Measure Theory-LebesgueMeasurable

[WannaBe22]

Homework Statement

Is there any non-Lebesgue-Measurable set A in R such as A contains all rational numbers?

Homework Equations

The Attempt at a Solution

I've tried assuming that this is true... If such a set exists, then both $$A$$ and $$A^c$$ aren't countable... I've tried looking at $$A^c$$ ... It's a non-countable set containing only non-rational numbers...I can't find any contradiction from this fact...
[I'm pretty sure the answer to the given question is no...]

Hope you'll be able to help me

Thanks !

 

[ystael]

Don't try to construct a nonmeasurable set in such a way that it contains all of $$\mathbb{Q}$$. Instead, approach the problem in two parts: first get yourself a nonmeasurable set, then try to figure out whether you can make it contain $$\mathbb{Q}$$ and still be nonmeasurable.

 

[WannaBe22]

Well...After reading your guidance , I've tried taking the "Vitali Set" (The subset of
[-0.5,0.5] which, for each real number r, contains exactly one number v such that v-r is rational )... If we'll denote this set as P, then we need to consider $$P \cup Q$$.
P is nonmeasurable and contains excatly one rational numbers...I can't figure out whether our "union set" is measurable or not... We've added to P a set of measure 0...Does the new set is measurable or not?

Hope you'll be able to continue your guidance...

Thanks !

 

[ystael]

You have reduced the problem to exactly the question you need to answer. Let $$A \subset \mathbb{R}$$ be any set, and $$Z \subset \mathbb{R}$$ be a set of measure zero. How is the measurability or nonmeasurability of $$A \cup Z$$ related to that of $$A$$? (What test for measurability should you be using to answer this question?)

 

[WannaBe22]

Well... $$Z$$ is obviously measurable...If $$A$$ is also measurable then $$A \cup Z$$ is also measurable... Hence, if $$A \cup Z$$ is nonmeasurable , $$A$$ must be nonmeasurable... So, if we take a nonmeasurable set $$A$$, containing all rational numbers, $$A-Q$$ must also be nonmeasurable...But $$A-Q$$ is a set contained in the irrational numbers set... Is it a contradiction?
As you can see, I'm pretty much stuck... Hope you'll be able to give me some further guidance

Thanks !