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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 $B_n$?

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I fixed two typos in your post: 1) in the definition of subadditivity there should be a union instead of an intersection; 2) in the definition of $B_n$ we should take $A^\prime_n$ and exclude all past $A^\prime_m$ (i.e. $m=1,\dots,n-1$) in order to make $B_n$ disjoint. I don't know where these typos come from, I can take a look on the book only in two days. – Ilya Feb 26 '12 at 22:30
Oh, so I'm right, there is a typo in the book. You can download it from the link in my post. The author put it online. Can you change the typo back and make your comment an answer? Without that typo, my question seems to become confusing. – ablmf Feb 26 '12 at 22:42
Richard Durrett is a world class probabilist, but his books have lots of typos. His details always need to be double checked. – Byron Schmuland Feb 26 '12 at 23:57
Done, you're right that it might be confusing. Nevertheless, two typos you made by yourself. – Ilya Feb 27 '12 at 7:56
We searched on this site and foudn Richard Durrett's book is recommended. But if that book has many this kind of problem, what alternatives should I have a look? – ablmf Feb 27 '12 at 16:29
up vote 2 down vote accepted

By your request:

  1. 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.

  2. 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).

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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 $j=\min_{n\in\mathbb N}\{x\in A'_j\}$.

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