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I'm having trouble understanding this proof:

Lemma: Let $\lambda$ be the Lebesgue measure restricted to the Borel sets of $[0,1]$. Then there exists a decreasing sequence $(E_n)_{n\ge0}$ of subsets of $[0,1]$ such that each of them is $\lambda$-thick and $\bigcap E_n = \emptyset$.

Proof: Let $E_n = [0,1]\cap\{f\ge n\}$, where $f$ is a noncontinuous solution $f:\mathbb{R} \to \mathbb{R}$ of $f(x+y) = f(x) + f(y)$.

A continuous solution of that functional equation is a line through the origin.

Let $\sim$ be the equivalence relation on $\mathbb{R} \setminus \{0\}$ such that $x \sim y \leftrightarrow \exists q\in\mathbb{Q}$ s.t. $x = qy$.

Let $A$ be a set of representatives from each equivalence class. Let $g:A\longrightarrow\mathbb{R}$ be a non-constant "assignment of slopes". Let $f:\mathbb{R} \setminus \{0\} \to \mathbb{R}$ s.t. $x \mapsto x\cdot g(y)$ where $x \sim y\in A$, let $f(0) = 0$.

To put it roughly, a noncontinuous solution is a stack of lines through the origin.

a) Is this description of a noncontinuous solution correct?

No it's not! As TCL helped me realize, this construction does satisfy the desired functional equation only for $x$ and $y$ in the same equivalence class. I need to require something more from g.

b) Why does the proof work? I fail to see why are the $E_n$ measurable, or why are they $\lambda$-thick.

They don't need to be measurable! In fact, the concept of thickness is introduced to get around measurability issues.

Let's construct $f$ this way: let $H$ be an Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$, and $a\in H$. Let $g(x) = x$ for $x\in H \setminus \{a\}$ and $g(a) = 0$. Let's extend this function to $f:\mathbb{R} \to \mathbb{R}$ by linearity using the Hamel basis.

Using this $f$ we have that $E_0 = [0,1]$, $E_1 = \{1\}$, which clearly isn't $\lambda$-thick. Where's the flaw?

The flaw is that $E_1 \ne \{1\}$ because... Hamel bases don't work that way :) See the comments for details.

I give up. I will ask my professor about this.

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I don't think they're measurable: a discontinuous additive function is not measurable. –  Akhil Mathew Dec 6 '10 at 14:45
    
For the noncontinuous function $f$ described above, I don't see why it should satisfy $f(x+y)=f(x)+f(y)$. –  TCL Dec 6 '10 at 14:49
    
@Jens.I am not questioning about measurability. I am saying the function described does not satisfy $f(x+y)=f(x)+f(y)$. –  TCL Dec 6 '10 at 15:23
2  
Jacopo, it is not hard to show that $f:\mathbb{R}\longrightarrow\mathbb{R}$ is additive if and only if it is $\mathbb{Q}$-linear (seeing $\mathbb{R}$ as $\mathbb{Q}$-vector space). So you were already on the right track. Can you go on now? –  j.p. Dec 6 '10 at 15:44
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Why do you think $E_1 = \{1\}$? Assume, WLOG, your $a$ is between 0 and 1, and not rational. Then there exists a sufficiently small rational $q$ such that $0 < 1+q - a < 1$. But your $f$ acting on $f(1+q - a) = 1+q > 1$... –  Willie Wong Dec 7 '10 at 14:31
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1 Answer

up vote 2 down vote accepted

Edit: I think my previous answer can be adapted to the case we are interested in. So let $E_n=\{ f \geq n\}$. Let $m$ be a positive integer. If $s = \sup_{F \subset [-m,m] \setminus E_n} \lambda(F)$ (the sets $F$ are taken measurable here), then there is a measurable set $F \subset [-m,m] \setminus E_n$ such that $\lambda(F)=s$ (take a sequence and a union). We are going to prove that for any real M, for almost all $x \in F$, $f(x)<-M$, that is, there is a measurable $A(M) \subset F$ of zero measure such that $f(x)<-M$ for all $x \in F \setminus A(M)$. For this, it is enough to find, for every positive integer $k$, $A_k(M) \subset F$ such that $f(x)<-M$ for all $x \in F \setminus A_k(M)$ and $\lambda(A_k(M)) \leq 1/k$. Since $f$ is not continuous and $\mathbb{Q}$-linear, there exists $a_k \in \mathbb{R}$ such that $|a_k| \lt 1/(2k)$ and $f(a_k)<-n-M$. Then $a_k+F \subset [-m-1/(2k), m + 1/(2k)] \setminus E_n$, so $\lambda((a_k+F) \cap [-m,m] \setminus F) = 0$ and consequently, $\lambda(F \setminus (a_k+F)) \leq 1/k$, so we can set $A_k(M) = F \setminus (a_k+F)$.

Now since $f(x)<-M$ for all $x \in F \setminus A(M)$, $F = \cup_{M \in \mathbb{N}} A(M)$ has measure $0$.

We have shown that $E_n$ is $\lambda$-thick in $[-m,m]$ for all $m$, so $E_n$ is $\lambda$-thick in $\mathbb{R}$, and this is stronger than $E_n$ being $\lambda$-thick in $[0,1]$.

Old answer: This is not exactly what you want, but here is a similar proof of your lemma. Define $E_n = \{ |f| \geq n \}$. Then $E_n$ is $\lambda$-thick (in $\mathbb{R}$). We need to show that if $F$ is measurable and does not intersect $E_n$, then $\lambda(F)=0$.

Assume otherwise, and let $m$ be big enough so that $F \cap [-m,m]$ has positive measure. Since $f$ is not continuous and $\mathbb{Q}$-linear, $f_{|[0,1]}$ is not bounded, so there is a sequence $(a_k)_{k \geq 0}$ of elements of $[0,1]$ such that $|f(a_{k+1})| > |f(a_k)|+2n$. Then $\cup_{k \geq 0} \left( a_k + (F \cap [-m,m]) \right)$ is included in $[-m,m+1]$ and its measure is equal to $\sum_{k \geq 0} \lambda (F \cap [-m,m]) = + \infty$, this is because the $a_k + F$ are disjoint (look at the images by $f$), so we get a contradiction.

So $E_n$ is $\lambda$-thick, and $E_n \cap [0,1]$ is $\lambda$-thick in $[0,1]$.

Maybe there is a way to deduce your statement from this one.

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Despite not being exactly what I asked for, this answer deserved some recognition. I don't have enough reputation to vote you up. –  Jacopo Notarstefano Dec 9 '10 at 9:11
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