# Show that $(x_{n})_{n}$ is dense in $[0,1]\setminus A$ then it is also dense in $[0,1]$

Let $$B([0,1])$$ be the borel set on $$[0,1]$$ and suppose that $$\mu$$ is a finite measure on $$B([0,1])$$. Then, define $$A=[0,1]\cap \left \{ x;\mu(\left \{ x \right \})>0 \right \}$$. I have shown that if $$A$$ is countable, then $$[0,1]\setminus A$$ is a separable metric space (because it is a subset of a separable metric space). Now if $$(x_{n})_{n}$$ is a dense sequence in $$[0,1]\setminus A$$, then I want to show that it is also dense in $$[0,1]$$. How can I prove it?

• Let $K$ be the limit points of $(x_n)_{n}$ in $[0,1]$, which is in particular closed. If $K\neq [0,1]$, then $U=[0,1]\setminus K$ is open and non-empty. This implies that there is an interval $(a,b)\subset U$ with $a\neq b$. We must have $(a,b)\subset A$, since all the points of $[0,1]\setminus A\subset K$. You already proved that $A$ is countable, but $(a,b)$ is uncountable.
– plop
Nov 29, 2021 at 15:12

If $$I=(a,b)$$ is an open interval in $$[0,1]$$ then $$I \setminus A$$ is uncountable so in particular a non-empty open subset of $$[0,1]\setminus A$$ and so contains some $$x_n$$ etc.
The complement of a countable subset of $$\mathbb R$$ is dense, as explained here. Thus, $$[0,1]\setminus A$$ is dense in $$[0,1]$$.