CounterExampleToTonellisTheorem 2008-07-27 22:35:23 2008-07-27 22:35:23 Example Tonelli's theorem \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \newcommand{\reals}{\mathbb{R}} \newcommand{\natnums}{\mathbb{N}} \newcommand{\cnums}{\mathbb{C}} \newcommand{\znums}{\mathbb{Z}} \newcommand{\lp}{\left(} \newcommand{\rp}{\right)} \newcommand{\lb}{\left[} \newcommand{\rb}{\right]} \newcommand{\supth}{^{\text{th}}} \newtheorem{proposition}{Proposition} \newtheorem{definition}[proposition]{Definition} \newtheorem{theorem}[proposition]{Theorem} The following observation demonstrates the necessity of the $\sigma$-finite assumption in Tonelli's and Fubini's theorem. Let $X$ denote the closed unit interval $[0,1]$ equipped with Lebesgue measure and $Y$ the same set, but this time equipped with counting measure $\nu$. Let \[ f(x,y) = \left\{ \begin{array}{cl} 1 & \mbox{ if } x=y,\\ 0 & \mbox{ otherwise}. \end{array}\right. \] Observe that \[ \int_Y \left( \int_X f(x,y) d\mu(x)\right) d\nu(y) = 0,\] while \[ \int_X \left( \int_Y f(x,y) d\nu(y)\right) d\mu(x) = 1.\] The iterated integrals do not give the same value, this despite the fact that the integrand is a non-negative function. Also observe that there does not exist a simple function on $X\times Y$ that is dominated by $f$. Hence, \[ \int_{X\times Y} f(x,y) d (\mu(x)\times \nu(y) = 0.\] Therefore, the integrand is $L^1$ integrable relative to the product measure. However, as we observed above, the iterated integrals do not agree. This observation illustrates the need for the $\sigma$-finite assumption for Fubini's theorem.