Could you help me with the following exercise?
Consider $X=Y=[0,1]$ with Lebesgue measure $m$ on $X$ and counting measure $\omega$ on $Y$.
Let $f:X \times Y \rightarrow \mathbb{R}$ and $f(x,y)= \begin{cases} 1, \ \ \ \ \ \text{for} \ x =y \\ 0 \ \ \ \ \ \ \ \text{elsewhere}\end{cases}$
Could you tell me why $\int_X f(\cdot, y) \ d m=0$ and $\int_Y f(x, \cdot) d\omega = 1$?
Best Answer
For a fixed $y$, $\int_X f(x,y)\mathrm d\lambda(x)=\lambda(\{y\})$ because $f(x,y)=\chi_{\{y\}}(x)$, where $\chi$ denotes the characteristic function. The Lebesgue measure of a singleton is $0$, hence $\int_X f(x,y)\mathrm d\lambda(x)=0$.
A similar computation gives $\int_Y f(x,y)\mathrm d\mu(y)=\mu(\{x\})=1$, where $\mu$ is the counting measure.
This show that Fubini's theorem is not necessarily valid when one of the measures is not $\sigma$-finite (here the counting measure on the uncountable set $[0,1]$).