Bernstein construction

[Sebastian Poloczek]

>In article <38FD981A...@mathematik.uni-mainz.de>, Sebastian
>Poloczek <le...@mathematik.uni-mainz.de> wrote:

>> I am looking for a subset A of [0,1] that fulfills the following
>> conditions:
>> 1) A has Lebesgue outer measure 1.
>> 2) A has Lebesgue inner measure 0.
>>
>> That means (I suppose):
>> 1) inf( \lambda(B); A \subset B, B \in Borel-Sigma field) =1
>> 2) sup( \lambda(B); B \subset A, B \in Borel-Sigma field) =0
>>
>> Does such a subset exist?
>>
>>

>According to the Axiom of Choice, yes. For example, Bernstein
>constructed (so to speak) such a set by transfinite
>induction... The result is a set A such that every
>uncountable closed set F meets both A and its complement.

>--
>Gerald A. Edgar ed...@math.ohio-state.edu

Where can I find this construction?

Thanks for help

Sebastian

[G. A. Edgar]

Briefly, it goes like this: Let c be the cardinal of the continuum.
Prove first that the number of closed sets in [0,1] is c. Let G be the
least ordinal of power c. (Use AC.) Enumerate the uncountable closed
sets using G, that is { F(g) : g in G} is the set of all uncountable
closed sets in [0,1]. Now proceed recursively. When you reach
the index g in G, choose a(g) and b(g) in F(g), different from
each other, and different form all a(g'), b(g') with g' < g. We can
do this since the set F(g) has power c, but the set of all the a(g'),
b(g') already chosen has power less than c. Bernstein's set is
A = { a(g) : g in G}. Every uncountable closed set is of the
form F(g) for some g, so it meets both A (in the point a(g)) and
the complement of A (in the point b(g)).

A has outer measure 1 and inner measure 0. And A has the same
property for any continuous measure. And it has the corresponding
property for category: G - A and G intersect A both have second
category for any nonempty open set G.