# Probability with Martingales proof

I am working with the book Probability with Martingales'' by David Williams. In the class that I am taking, we cover the following proof.

Consider $\mu$ a measure on $(S,\Sigma)$ where $S$ is a set (not necessarilty a sample space) and $\Sigma$ is a $\sigma-$algebra on this set. We assume that $\mu$ is countably additive, i.e. $$\mu(\emptyset)=0$$ If $\{A_{i}:i\in\mathbb{N}\}$ are a sequence of disjoint measurable sets, then $$\mu(\bigcup_{i}A_{i})=\sum_{i}\mu(A_{i})$$

Finally, we form the triple $(S,\Sigma,\mu)$ a measure space. We have the following \textbf{\uline{proposition}}:

If $A_{n}\in\Sigma,$ $n\in\mathbb{N},$ then $$\mu(\bigcup_{n=1}A_{n})\leq\sum_{n=1}^{\infty}\mu(A_{n})$$

The proof is given as follows:

Set $B_{1}=A_{1}$and $B_{n}=A_{n}\backslash (\bigcup_{j=1}^{n-1}A_{j})$; $B_{n}'s$ disjoint. As such, we can use additivity and obtain: $$\mu(\bigcup_{n=1}B_{n})=\sum_{n=1}^{\infty}\mu(B_{n})$$ Since $B_{n}\subseteq A_{n};$ $\mu(B_{n})\leq\mu(A_{n})\leq$$\sum_{n=1}^{\infty}\mu(A_{n})$

I am having some troubles with this seemingly simple proof (I think I am missing a step):

1. The proposition itself does not specify the disjoint property \emph{anywhere }and it seems crucial in the proof. The proof does not seem complete as it does not cover cases where the disjointedness property can be dropped.
2. Even if we take this at face value, I don't understand how it makes sense whilst proving to construct a specific example of a set and prove the property for that set, and them assume that the property is true for all sets. I have seen it done in this course again and again- to prove a proposition, we construct special cases and prove the proposition for those cases. How can we assume that the proposition would hold true for all cases? In other words, when is the construction of a specific case permissible as part of a general proof?
• Yes it was. I corrected it Commented Sep 13, 2016 at 20:18
• should your $B_n$ be $B_n = A_n \backslash (\cup_{j=1}^{n-1} A_j)$? Commented Sep 13, 2016 at 20:23

Hopefully we agree that $B_i$ are disjoint, so the stated equality holds. They neglect to mention that $\cup_n A_n=\cup_n B_n$, which gives you $\mu(\cup_n A_n)=\mu(\cup B_n)$.
The second observation is that $B_n\subset A_n$. So they're going to use $\mu(B_n)\leq \mu(A_n)$ for each term in the sum. If this isn't clear, write $A_n=B_n^c\cup B_n$ and observe that the two sets on the right are disjoint.
• Thanks..I'll have to look closer at your answer. What I don't understand is why is one allowed to create such a special case of subsets of $A$ to prove a much more general statement? Commented Sep 13, 2016 at 20:14
• If you agree that the two families are equal in union, then that justifies it being "allowed". This is actually a common technique to write a union as a disjoint union. Each $B_n$ takes away all prior $A_i$. It's easiest if you draw a Venn Diagram to convince yourself. Commented Sep 13, 2016 at 20:17
• @KwameBrown The motivation for rewriting the union of the $A_{n}$ as the union of the $B_{n}$ is this: we know if a countable collection of sets $\{C_{n}\}$ is pairwise disjoint, then the measure of the union is the sum of the measures, i.e., $\mu(\bigcup \limits_{n=1}^{\infty} C_{n}) = \sum \limits_{n=1}^{\infty} \mu(C_{n})$. If $\{C_{n}\}$ is not a pairwise disjoint collection of sets, we don't have this equality. We only have $\mu(\bigcup \limits_{n=1}^{\infty} C_{n}) \leq \sum \limits_{n=1}^{\infty} \mu(C_{n})$. In the proof, they want to use the equality, not inequality... Commented Sep 13, 2016 at 20:21
• @KwameBrown ...So they defined the sets $B_{n}$ so that they are pairwise disjoint and you have that inequality. So basically the proof is: we want to show $\mu(\bigcup \limits_{n=1}^{\infty} A_{n}) \leq \sum \limits_{n=1}^{\infty} \mu(A_{n})$. Well, $\mu(\bigcup \limits_{n=1}^{\infty} A_{n}) = \mu(\bigcup \limits_{n=1}^{\infty} B_{n}) = \sum \limits_{n=1}^{\infty} \mu(B_{n}) \leq \sum \limits_{n=1}^{\infty} \mu(A_{n})$ (since $B_{n} \subseteq A_{n}$ implies $\mu(B_{n}) \leq \mu(A_{n})$. Commented Sep 13, 2016 at 20:24