\mu(x) = \nu([0,1]) \cdot \int_{[0,1]} f(x,y) d\mu(x) = +\infty \cdot (\mu([0,1]) \cdot 1) = +\infty $$

Third and finally,

$$ \int_D f(x,y) d\lambda(x,y) = \chi_D \cdot (\mu([0,1]) \cdot \nu([0,1])) = \chi_D \cdot (1 \cdot +\infty) = +\infty $$

Now I'm pretty sure I've done something wrong above because the point of this exercise -- I thought -- was to demonstrate that even though these three values don't equal, there is no contradiction here between this and Fubini's Theorem nor Tonelli's Theorem (since neither of their two hypotheses have been satisfied in the above setting). Am I wrong here or am I wrong above?

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The inside integral depends on $x$ or $y$, thus, you cannot set the double integral as the "product" of the two integrals... – N. S. Dec 20 '12 at 16:16
up vote 3 down vote accepted

For each $x$, there is one and only one $y$ for which $f_x(y) = f(x, y) = 1$; namely $y = x$. Otherwise, $f_x(y) = f(x, y) = 0$. We have:

$$\int_{[0,1]} f(x,y) \, d\nu(y) = \int_{\{x\}} 1 \, d\nu(y) = \nu(\{x\}) = 1$$


$$\int_{[0,1]} d\mu(x) \int_{[0,1]} f(x,y) \, d\nu(y) = \mu([0,1]) = 1$$

On the other hand:

$$\int_{[0,1]} f(x,y) \, d\mu(x) = \int_{\{y\}} 1 \, d\mu(x) = \mu(\{y\}) = 0$$


$$\int_{[0,1]} d\nu(y) \int_{[0,1]} f(x,y) \, d\mu(x) = 0$$

As for (3), how do you define the product measure when the measures aren't both $\sigma$-finite? Some references only define it when the measures are both $\sigma$-finite, and it's uniquely determined only when so.

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In the first computation, for any given $x\in[0,1]$, we have that $f(x,y)=1$ if $y=x$, and $0$ otherwise. Therefore, $$\int_{[0,1]}f(x,y)\,d\nu(y)=\int_{[0,1]}\chi_{\{x\}}(y)\,d\nu(y)=\nu(\{x\})=1.$$ As user N.S. points out above, you also have changed an iterated integral to a product of two separate integrals, which is not valid because the inner term depends on $x$. Look again at the statement of Fubini's theorem.

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