0
$\begingroup$

I am studying for my midterm review. Could someone give me hints on how to start this question?

Useful theorem: Vitali Any set of real numbers with positive outer measure contains a subset that fails to be measurable.

Thanks for the help!

1 Answers 1

1

To be honest, I'm not sure how to make use of the theorem - there may be a way to apply it, but it looks harder than a much more straightforward approach.

Three facts are important: (1) There is a non-measurable set. (2) All countable sets are measurable. (3) The union of measurable sets is measurable. [If you don't have all three of these yet, then this approach might not be workable for you.]

So, for example: take any non-measurable set $A$. What happens if you take away one point?

  • 0
    Thanks Reese. Here's what I have: Suppose we have a non-measurable set A. We know that a countable set => measurable. Then by contraposition, not measurable => not countable. We also know that if we have a countable collection of measurable sets, then the union of the countable collection is also measurable. So by contraposition, take A = UAi which is non-measurable, then we will have an uncountable collection of non-measurable sets. – 2017-02-05
  • 0
    @Wilson The fact that a countable union of measurable sets is measurable doesn't automatically mean that every nonmeasurable set can be decomposed into uncountably many nonmeasurable sets; it means that if $A = \bigcup A_i$ is nonmeasurable, then *either* the union is uncountable *or* at least on $A_i$ is nonmeasurable. That's not what you want. Instead, try deleting one point from $A$, and try to show that the resulting set is not measurable. How many different sets can you get this way? – 2017-02-05