关于《Probability with Martingales》中集合代数与可测集的技术问询

Let's break down your questions one by one—they’re tightly linked to the core building blocks of measure theory, which is the backbone of modern probability.

1. Why does a set algebra emphasize "finite" closure? What happens if we drop this restriction? #

First, let’s anchor on the definition you pulled from the book:

集合$S$上的代数是$S$的子集族，对有限个集合运算封闭

When we say "closed under finite operations," we mean the family is closed under finite unions, finite intersections, and complements (though technically, closure under complements + finite unions is enough to cover intersections via De Morgan’s laws).

The "finite" qualifier is critical to distinguish a set algebra from a stricter, more specialized structure called a σ-algebra (sigma-algebra). Here’s why the line matters:

• If we remove the "finite" rule and require closure under countably infinite unions/intersections, we’re no longer talking about an algebra—we’ve jumped straight to a σ-algebra. These structures are far more restrictive because they force the family to include all limit sets of countable sequences of sets.
• Let’s use a concrete example: take the set of integers $\mathbb{Z}$. Consider the family $\mathcal{A}$ made up of all finite subsets of $\mathbb{Z}$ and their complements. This is a valid set algebra:
  • Finitely many finite subsets union to another finite subset (so it’s in $\mathcal{A}$).
  • The complement of a finite subset is an infinite set (which is the complement of a finite set, so it’s also in $\mathcal{A}$).
    But $\mathcal{A}$ fails closedness under countable infinite unions: the union of ${0}, {1}, {2}, ...$ is the set of non-negative integers, which is neither finite nor the complement of a finite set. So this set isn’t in $\mathcal{A}$.

If we didn’t restrict to finite operations, we’d erase the difference between algebras and σ-algebras. Algebras are useful as simple, intuitive building blocks—we often start with a basic algebra (like all intervals on the real line) and then generate the smallest σ-algebra containing it (the Borel σ-algebra), which is what we use to define probabilities.

2. What does "there are always enough measurable sets in probability" mean, and is it connected to the first question? #

Absolutely—this ties directly to the algebra/σ-algebra distinction and how we define valid probability measures.

First, why aren’t all sets measurable? Using the axiom of choice, we can construct "pathological" sets (like the Vitali set on the real line) that can’t be assigned a consistent measure without breaking the countable additivity rule—this is a non-negotiable property for probabilities (think about adding up the odds of infinitely many disjoint events). Measures only work on σ-algebras (not just algebras) because countable additivity relies on the σ-algebra’s closure under countable operations.

Now, the phrase "enough measurable sets" translates to:

• The σ-algebras we use in practice (like the Borel σ-algebra on $\mathbb{R}$, or product σ-algebras for random vectors) include every set we’d ever need to model real-world events.
• For example: "the random variable $X$ is less than 10," "a coin flip sequence has infinitely many heads," "the temperature stays between 20°C and 30°C all week"—all these are measurable sets. They’re built from countable unions/intersections of basic sets (intervals, singletons, etc.) that live in the σ-algebra.
• Non-measurable sets are weird, abstract constructs that never come up in day-to-day probability. They don’t correspond to any event we’d actually want to assign a probability to, so we don’t need to worry about them.

The link to the first question is this: algebras are the starting point, but we need σ-algebras to define valid probabilities. Even though σ-algebras don’t include every possible set, they include everything we need to do practical probability theory.

内容的提问来源于stack exchange，提问作者user30614