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?
Best Answer
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).