关于Durrett《Probability: Theory and Examples》鞅收敛定理的有理数疑问

Hey there, great question—this is a classic "bridge" between discrete dense sets and general real numbers in measure theory, and it’s totally normal to get stuck here at first. Let’s break down the reasoning step by step, tying it directly to the fundamentals you’re missing.

First, let’s recap the context from Durrett’s proof: when proving the martingale convergence theorem, we often focus on showing that the set

{ limsup Xₙ > b > a > liminf Xₙ }

has measure 0. Durrett starts by establishing this for all rational pairs (a < b), then extends it to all real numbers. Here’s why the rational case is sufficient, and the measure theory backing it:

1. Rational Numbers Are Dense in (\mathbb{R}) #

The first critical fact is that every real number can be approximated arbitrarily closely by rational numbers. For any real (a), there’s a sequence of rationals (a_k \uparrow a) (increasing to (a)) and another (a'_k \downarrow a) (decreasing to (a)). The same holds for any real (b). This density lets us "cover" real-number conditions with rational-number ones.

2. Measurable Sets and Countable Operations #

Measure theory is built to handle countable unions/intersections smoothly, which is exactly what we use here. Let’s formalize the set we care about for real (a < b):

• The set ({ \limsup Xₙ > b }) is equivalent to the countable union of sets ({ \limsup Xₙ > q }) where (q) is a rational number greater than (b). If the limsup is greater than (b), it must be greater than some rational (q) between (b) and the limsup, and vice versa.
• Similarly, ({ \liminf Xₙ < a }) is the countable union of ({ \liminf Xₙ < q }) where (q) is a rational number less than (a).

Putting these together, the set ({ \limsup Xₙ > b \text{ and } \liminf Xₙ < a }) equals the countable union of sets of the form:

{ limsup Xₙ > q₁ and liminf Xₙ < q₂ }

where (q₁, q₂) are rationals with (q₂ < a < b < q₁).

3. Measure’s Countable Subadditivity and Zero-Measure Sets #

Here’s the key payoff: Durrett already proves that for every rational pair (q₂ < q₁), the set above has measure 0. Now, measure has a core property called countable subadditivity: for any countable collection of measurable sets (A₁, A₂, ...),

P(∪ₖ Aₖ) ≤ Σₖ P(Aₖ)

Since each (P(Aₖ) = 0), the sum on the right is 0. And because probability measures are non-negative, this means (P(∪ₖ Aₖ) = 0).

In short: we cover the real-number case with a countable collection of rational-number cases we already know are zero-measure, and measure theory guarantees the union of these zero-measure sets is still zero-measure.

Bonus: Why Start with Rationals? #

You might wonder why Durrett doesn’t just start with real numbers. The answer is practical: rationals form a countable set, which makes constructing sequences or applying martingale inequalities (like Doob’s maximal inequality) far easier. For example, when defining stopping times or counting "crossings" of the martingale between (a) and (b), working with a countable dense set avoids technical headaches around uncountable collections (which don’t play nicely with measure theory’s countable axioms).

内容的提问来源于stack exchange，提问作者Quasar