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?
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$\begingroup$ 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. $\endgroup$– plopNov 29, 2021 at 15:12
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2 Answers
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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.
answered Nov 29, 2021 at 16:20
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The complement of a countable subset of $\mathbb R$ is dense, as explained here. Thus, $[0,1]\setminus A$ is dense in $[0,1]$.
The rest follows from the transitivity of the denseness property.
