Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

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).

share|improve this answer
    
Thanks, there was actually no sigma-algebra specified in the exercise as it was given. Which is as you say a bit strange. –  Erik Mar 20 '13 at 21:40
    
@Erik: Yes, just assume that it's the Borel sigma-algebra on each of them. That is fine. –  Stefan Hansen Mar 20 '13 at 21:41
    
Yes, that was probably the idea. Thanks, now it's clear. I will give the first part a shot tomorrow. –  Erik Mar 20 '13 at 21:46
    
And yes it is a contradiction to Fubini. Is it because $\nu$ is not $\sigma$-finite? –  Erik Mar 20 '13 at 21:47
1  
I agree with that guess. I forgot to give you a vote before, done now. –  Martin Mar 21 '13 at 9:37
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.