Subset of $[0,uno]$ with Lebesgue measure uno I was proving that there exists a closed set in $[0,1]$ such that its interior is empty and with measure $(0,1)$, with this I had no problem, so I will not write the details of that.
However I had a question, a CLOSED subset of $[0,1]$ such that its interior is empty, can have measure $1$?, I came to this conclusion but I don't know if I am right.
Suppose that $I$ is in $[0,1]$ is closed and $m(I)=1$ and with an empty interior, then
$1=m(I)<m(I)+m([0,1])=m_\ast(I)+m_\ast([0,1])\leq m_\ast(I\bigcup [0,1])=m_\ast([0,1])=1$
then $1<1$.
However, I feel it is wrong what I did, plus I didn't use the fact that $I$ has an empty interior.
I would be grateful if you could guide me with this
 A: I'll provide a lemma that mostly solves the problem, and leave it as a hint:
Lemma: if $x \notin X$, then there is some $\epsilon > 0$ such that $B(x, \epsilon)$ is disjoint from $X$.
Proof: Suppose otherwise, that for all $\epsilon > 0$, there is a $y \in B(x, \epsilon) \cap X$. Then, for $\epsilon = 1/n$, let $y_n$ be such a $y$. We have $d(y_n , x) < 1/n$ so $y_n \rightarrow x$. As $X$ is closed, contradicting $x \notin X$.
A: Suppose $A=\overline A\subsetneqq [0,1].$ There exists $x\in (0,1)\setminus A$, otherwise $A=\overline A\supseteqq \overline {(0,1)}=[0,1].$ Since $x\not\in\overline A$ and $0<x<1$ there exists $r\in (0,\min (x,1-x))$ such that $A\cap (x-r,x+r)=\emptyset.$
Now $A$ and $(x-r,x+r)$ are disjoint measurable subsets of $[0,1]$ so $1=m([0,1])\ge m(A)+m((x-r,x+r))=m(A)+2r.$
Therefore $m(A)\le 1-2r<1.$
A: To summarize most of the comments...
Suppose $I \subseteq [0,1]$ is a closed set with measure $1$.
Then $(0,1) \setminus I$ is an open set with measure $0$, and since every nonempty open set has positive measure, this implies $(0,1) \setminus I = \emptyset$, or in other words, $(0,1) \subseteq I$.  As $I$ is closed, $[0,1] = \operatorname{cl}(0,1) \subseteq I$.  Therefore $I = [0,1]$.
By assuming $I \subseteq [0,1]$ is a closed set with measure $1$, we've shown $I=[0,1]$.  That is, this shows the only closed subset of $[0,1]$ with measure $1$ is $[0,1]$ itself.
Now, if you ask whether there is a closed subset of $[0,1]$ with measure $1$ that also satisfies property $(*)$ (for any property you want), you just have to ask whether $[0,1]$ satisfies $(*)$.

*

*Is $[0,1]$ a proper subset of $[0,1]$?  No.  Then there is no proper closed subset of $[0,1]$ with measure $1$.

*Does $[0,1]$ have empty interior?  No.  Then there is no closed subset of $[0,1]$ with measure $1$ and empty interior.

