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Let $X=Y=\mathbb{R}, \mathcal{B}$ the Borel $\sigma$-algebra.

Consider the function $f(x,y)=\begin{cases} 1 & \text{ if } x\geq 0 \text{ and } x\leq y<x+1 \\ -1 & \text{ if } x\geq 0 \text{ and } x+1\leq y<x+2 \\ 0 & \text{ if } else \end{cases}$.

Calculate $\int \int f(x,y) dy dx$ and $\int \int f(x,y) dx dy$.

Are they equal? Why is that example not a contradiction to Fubini?

My attempt: I know that if $E \in \mathcal{A} \times \mathcal{B}$ is a measurable rectangle. Then one can define $h(x)=\mu_1(E_x)$ and $k(y)=\mu_2(E^y)$, where

$E_x:=\{y \in Y:(x,y) \in E\}$ and $E^y:=\{x \in X:(x,y) \in E\}$.

Now if $\mu_1,\mu_2$ are $\sigma$ finite, we have $\int h(x) \mu_1(dx)=\int k(y) \mu_2(dy)$.

In my notes it is stated that we can further wirte the above equality as (which I don't understand why): $\int( \int \chi_E(x,y) \mu_2(dy)) \mu_1(dx)=\int (\int \chi_E(x,y) \mu_1(dx)) \mu_2(dy)$

I would be very happy if someone could explain, how this works.

Peter
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1 Answers1

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Too long for a comment.

  • Fubini's theorem requires that $|f|$ is integrable over the entire $\mathbb R^2\,.$ This is obviously not the case with your $f\,.$

  • About the iterated integrals they want you to calculate. If I am not mistaken: \begin{align} \textstyle\int_0^a\int_0^af_+\,dy\,dx&=\textstyle\int_0^{a-1}\int_x^{x+1}\,dy\,dx+\int_{a-1}^a\int_x^a\,dy\,dx=a-1+\int_{a-1}^a(a-x)\,dx\\[2mm] &=a-\tfrac{1}{2}\,,\\[2mm] \textstyle\int_0^a\int_0^af_-\,dy\,dx&=-\textstyle\int_0^{a-2}\int_{x+1}^{x+2}\,dy\,dx-\int_{a-2}^{a-1}\int_{x+1}^a\,dy\,dx=-a+2-\int_{a-2}^{a-1}(a-x-1)\,dx\\[2mm] &=-a+2-\tfrac{1}{2}\,,\\ \textstyle\int_0^a\int_0^af\,dx\,dy&=1\,. \end{align} Can you proceed?

Kurt G.
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  • First, thank you. I do know how to use the Riemann integral to calculate this, but how would I write it down in a "measure theory" way. I am new to product measures, so I am a little bit confused on how to do that (in a measure theory way). – Peter Dec 06 '23 at 08:54
  • In the case of that simple function Riemann integral is the same as Lebesgue. We do not need measure theory to calculate any of those integrals. Draw the region of integration. Even a look at that is enough to know the results. – Kurt G. Dec 06 '23 at 10:41
  • Thank you very much – Peter Dec 06 '23 at 18:42
  • For the other integral direction I get the result $0$, is that correct? – Peter Dec 07 '23 at 06:39
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    @Peter . Can't be. When we integrate over the bounded square $[0,a]\times[0,a]$ then $|f|$ is clearly integrable and Fubini applies. I am not sure what that homework is trying to achieve with this function. A better example might be this. – Kurt G. Dec 07 '23 at 07:46