The subadditivity of a measure. I'm reading Probability: Theory and Examples by Rick Durrett.  Theorem 1.1.1 states that

Let $\mu$ be a measure on $(\Omega, \mathcal F)$
(i) monotonicity. If $A \subset B$ then $\mu(A) \le \mu(B)$.
(ii) subadditivity. If $A \subset \cap_{m=1}^\infty A_m$, then $\mu(A) \le \sum_{m=1}^\infty \mu(A_m)$
Proof. (i) Let $B − A = B \cap A^c$ be the difference of the two sets ...

The proof of (ii) given in the book is like this

Let $A_n' = A_n \cap A$, $B_1 = A_1'$ and for $n > 1$, $B_n = A_n' − \cap_{m=1}^{n−1} (A_m')^c$ . Since the $B_n$ are disjoint and have union $A$ we have using (i) of the definition of measure, $B_m \subset A_m$ , and (i) of this theorem
$$\mu(A) = \sum_{m=1}^\infty \mu(B_m) \le \sum_{m=1}^\infty \mu (A_m)$$

I can't understand why

the $B_n$ are disjoint and have union $A$

Is there a typo in the definition of $B_n$?
 A: By your request: 


*

*in the definition of subadditivity, there should be $A\subset\bigcup\limits_{m=1}^\infty A_m$. Indeed, for an intersection we have rather obvious bound:
$$
\mu\left(\bigcap\limits_{m=1}^\infty A_m\right)\leq\inf\limits\limits_{m\in \mathbb N}\mu(A_m)
$$
just by the monotonicity of the measure.

*in the definition of $B_n$ we should take $A^\prime_n$ and exclude all previous $A^\prime_m$ (i.e. $m=1,…,n−1$) in order to make sets $B_n$ disjoint and apply countable additivity of $\mu$. I.e. we should have $B_1 = A^\prime_1$ and
$$
B_n = A^\prime_n - \bigcup\limits_{m=1}^{n-1}A^\prime_m.
$$
Now, about typos - comparing with the pdf file linked there are two typos coming from you (in $1.$ in the book shte union used, you've typed the intersection) and the same typo you made in the definition of $B_n$ when typed the OP. However, there is one typo coming from the book: there $(A^\prime_m)^c$ is used although the complement here is not needed.
An example: suppose $A = A_1\cup A_2$ - then we have $A^\prime_i = A_i$ and $B_1 = A_1$. For $B_2$ we would like to have $B_2 = A_2-A_1 = A_2\cap(A_1)^c$ but if we use the formula with a complement as in the book we obtain $B_2 = A_2 - A_1^c = A_2\cap A_1$ so either the disjointness may fail (when $A_1\cap A_2\neq \emptyset$) or that union of $B_n$ is $A$ (when $A_1\cap A_2 = \emptyset$ and $A_1,A_2$ are non-empty).
A: I'm not sure there is a typo. As it stands, the $B_n$ are disjoint since if $n_1<n_2$ then $B_{n_1}\subset A'_{n_1}\subset \bigcup_{j=1}^{n_2-1}A'_j$ so $B_{n_1}\cap B_{n_2}=\emptyset $ and
$$\bigcup_{n\geq 1}B_n=\bigcup_{n\geq 1}A'_n=\bigcup_{n\geq 1}A_n\cap A=A\cap \left(\bigcup_{n\geq 1}A_n\right)=A$$
since $A\subset \bigcup_{n\geq 1}A_n$. (the first equality follows from $B_n\subset A'_n$ and if $x \in \bigcup_{n\geq 1}A'_n$ then $x\in B_j$ where $j=\min_{n\in\mathbb N}\{x\in A'_j\}$.
