Let $m$ be the Lebesgue measure on $[0,1]$, and $\nu$ be the counting measure on $[0,1]$. Show that the diagonal

(1) $$D = \{ (x,x), x \in [0,1] \} $$

is measurable with respect to $m \times \nu$, but if $\chi_{D}$ denotes its characteristic function then

(2) $$ \int_{[0,1]}\int_{[0,1]} \chi_{D}(x,y)\,dm(x)\,d\nu(y) \neq \int_{[0,1]} \int_{[0,1]} \chi_{D}(x,y)\,d\nu(y)\,dm(x).$$

Ok, so I do not have so much experience of product measures. Let's start with showing that is a measurable set.

What does it take for $D$ to be measurable w.r.t. $m \times \nu$ ? Must it lie in a sigma-algebra, (which is not given in the exercise)? Or what do I have to show?

And how about (2)? My guess is that it has something to with that the counting measure is infinite on $[0,1]$.

Shall it be seen as a integral in $\mathbb{R}^{2}$ and $D$ as the line $y = x$?

Then the lebesgue measure of a straight line in $\mathbb{R}^{2}$ is always zero and the left integral would be $0$?

On the other hand, in the right integral we have an inner integral w.r.t. the counting measure which is infinite so that one equals $\infty$?

On this one I am quite stuck and any help and explanation of what is going on is appreciated.

/ Erik


Being measurable w.r.t. $m\times\nu$ doesn't make sense, and furthermore you don't even use the product measure in this exercise.

Instead, you should have specified which sigma-algebra you equip $[0,1]$ with - both when you're speaking of $m$ and when you're speaking of $\nu$. You could for example consider the measure-space $([0,1],\mathcal{E},m)$ and $([0,1],\mathcal{F},\nu)$, where $\mathcal{E}=\mathcal{B}([0,1])$ is the Borel sigma-algebra on $[0,1]$ and $\mathcal{F}$ could be $\mathcal{B}([0,1])$ or even the power set $\mathcal{P}([0,1])$. But let us assume that $\mathcal{F}=\mathcal{B}(\mathbb{R})$ since this is the smallest of the two.

Now you should show that $$D\in\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})=\mathcal{B}(\mathbb{R}^2),$$ i.e. that $D$ belongs to the product-sigma-algebra of $[0,1]\times [0,1]$. One strategy for that is to show that $D$ is closed in $\mathbb{R}^2$. This ensures that the sections $D_x=\{y\in\mathbb{R}\mid (x,y)\in D\}$ and $D_y=\{x\in\mathbb{R}\mid (x,y)\in D\}$ belongs to $\mathcal{B}(\mathbb{R})$.

For (2) you just evaluate the inner integrals first: For a fixed $y\in [0,1]$ we have that $\chi_D(x,y)=\chi_{D_y}(x)=1$ if and only if $x=y$ and zero otherwise. Therefore,

$$ \int_{[0,1]}\chi_D(x,y)\,m(\mathrm dx)=\int_{[0,1]}\chi_{D_{y}}(x)\,m(\mathrm dx)=m(\{y\})=0, $$ for all $y\in [0,1]$.

For the right-hand side we have that for a fixed $x\in [0,1]$: $$ \int_{[0,1]}\chi_D(x,y)\,\nu(\mathrm dy)=\int_{[0,1]}\chi_{D_{x}}(y)\,\nu(\mathrm dy)=\nu(\{x\})=1. $$

Is this a contradiction to Tonelli/Fubini's theorem? (This is probably the key point of the exercise).

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  • $\begingroup$ Thanks, there was actually no sigma-algebra specified in the exercise as it was given. Which is as you say a bit strange. $\endgroup$ – El_Loco Mar 20 '13 at 21:40
  • $\begingroup$ @Erik: Yes, just assume that it's the Borel sigma-algebra on each of them. That is fine. $\endgroup$ – Stefan Hansen Mar 20 '13 at 21:41
  • $\begingroup$ Yes, that was probably the idea. Thanks, now it's clear. I will give the first part a shot tomorrow. $\endgroup$ – El_Loco Mar 20 '13 at 21:46
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    $\begingroup$ Yes, you can't apply Fubini because $\nu$ is not $\sigma$-finite, and hence it is not a contradiction to Fubini :) $\endgroup$ – Stefan Hansen Mar 20 '13 at 21:48
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    $\begingroup$ I agree with that guess. I forgot to give you a vote before, done now. $\endgroup$ – Martin Mar 21 '13 at 9:37

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