Let $X$ be a complete separable metric space and let $\lambda$ be probability measure on $(X, \mathcal{B}(X))$ where $\mathcal{B}(X)$ is the Borel sigma algebra of $X$. I know that $$ \lambda(B) = \sup \{ \lambda(K) \mid K \subset B, \, K \text{ compact} \} $$ for every $B \in \mathcal{B}(X)$.
I would like to know under which conditions the following is true:
Given $B \in \mathcal{B}(X)$, there exist a family $\{K_n\}_{n \ge 1}$ of compact subsets of $B$ and a measurable subset of $B$, call it $N$, s.t. $$ \bigcup_{n \ge 1} K_n \cup N = B, \quad \lambda(N)=0, \quad \lambda(K_i \cap K_j)=0 \text{ if } i \ne j.$$
Hint. Define $K_n$ recursively. Let $K_0 = \varnothing$, and by your "I know that" clause choose compact $K_{n+1} \subseteq B \setminus \bigcup_{j=0}^n K_j$ with $\lambda(K_{n+1}) \ge \frac12\lambda(B \setminus \bigcup_{j=0}^n K_j)$. At each stage you get at least half of what is left.