I have found the following example in a book, where the example emphasizes the importance of the fact, that the $\left(X,\mathscr{A},\mu_{1}\right)$ and $\left(Y,\mathscr{B},\mu_{2}\right)$ are have to be $\sigma$-finite measure spaces in the Fubini theorem.
Let me define the following measure spaces: $$\left(X,\mathscr{A}\right)=\left(Y,\mathscr{B}\right)=\left(\left[0,1\right],\mathcal{B}\left(\left[0,1\right]\right)\right),$$ where $\mathcal{B}\left(\left[0,1\right]\right)$ denotes the Borel sets of $\left[0,1\right]$. Let $\mu_{1}$ be the counting measure, $\mu_{2}$ be the Lebesgue measure. Let $f$ be the $\chi_{\Delta}$ characteristic function of $\Delta$, where $\Delta$ denotes the diagonal of the $\left[0,1\right]\times\left[0,1\right]$ unit square. In this case $$0=\int_{X}\int_{Y}fd\mu_{2}d\mu_{1}\neq\int_{Y}\int_{X}fd\mu_{1}d\mu_{2}=1.$$ I don't understand the example, because I don't understand how can we integrate $f$ over $X=\left[0,1\right]$ with the counting measure. How can we interpret it?
Best Answer
Be careful, as you are not integrating $f$ over $X=[0, 1]$. Instead, for every $b \in Y$, you are integrating $f_b(x)=f(x, b)$ over $X=[0, 1]$, as this is an iterated integral. (And then of course you are integrating $b \mapsto \int_X f_b \,\mathrm{d}\mu_1$ over $Y$.)
Now, this integral is not so difficult to compute: as $f_b(x)=1_{\{b\}}$ is the indicator function at $b$, we have $$ \int_X f_b \,\mathrm{d}\mu_1 = \int_X 1_{\{b\}} \,\mathrm{d}\mu_1 = \mu_1(\{1\})=1. $$