I can’t understand how to derive the following conclusion from
Lemma (Ulam) Let $(X, \tau)$ be a Polish space and $\mu$ a positive finite Borel measure on $X$. Then for every $\epsilon >0 $ there exists a compact set $K =K(\epsilon) \subset X$ s.t. $ \mu(X \setminus K) < \epsilon$
Then I have the following remark:
Let $(X,d)$ be a complete separable metric space and let $\mu$ be a probability on $X$. Let $A \subset X$ be a Borel set s.t. $\mu( X \setminus A) = 0 $. Then $A$ is $\sigma$ compact.
For me $\sigma$ compact means being the union of (at most) countable compact sets.
How can it be true? Even if the statement is not precise and it means
there exists a family $\{ K_n \}_{n \in \mathbb{N}} $ of compact set each one contained in $A$ s.t. $$ \mu ( A \setminus \bigcup_n K_n) =0 $$
how can I prove it?
I know I can take (thanks to Ulam lemma) a sequence of compact sets $C_n$ s.t. $\mu(X \setminus C_n ) = \mu( A \setminus C_n) <1/n$ and take $$K_n := \bigcup_{k=1}^n C_k$$
Then the $K_n$ are compact sets and $$ \mu ( A \setminus \bigcup_n K_n) =0 $$
But they are not contained in $A$. Maybe I can take $Q_n = C_n \cap A$ but now they are not compact any more. Unless $A$ is closed (and it is in the case $A= \text{supp} \mu$).
Am I wrong?