0
$\begingroup$

Let $\mu^n$ be the n-dimensional Lebesgue measure. I want to show that the transformation $\mu^n: \mathcal{P}(\Omega)\rightarrow [0,\infty]$ doesn't exist. In other word I want to use Vitali sets to demonstrate that there are sets which aren't measurable.

I spoke with a friend of mine who said that we might use the Vitali sets for $\mathbb{R}$ we got in a proof and just attach to them a line in so that we get a similar set in $\mathbb{R}^2$.

My first question would be: How can I show that this new "line"-set isn't measurable?

When we go further then we might deduce for a given dimension n that we attach to the points of a Viatli set lines and have sets which aren't measurable. Is this correct?

$\endgroup$
  • $\begingroup$ I have no idea what you are saying. What is $\Omega$? What is a line set? $\endgroup$ – mathworker21 Jul 18 '18 at 11:34
  • $\begingroup$ $\Omega$ is in this case $\mathbb{R}^n$. Imagine that you have a Vitali set for $\mathbb{R}$ in $\mathbb{R}^2$ this can be located on the x-axis. If you take lines which are orthogonal to the x-axis and intersect with the x-axis in a point of the chosen Vitali set you get what I called a line-set. $\endgroup$ – Rico1990 Jul 18 '18 at 11:41
  • $\begingroup$ Have you tried picking representatives from $\Bbb{R^n/Q^n}$? $\endgroup$ – Asaf Karagila Jul 18 '18 at 12:07
  • $\begingroup$ Thank you for your answer. $\endgroup$ – Rico1990 Jul 19 '18 at 10:14
1
$\begingroup$

For concreteness, let's first sketch a Vitali-based proof that not all sets of reals are measurable:

Let $A_1$ be a subset of $[0,1]$ that contains exactly one representative for each equivalence class in $\mathbb R/\mathbb Q$. The set $$ B_1 = \bigcup_{q\in\mathbb Q\cap[-1,1]} (A_1+q) $$ then satisfies $$ [0,1] \subseteq B_1 \subseteq [-1,2] $$ so if it is measurable its measure must be between $1$ and $3$. But it is a disjoint union of countably many translated copies of $A_1$. This means that $A_1$ cannot have measure $0$ (because then $B_1$ would have measure $0$ too), nor can it have measure $>0$ (because then $B_1$ would have infinite measure). So $A_1$ is not measurable.

In two dimensions you can simply set $$ A_2 = A_1 \times [0,1]$$ $$ B_2 = \bigcup_{q\in\mathbb Q\cap[-1,1]} (A_2+\langle q,0\rangle) = B_1 \times [0,1] $$ and then repeat the same argument: $B_2$ should have measure between $1$ and $3$, but that cannot be a countably infinite sum of identical terms.

The generalization to higher dimensions should now be clear.

$\endgroup$
  • $\begingroup$ Which equivalence relation did you use in your proof? We used groups $I_x \lbrace z \in (0,1) | z - x \in \mathbb{Q} \rbrace$ and chose for each a representative, what gave us the Vitali set. Then we continued like you did. Can you explain why we get the measure between 1 and 3 in the 2-dimensional case? I suppose that this is derived by the product measure, right? $\endgroup$ – Rico1990 Jul 19 '18 at 10:14
  • $\begingroup$ @Rico1990: Your Vitali set is the same as my $A_1$, just described using (very slightly) different words, and I use the closed unit inverval intstead the open one, which matters not at all. --- In the two-dimensional case we have $[0,1]\times[0,1]\subseteq B_2 \subseteq [-1,2]\times[0,1]$ and those two rectangles have measure $1$ and $3$. This is directly derived from the specification of the Lebesgue measure: The unit square must have measure $1$ and the measure is invariant under translations. (The long rectangle is a sum of three squares). $\endgroup$ – Henning Makholm Jul 19 '18 at 10:46
  • $\begingroup$ Ok, thank you again. In case further questions appear I'll give a signal. $\endgroup$ – Rico1990 Jul 19 '18 at 17:05
0
$\begingroup$

Let $V$ be a Vitali set and consider $V \times \mathbb R^{n-1}$. Then $$\mathbb R^n = \bigcup_{q \in \mathbb Q} (V+q) \times \mathbb R^{n-1}$$ and it's obvious that each $(V+q) \times \mathbb R^{n-1}$ is not measurable.

$\endgroup$
  • $\begingroup$ Thank you for your answer. $\endgroup$ – Rico1990 Jul 19 '18 at 10:14
  • $\begingroup$ Can you explain for me why $(V+q)×\Bbb{R}^{n−1}$ is not measurable.? $\endgroup$ – 129492 Mar 23 at 15:31
  • $\begingroup$ @129492. Do you know what a Vitali set is and why such is not measurable? $\endgroup$ – md2perpe Mar 23 at 15:38
  • $\begingroup$ Vitali set in $R$ is $[0;1]$ and to show that is not a measurable we define an equivalence relation on S... Right ? I read it in Real_Analysis__Measure_Theory__Integration__and_Hilbert_Spaces__Princeton_Lectures_in_Analysis___Volume_3_.pdf $\endgroup$ – 129492 Mar 23 at 15:45
  • $\begingroup$ @129492. To construct $V$ we define an equivalence relation on $[0,1)$. To show that $V$ is not measurable, we take a countable union $\bigcup_k V_k$ of translations modulo 1 of $V$ such that the union is all of $[0,1)$. Then, since Lebesgue measure is countably additive and translation invariant, we must have $$1 = m([0,1)) = m(\bigcup_k V_k) = \sum_k m(V_k) = \sum_k m(V) = \infty \times m(V).$$ The last expresion is either $0$ (if $m(V)=0$) or $\infty$ (if $m(V)>0$). Contradiction! $\endgroup$ – md2perpe Mar 23 at 16:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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