Lebesgue-unmeasurable sets

[Anonymous No.6755323]

So, I've started graduate level Real Analysis and am currently studying measure theory and the Lebesgue measure in particular. Having done some searching around the web, the only examples of sets that I've come across that cannot be measured by the Lebesgue measure are Vitali sets.

Does anyone have any other interesting examples of unmeasurable sets (in particular ones which don't rely on factor groups or the axiom of choice)?

[Anonymous No.6755342]

The most common Vitali set used in examples it would seem is constructed in the following way:

Define a relation <span class="math">x \approx y[/spoiler] where <span class="math">x \approx y \Leftrightarrow x - y \in \mathbb{Q} [/spoiler]

Define the factor group <span class="math">\mathbb{R} / \approx [/spoiler] and by invoking the axiom of choice, pick one element from each equivalence class that is in the range of [0,1].

You can see that by translating(modulo 1) this by each rational number in [0,1], you will cover the entire interval [0,1] with disjoint sets of the same measure, however if their measure were positive, the infinite sum of the measures would be infinite, or if their measure were zero it would be total sum of zero., neither of which are equal to 1, so it follows that it could not have been measured in the first place.

[Anonymous No.6755386]

>ones which don't rely on the axiom of choice
>ones which don't rely on the axiom of choice
>ones which don't rely on the axiom of choice
Son, you better read up on immeasurable sets!

[Anonymous No.6755745]

Well, yea. I did read somewhere that the axiom of choice is necessary, though at the same time I've seen other sources citing that you can use weaker assumptions such as free ultrafilters to construct non-measurable sets. As I had seen the method of constructing Vitali sets (or other similar sets using similar construction methods) I was hoping to be pointed in the direction of a <span class="math">different[/spoiler] method of construction so that I might not have to continually refer to the same example over and over again.

[Anonymous No.6756214]

>>6755745

Sets that aren't Lebesgue measurable cannot be constructed without the Boolean Prime Ideal theorem, which is not inferred from ZF and lies between ZF and ZFC (ZF + axiom of choice). It is equivalent to the ultrafilter lemma, both implied by the axiom of choice.

In essence, sets that are not Lebesgue measurable cannot be "constructed", they rely on an argument which uses something strange that can't be pinned down (the existence of a boolean prime ideal, an ultrafilter to extend a particular filter, or a choice function on a strange collection of sets).

Vitali sets are the only thing you're going to find. I know you're thinking "hey wouldn't it be cool to talk about another kind of non-Lebesgue set", but the answer to that question is no. Vitali sets are talked about as a way to confirm that non-Lebesgue measurable sets exist in ZFC. However...

Nobody cares about non-Lebesgue measurable sets in ZFC. Seriously. Since anything you ever describe is described in ZF, all those sets and functions will be measurable. Non-Lebesgue measurable sets aren't just pathological, but they might not exist because...

Also, ZF is consistent with NOT choice, meaning you can append things like the axiom of dependent choice (much weaker) and the Baire property and make models where every subset of the reals is Lebesgue measurable. In fact, some models of this type of system are INCREDIBLY NICE, notably that L^1 and L^Infinity are reflexive (this is impossible with axiom of choice).

[Anonymous No.6756225]

>measure theory
>2014

not even once. my condolences OP

[not OP No.6756294]

>>6756214
thx for the answor.

>notably that L^1 and L^Infinity are reflexive
I'm 13 and what is this?