If $U_1,U_2,\dots$ are open subsets of $[0,1]$, then either prove or disprove the statements Suppose $U_1,U_2,\dots$ are open subsets of $[0,1]$. In each case, either prove the statement or disprove it.
(a) If $m(\cap_{n=1}^\infty U_n)=0$, then for some $n ≥ 1$, we have $m(\overline{U_n})<1$, where $m$ is Lebesgue measure and $\overline{U_n}$ is the closure of $U_n$ in the usual topology on $[0, 1]$.
(b) If $\cap_{n=1}^\infty U_n = \emptyset$, then for some $n\geq 1$, the set $[0,1] \setminus U_n$ contains a nonempty open interval.

Thoughts.
I think (a) is FALSE. i.e we can produce some open subsets $U_n$ such that $m(\cap_{n=1}^\infty U_n)=0$ and  $m(\overline{U_n})=1$
For (b) I think it is TRUE. Suppose not. i.e. For each $n\geq 1$ if the closed set $[0,1]\setminus U_n$ does not contain a nonempty open interval then $m([0,1]\setminus U_n)=0$, so $m(U_n)=m([0,1])=1$ for all $n\geq 1$. So $\cap_{n=1}^\infty U_n\neq \emptyset$, contradiction. I feel that I skip some details.
Thanks for any comments/ideas/answers.
 A: a): Let $(r_k)$ be an ennumeration of rationals in $[0,1]$. Let $U_n$ be the union of the intervals $(r_k-\frac 1 {2^{n+k+1}},r_k+\frac 1 {2^{n+k+1}})$ over $k \geq 1$. Then $m(U_n) \leq \frac  1{2^{n}}$ so $m (\cap U_n)=0$. Since $U_n$ contains all the rationals it is dense. So $m(\overline {U_n})=1$ for all $n$.
b) $[0,1]=\cup_n U_N^{c}%$. Baire Category Theorem implies that $U_n^{c}$ has non-empty interior for some $n$.
A: For (a): Right. A "fat Cantor set" is a perfect $C\subset [0,1]$ (see below for def'n of perfect) such that $[0,1]\setminus C$ is dense in $[0,1]$ and $m([0,1]\setminus C )<1.$ The value $m([0,1]\setminus C )$ can be any member of $(0,1).$
For $n\ge 1$ let $C_n$ be a fat Cantor set with $m(C_n)=1-1/(n+1).$ Let $U_n=[0,1]\setminus C_n.$  Then $\overline U_n=[0,1]$ so $m(\overline U_n)=1$ for every $n$. But $m(\cap_{j\in\Bbb N}U_j)=0$ because $m(\cap_{j\in\Bbb N}U_j)\le m(U_n)=1/(n+1)$ for every $n\in \Bbb N.$
A perfect subset of $[0,1]$ is non-empty and closed and, when considered as a subspace, has no isolated points.
