June 29, 2026
A common trick I've seen while studying measure-theoretic probability is to convert a countable union into a disjoint countable union. I think it recurs enough that it's worth proving here:
Let \(\{B_k\}_{k=1}^\infty\) be a collection of sets that are not disjoint. Let \(A_1 = B_1\) and \(A_k = B_k \setminus \cup_{j=1}^{k-1} B_j\) for \(k > 1\). Then:
We first prove (1). First, suppose that \(x \in \cup_{k=1}^\infty A_k\). Then, \(x \in A_j\) for some \(j \in \mathbb{N}\). Since \(A_j = B_j \setminus \cup_{k=1}^{j-1} B_k\), \(x \in B_j\). Then, \(x \in \cup_{k=1}^\infty B_k\). Consequently, \(\cup_{k=1}^\infty A_k \subset \cup_{k=1}^\infty B_k\). Now, we show that \(\cup_{k=1}^\infty B_k \subset \cup_{k=1}^\infty A_k\). Let \(x \in \cup_{k=1}^\infty B_k\). So, \(x \in B_j\) for some \(j \in \mathbb{N}\). There are two cases:
In the first case, \(x \in A_j\), so \(x \in \cup_{k=1}^\infty A_k\). In the second case, we do the following: define \(S = \{k \in [j-1]: x \in B_k\}\). Then, since \(S\) is finite, we can compute \(i = \min S\). So, \(x \in B_i \setminus \cup_{k=1}^{i-1} B_k\) since \(i\) is the smallest index for which \(x\) is contained in one of the \(B_k\)'s. Then, \(x \in A_i\), so \(x \in \cup_{k=1}^\infty A_k\). In either case, \(\cup_{k=1}^\infty B_k \subset \cup_{k=1}^\infty A_k\).
This proves (1). Now we prove (2). WLOG, suppose \(m < n\). We can write:
\begin{align*} A_m \cap A_n &= \Bigl(B_m \cap \bigcap_{k=1}^{m-1} B_k^c\Bigl) \cap \Bigl(B_n \cap \bigcap_{k=1}^{n-1} B_k^c \Bigl) \end{align*}Since \(m < n\), \(B_m^c\) is contained in the intersection \(\bigcap_{k=1}^{n-1} B_k^c\), so the intersection is empty, i.e, \(A_m \cap A_n = \emptyset\).