Subsets of $[0,1]$

Suppose we have a closed subset $A\subset[0,1]$ that is not equal to $[0,1]$. Is it possible $mA=1$? Suppose you have an open subset $B\subset[0,1]$ that is dense in $[0,1]$. Is it possible that $mB<1$? Where $mA$ is the Lebesgue measure of some set $A$ of real numbers.

I've spent the last 4 or so hours thinking about what what the possibilities are here. Rationals and Irrationals and Cantor sets oh my!!! Truth be told, I would love to have an intuitive feeling for the intricate structure of all the real numbers in $[0,1]$ but alas I feel I am no closer to this goal than when I started looking at it a couple years ago.

When I look at these measurable sets problems Im always trying to construct examples in my head or to construct some sketches in an effort to intuit whats going on but often run into the brick wall that is the actual complexity of the real numbers. Anyone have a suggestion on how to build intuition here?

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What's $m A$?${}{}{}$ – Sanath K. Devalapurkar Apr 16 '14 at 1:28
@SanathDevalapurkar It's standard notation for Lebesgue measure. – user61527 Apr 16 '14 at 1:28
@T.Bongers Forgive me - I'm a self-learner, and am not used to the conventional math notation. – Sanath K. Devalapurkar Apr 16 '14 at 1:29

The first option is not possible: Note that if $A$ is a closed subset of $[0,1]$, $A$ is also a closed subset of $\mathbb{R}$. Then $A^c$ contains an interval inside $[0,1]$, and open intervals always have positive measure. Hence

$$1 = m A + mA^c > mA$$

For the second question, there are dense open subsets of $[0,1]$ with arbitrarily small measure: Put an interval of length $2^{-n} \epsilon$ around the $n$th rational number in $[0,1]$ according to your favorite enumeration. (Putting intervals around a countable, dense subset is a useful trick sometimes).

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I suppose when the ability to visualize no longer exists, we cling to the rules alone. Thx for the feedback. – JEM Apr 16 '14 at 4:28
@JEM Yes, intuition is hard in these spaces. If you can somehow relate the sets back to easier topological ideas (open set == union of intervals), it can sometimes be helpful. – user61527 Apr 16 '14 at 4:34
1. We can't have $m(A)=1$, since $m(A)=1-m(A^c)$ and $A^c$ is a nonempty open set, hence contains some open interval $(a,b)$ so $m(A^c)\ge b -a$.

2. Yes. Consider the set $$A=[0,1]\cap \bigcup\limits_{n=1}^\infty(q_n-\epsilon2^{-n},q_n+\epsilon2^{-n})$$ where $(q_n)$ is an enumeration of the rationals in $[0,1]$, which is open and dense since it contains the rationals, but has measure at most $\epsilon$.

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It will not have the measure $\epsilon$ due to overlapping intervals (think of the rationals which lie inside the very first interval, for example). It will, however, have measure $\mu$ with $0<\mu<\epsilon$. In addition, you have to intersect with $(0,1)$ because otherwise it may not be a subset of $[0,1]$. – celtschk Apr 16 '14 at 6:48
@celtschk Good points both. – Alex Becker Apr 16 '14 at 6:53
Note that you have to intersect with $(0,1)$, not $[0,1]$ because an open set was requested. – celtschk Apr 16 '14 at 6:55
@celtschk I interpreted that as an open set in $[0,1]$ (with the subspace topology). – Alex Becker Apr 16 '14 at 6:55
OK, I interpreted it as subset of the $[0,1]$ interval in $\mathbb R$ (with the topology of $\mathbb R$). Anyway, when intersecting with the open interval, it's true for both. – celtschk Apr 16 '14 at 6:58