existence of a subset that carries all measure of a chain Suppose $m: \mathcal{P}(X) \to [0, 1]$ is a $\sigma$-additive measure on all subsets of $X$ and $\mathcal{L} = \{A_\alpha: \alpha < \beta\}$ is a well ordered by inclusion chain of subsets of $X$ such that for all $\alpha < \beta$, $m(A_\alpha) \leq 1/2$. Prove that there exists a subset $B \subset X$ such that $m(B) \leq 1/2$ and $m(A_\alpha - B) = 0$ for all $\alpha < \beta$.
I was thinking along the lines: let $V_\alpha = A_\alpha - \bigcup_{\gamma < \alpha} A_\gamma$. The sets $V_\alpha$ are pairwise disjoint, so only countably many of them can have positive measure, let these be $V_{\alpha_1}, V_{\alpha_2}, \ldots$. Let $B_n = \bigcup_{i = 1}^n V_{\alpha_i}$ and $B = \bigcup_{n < \omega} B_n$. Obviously we have $B_n \subset A_{\alpha_n}$, so $m(B_n) \leq 1/2$, thus by continuity of measure, $m(B) \leq 1/2$. The set $B$ should intuitively carry al of the measure of the chain $\mathcal{L}$ but I have a hard time showing that this is actually the case: the problem I see is that it may happen that for some ordinal $\gamma < \beta$ with cofinality strictly larger than $\omega$, $m(A_\gamma) > 0$, but $m(A_\theta) = 0$ for all $\theta < \gamma$, and also $m(V_\theta) = 0$ for all $\theta \leq \gamma$ -- the measure jumps at $A_\gamma$, but my subsets $V_\theta$ for $\theta \leq \gamma$ does not detect it. Can this situation actually happen? If not, why? If so, how to fix that problem?

The sequence $\langle m(A_{\alpha}): \alpha < \beta \rangle$ is a monotonically increasing $\beta$-sequence of reals hence it has at most countably many reals. Let $\theta$ be the supremum of a countable set of stages at which all these reals occur. Take $B = \bigcup_{\alpha < \theta} A_{\alpha}$, if $\theta = \beta$, otherwise $B = A_{\theta}$. If $\alpha < \theta$, $A_{\alpha} \backslash B$ is empty. If $\theta \leq  \alpha < \beta$, $m(A_{\alpha}) = m(A_{\theta})$ so $m(A_{\alpha} \backslash B) = 0$.
