If a set $A\subseteq[0,1]$ contains a large subinterval of every interval, must $A$ have full measure? Suppose that $A$ is a subset of $[0,1]$ such that $A$ is open in $[0,1]$, and there exists fixed constants $C, \mu > 0$ with $1\leq \mu < 2$ so that whenever $I$ is a sub-interval of $[0,1]$, say $[a,b]$, then $(A \cap I)$ contains an open subinterval of length at least $C(b-a)^{\mu}$.
Does $A$ have full measure in $[0,1]$?
Note that when $\mu = 1$ the result is true by the Lebesgue density theorem. I chose $\mu < 2$ because I did not want "trivial" counterexamples (not that I know of any as of right now). This is not an assignment problem.
 A: This is not true for any $\mu>1$; in fact, you can find a single set $A$ which is a counterexample for all $\mu>1$ at once.  Consider the following variation of (the complement of) the Cantor set.  Say a number $x\in[0,1]$ is $n$-good if all of the digits in the base $3$ expansion of $x$ from the $n^2$th digit to the $((n+1)^2-1)$st digit are $1$.  (If $x$ has two different base $3$ expansions, we require this to be true of both of them.)  Let $A$ be the set of numbers that are $n$-good for some $n\in\mathbb{N}$.
Note first that $\lambda(A)<1$, where $\lambda$ is Lebesgue measure.  To prove this, observe that a number $x\in[0,1]$ has probability $1/3^{(n+1)^2-n^2}=1/3^{2n+1}$ of being $n$-good, and these events are independent over different values of $n$.  So, $\lambda([0,1]\setminus A)=\prod_n(1-1/3^{2n+1})$.  Since the sum $\sum_n 1/3^{2n+1}$ converges, the product $\prod_n(1-1/3^{2n+1})$ converges to a value greater than $0$, so $\lambda([0,1]\setminus A)>0$ and $\lambda(A)<1$.
However, I claim this set $A$ satisfies your condition with respect to any $\mu>1$.  Indeed, suppose $[a,b]\subseteq [0,1]$.  Let $m$ be minimal such that $[a,b]$ contains an interval of the form $[k/3^m,(k+1)/3^m]$ (where $k,m\in\mathbb{N}$); note that $1/3^m\geq \frac{b-a}{9}$.  Take $n$ minimal such that $n^2>m$.  Then $[a,b]$ contains an interval of length $1/3^{(n+1)^2}$ consisting of $n$-good numbers (pick a string of first $(n+1)^2$ digits to ensure the number is in $[k/3^m,(k+1)/3^m]$ and is also $n$-good, and the set of numbers that start with those digits is an interval of length $1/3^{(n+1)^2}$).  Now $$(n+1)^2\leq (\sqrt{m}+2)^2\leq m+C_0\sqrt{m}=m(1+C_0/\sqrt{m})$$ for some constant $C_0$.  For any fixed $\mu>1$, we can then find a constant $C_1>0$ such that $$1/3^{m(1+C_0/\sqrt{m})}\geq C_1/3^{m\mu}$$ for all $m$ (since $1+C_0/\sqrt{m}<\mu$ for $m$ large) and so $$1/3^{(n+1)^2}\geq  C_1 (1/3^m)^\mu\geq C_2 (b-a)^\mu$$ for some constant $C_2>0$.
