Decomposing a general complete seperable metric space into a meager set and a null set Suppose $X$ is Polish, i.e, a seperable and completely metrizable space. And let $\mu$ be a Borel probability measure on $X$. Furthermore suppose $\mu[\{x\}] = 0$ for every isolated point $x \in X$. Then $X = M \cup N$ where $M$ is meager and $N$ is measure $0$.
This feels simple but I'm having a brain fart.
Firstly, thanks to separability there can only be countably many isolated points. All we really need then is the ability to take neighborhoods of arbitrarily small measure around any non-isolated point. Then the result follows fairly easily by reproducing in $X$ the proof that $\mathbb{R}$ is a union of  meager and a null set.
However, I'm not sure if this is true and I don't have many ideas otherwise. I think I'm overlooking something simple.
 A: $\newcommand{\N}{{\mathbb{N}}}$
Lemma 1.  The set
$$
  D = \big\{x\in  X: \mu (\{x\}) = 0 \big\}
  $$
is dense in $X$.
Proof.  First of all notice that, for all finite subsets $F\subseteq X$, one has that
$$
  \sum_{x\in F} \mu (\{x\}) \leq  \mu (X)=1.
  $$
This implies that the (possibly uncountably indexed) series
$$
  \sum_{x\in X} \mu (\{x\})
  $$
converges, so it has at most countably many nonzero terms, that is, $X\setminus D$ is countable.
Now suppose by contradiction that some point $a\in X$ is not in the closure of $D$.  Then there exists some $r>0$ such that
the ball  $B_r(a)$ is contained in $X\setminus D$.  Needless to say, $ B_r(a)$  is therefore countable.
We next wish to find some positive number $s<r$ such that the  sphere
$$
  S_r(a) = \{x\in  X: d(x, a)=s\}
  $$
becomes  empty.  This is in fact easy  since the interval $(0,r)$ is uncountable and $B_r(a)$ is countable.
Once such an $s$ is fixed, we deduce that the open ball of radius $s$ coincides with the corresponding closed ball, so
$B_s(a)$ is both open and closed.  Also  $B_s(a)\subseteq X\setminus D$.
Since $X$ is complete and $ B_s(a)$  is closed, we deduce that $ B_s(a)$  is also complete.  By  Baire's Theorem
(https://en.wikipedia.org/wiki/Baire_category_theorem) every complete countable metric space admits an isolated point.  It then follows that $B_s(a)$ contains at least one isolated point $x_0$, and because $B_s(a)$ is open as well, we have that
$x_0$ is also isolated relative to $X$.  By assumption $\mu (\{x_0\}) = 0$, but this contradicts the fact that $B_s(a)\subseteq X\setminus D$.
QED
Lemma 2.
Given any point $d$ in $D$, and any $\varepsilon >0$, there exists an open set $U$ containing $d$,   such that $\mu (U)<\varepsilon $.
Proof.
One has that
$$
  \{d\} = \bigcap_{n\in \mathbb N} B_{1/n}(d),
  $$
so
$$
  0 = \mu (\{d\}) = \lim_{n\to \infty } \mu \big (B_{1/n}(d)\big ),
  $$
and then  one can  find some $n$ such that  $\mu \big (B_{1/n}(d)\big )$ is as small as  needed.  QED

This said, let us proceed to the main argument.
It is well known that any subspace  of a separable metric space is also separable.  Therefore  $D$ is separable and we
may then find a countable subset
$$
  D_0 = \{d_k:k\in \N\}\subseteq D
  $$
whose closure contains $D$.   Since $D$ is dense, we deduce that the
closure of $D_0$ also contains $X$.  In other words,  $D_0$ is dense in $X$ as well.
For each pair $(k, n)$ of natural numbers,  use Lemma 2 to choose some open set $U_{k,n}$ cointaining $d_k$, and such that
$$
  \mu (U_{k,n})< {1\over 2^kn},
  $$
and define
$$
  V_n= \bigcup_{k\in {\bf N}} U_{k,n}.
  $$
We then have that
$$
  \mu (V_n)\leq  \sum_{k\in {\bf N}} \mu (U_{k,n}) \leq  \sum_{k\in {\bf N}} {1\over 2^kn} = {1\over n}.
  $$
Oberve that  $V_n$ is  open and also  dense, because it contains $D_0$,  so
$$
  N:= \bigcap_{n\in {\bf N}} V_n
  $$
is a dense $G_\delta $, by Baire, and clearly $\mu (N)=0$.  The complemrnt
$M:= X\setminus N$
is therefore a meagre set and we evidently have that $X=M\cup N$.

PS: After initially misreading the hypothesis, I hope this is now correct.
