Does $\mu(\{x\})=0$ imply non-atomic for Radon measure? I am wondering if the following statement is true.
Let $\mu$ be a Radon measure satisfying $\mu(\{x\})=0$ for all $x\in X$, then $\mu$ is non-atomic. (Non-atomic means that for all $B\in \mathcal{B}_X$, if $\mu(B)>0$, there exists $A\subseteq B$ in $\mathcal{B}_X$ s.t. $0<\mu(A)<\mu(B)$).
This question arise when I am trying to do problem 7.11 on Folland real analysis.

If $\mu$ is a Radon measure on $X$ s.t. $\mu(\{x\})=0$ for all $x\in X$, and $A\in \mathcal{B}_X$ satisfies $0<\mu(A)<\infty$. Then for any $\alpha$ such that $0<\alpha< \mu(A)$ there is a Borel set $B\subseteq A$ s.t. $\mu(B)=\alpha$.

There is a theorem says that all non-atomic finite measure satisfies this intermediate-value theorem. So basically I need show is $\mu$ restricted on $A$ is non-atomic. However, I $\mu$ is not assumed to be $\sigma$-finite. So $\mu$ restricted on $A$ is not necessarily a Radon measure. But I guess the proof idea might be similar.
There is post about problem 7.11. Folland's Real Analysis 7.11. However, I am just not quite convinced by the answer. First of all the proof of intermediate value theorem (of non-atomic finite measure) does not require Zorn's Lemma. Also, is it possible to have for all $x\in A$, there exists $U$ open containing $x$ such that $\mu(U\cap A)=0$? The definition of Radon measure does not assume positivity of measure on open sets. With this I feel like the proof in the link might break.
Thanks in advance!
 A: Theorem. Let  $\mu $ be a regular Borel measure on a topological space $X$.  If $\mu (\{x\})=0$, for every $x$ in $X$, then $\mu $ has no atoms.
Proof.  Suppose by contradiction that $A$ is an atom for $\mu $.  By regularity
$$
  \sup\{\mu (K): K\subseteq A, \ K \text{ is compact}\} = \mu (A)>0,
  $$
so there is at least one compact $K\subseteq A$ with nonzero measure.   Fixing such a $K$, observe that it is also an atom.
For each $x$ in $K$, again using regularity, we have
$$
  \inf\{\mu (U): \{x\}\subseteq U, \  U \text{ is open}\}  = \mu (\{x\}) = 0,
  $$
so we may choose an open set $U_x$, containing $x$, such that $\mu (U_x)< \mu (K)$.
We then obtain a cover $\{U_x\}_{x\in K}$ for $K$, which therefore admits a finite subcover, say  $\{U_{x_i}\}_{1\leq i\leq n}$.
Setting $V_i=U_{x_i}\cap K$,
observe  that $V_i\subseteq K$, and that $\mu (V_i)< \mu (K)$, so we deduce that $\mu (V_i)=0$, due to the fact that $K$ is an atom.
Moreover $K\subseteq \bigcup_{i=1}^nV_i$, so by subaditivity we get
$$
  \mu (K)\leq \sum_{i=1}^n\mu (V_i)=0,
  $$
a contradiction.
A: I have some ideas.. The following is a sketch.
WLOG, we can prove there exists $B\subseteq A$ s.t. $0<\mu(B)<\mu(A)$.
Let $U\supseteq A$ be an open set satisfying $\mu(U)$ is arbitrarily closed to $\mu(A)$. Apply outer regularity of $\mu$ to $U\backslash A$, we can find closed $F\subseteq A$ relative to $U$ s.t. $\mu(F)$ is arbitrarily closed to $\mu(A)$. Let's assume $\mu(F)=\mu(A)$. By inner regularity of $\mu$, take a compact $K\subseteq U$ which has measure arbitrarily closed to $\mu(U)$. By doing this, we can make $K\cap F$ has positive measure by exclusion-inclusion formula. Take $K\cap F$
Assume $\mu(K\cap F)=\mu(A)$. $K\cap F$ is compact, for every $x\in K\cap F$, there exists open $U$ s.t. $\mu(U)<\epsilon$. Then $U_{x}$ form a cover of $K\cap F$. It admits finite subcover. Clearly, it's impossible to have all $\mu(U\cap K\cap F)=0$. Let $V$ be one open set s.t. $\mu(V\cap K\cap F)>0$.  So take $(K\cap F)\backslash V$ will produce the desired $B$.
