There exist meager subsets of $\mathbb{R}$ whose complements have Lebesgue measure zero? Probem 27 of Chapter 5 in Folland's Real Analysis.
I have been asked to solve this using generalized Cantor sets. Something related to the construction outlined here https://en.wikipedia.org/wiki/Nowhere_dense_set 
Any help is appreciated, thank you.
 A: A "generalized" or "fat" Cantor set is constructed by removing open intervals from $[0,1]$ in such a way that the total removed length is less than $1$. The result is a nowhere-dense $C \subset [0,1]$ (since, anywhere you look, there's a missing interval) with positive Lebesgue measure. It's fairly quick to verify that removing the middle fourths at each step leaves behind something of measure $\mu(C) = 1/2$. 
Can you fill up all the measure in the unit interval by taking a clever union of fat Cantor sets?
There's also a clean solution here (they don't talk about Cantor sets, though) https://mathoverflow.net/a/43480
A: My favorite example is the following. Given a real $r$, let $Seq(r)$ be the infinite binary sequence gotten from looking at the binary expansion of $r$ after the decimal point (in case $r$ has two binary expansions, pick the mostly-zeroes one). Then the set $Rand$ of $r$ such that "$1$" appears in $Seq(r)$ half the time - that is, the lim sup of (proportion of first $n$ bits of $Seq(r)$ which are "$1$"s) and the lim inf of the same are each equal to ${1\over 2}$ - is almost all of $\mathbb{R}$: its complement has measure zero.
Meanwhile, the set $Gen$ of reals $r$ such that the liminf and limsup defined above are $0$ and $1$, respectively - that is, $r$ has "long" strings of $0$s and "long" strings of $1$s - is comeager. 
But $Gen\subseteq \mathbb{R}\setminus Rand$.
("Gen" and "Rand" stand for "Generic" and "Random," which are technical terms in mathematical logic corresponding to category and measure, respectively.)
