# Understanding Fubini's Theorem, and finding a counter example (measures)

I am having troubles understanding Fubini's Theorem, as well as product measures in general. I also have found a counter example which I am having troubles understanding.

Firstly, I have state the theorem as follows:

Let $(\Omega, \mathcal{F}, \mu)$ and $(\Omega ' , \mathcal{F}' , \nu)$ be complete $\sigma-$finite measure spaces. If $f: \Omega \times \Omega' \to \overline{\mathbb{R}}$ and $\int f d (\mu \otimes \nu)$ exists, then $$\int f d ( \mu \otimes \nu) = \int \left( \int f(x,y) d \nu(y) \right) \ d \mu (x)$$

Now, what I don't understand about this theorem statement is what does $\int f d ( \mu \otimes \nu)$ mean? I have never seen an intergral with respect to a product measure before, and I am not sure what it equals. I have defined the product measure as $\mu \otimes \nu ( E \times E' ) = \mu(E)\nu(E')$.

Now, I also have a counter example which is supposed to show that we require finiteness in the measure:

Let $\Omega = \Omega' = [0,1]$ Take $\mu$ to be the Lebesgue measure on $[0,1]$, and $\nu$ to be the counting measure on $[0,1]$. Now, since $[0,1]$ is uncountable, $\nu$ is not $\sigma -$finite.

Take $f(x,x) = 1$ on $0 \leq x \leq 1$ and $f(x,y) = 0$ for $x \not = y$. Then

$$\int f(x,y) \ d\nu(y) = 1, \text{ so } \int \left( \int f(x,y) \ d\nu(y) \right) \ d \mu(x) = 1$$

But $$\int f(x,y) \ d \mu(x) = 0, \text{ so } \int \left ( \int f(x,y) \ d\mu(x) \right) \ d\nu(y) = 0$$

Now from this counter example, I firstly don't understand:

1) Where is the author calculating $\int f d ( \mu \otimes \nu)$ in the theorem?

2) Where is the author using the fact that $\nu$ is not finite? Additionally, I don't understand the calculations that $\int f(x,y) \ d\nu(y) = 1$ and $\int f(x,y) \ d \mu(x) = 0$

I have been trying to understand this for many hours - any help would be appreciated. Thank you.

Ad 1): It is implicit in Fubini's theorem that if you are in a situation where it can be applied then the two iterated integrals (first with respect to $y$, then with respect to $x$, or the other way around) have the same value. In your exposition of the theorem you have considered just one of the two iterated integrals. In the counterexample the two different iterated integrals have different values, the reason being that an essential assumption of Fubini's theorem is not fulfilled. – The integral over $\Omega\times\Omega'$ with respect to $d(\mu\otimes\nu)$ has not been considered at all.
Ad 2): The author has computed these iterated integrals according to definition. Along each vertical stalk $\{x\}\times[0,1]$ there is exactly one point where $f(x,y)=1$, hence $\int_{\Omega'} f(x,y)\>d\nu(y)=1$. On the other hand, along each horizontal segment $[0,1]\times\{y\}$ the function $f$ is zero almost everywhere with respect to $\mu$, hence $\int_\Omega f(x,y)\>d\mu(x)=0$.