2
$\begingroup$

Consider the space $ (\Omega \times \Omega,\mathcal{A}\otimes \mathcal{A}, \mu \times \nu) $, where $ \Omega=\mathbb{N} $, $ \mathcal{A} $ is the collection of all subsets of $ \mathbb{N} $ and $ \nu=\mu $ denotes the counting measure, i.e., $ \mu(A)=\nu(A)=\# A $ for all $ A\in \mathcal{A} $. Consider the function $ f:\Omega\times\Omega\to\mathbb{R} $ given by: $$ f(x,y)= \begin{cases} x &\text{if } y=x \text{, } x\in \mathbb{N}\\ -x &\text{if } y=x+1 \text{, } x\in \mathbb{N}\\ 0 &\text{otherwise} \end{cases} $$

I compute two integrals: \begin{align*} \iint f \text{ } d\mu\text{ } d\nu&=\int\left( \int f_y d\mu \right) d\nu\\ &=\int\left( \sum_{x=1}^{\infty}f_y(x) \right) d\nu\\ &=\int\left( f_y(1)+\cdots+f_y(y-2)+f_y(y-1)+f_y(y)+f_y(y+1)+\cdots \right) d\nu\\ &=\int\left( 0+\cdots+0-(y-1)+y+0+\cdots \right) d\nu\\ &=\int 1 d\nu\\ &=\sum_{y=1}^{\infty} 1 \\ &=\infty. \end{align*} and \begin{align*} \iint f \text{ } d\nu\text{ } d\mu &=\int\left( \int f_x d\nu \right) d\mu\\ &=\int\left( \sum_{y=1}^{\infty} f_x(y) \right) d\mu\\ &=\int\left( f_x(1)+\cdots +f_x(x-1)+f_x(x)+f_x(x+1)+f_x(x+2)+\cdots \right) d\mu\\ &=\int\left( 0+\cdots +0+x+-x+0+\cdots \right) d\mu\\ &=\int 0 d\mu\\ &=\sum_{x=1}^{\infty}(0)\\ &=0+0+0+\cdots\\ &=0. \end{align*}

Why is not $$ \iint f \text{ } d\mu\text{ } d\nu = \iint f \text{ } d\nu\text{ } d\mu$$ in accordance with Fubini's theorem? Has it something to do with $f$ not being in $L^1(\mu\times\nu)$? If so, how can I show that this is the case?

  • 2
    Indeed, did you try to evaluate $\iint|f|d\mu d\nu$? What happened when you did? – 2017-01-11
  • 0
    I am not intierly sure about how to do that. Is $ |f(x,y)| $ simply the following? $$ |f(x,y)|=\begin{cases} x &\text{if } y=x \text{, } x\in \mathbb{N}\\ x &\text{if } y=x+1 \text{, } x\in \mathbb{N}\\ 0 &\text{otherwise} \end{cases} $$ – 2017-01-11
  • 0
    And is $ \iint|f|d\mu d\nu $ this? $$\begin{align*} \iint|f(x,y)|d\mu d\nu&=\int\left( \int |f_y| d\mu \right) d\nu\\ &=\int\left( \sum_{x=1}^{\infty}|f_y(x)| \right) d\nu\\ &=\int\left( 0+\cdots+0+(y-1)+y+0+\cdots \right) d\nu\\ &=\int (2y-1) d\nu\\ &=\sum_{y=1}^{\infty} (2y-1) \\ &=\infty. \end{align*} $$ – 2017-01-11
  • 0
    Fubini's theorem fails because $$\iint|f(x,y)|d\mu d\nu=\infty,$$ is that right? – 2017-01-11
  • 0
    Is $\iint|f(x,y)|d\mu d\nu<\infty$ the same as $f\notin L^1(\mu\times\nu)$? – 2017-01-11
  • 1
    Yes, that is correct. Except it would be more correct to say that Fubini's theorem doesn't apply since $\iint \lvert f(x,y)\rvert\,d\mu\,d\nu = +\infty$, rather than that it fails. – 2017-01-11
  • 2
    In this particular case, $\iint \lvert f(x,y)\rvert\,d\mu\,d\nu = +\infty$ is exactly the same as $f \notin L^1(\mu\times \nu)$, since all functions are $\mathcal{A}\otimes \mathcal{A}$-measurable. In other situations, $f\notin L^1(\mu\times\nu)$ might also be due to non-measurability of $f$. – 2017-01-11

0 Answers 0