Investigate why the set of rational numbers in [0,1] has Lebesgue measure zero but not Jordan volume zero | Step-by-Step Solution

What We're Solving:

We need to understand why the set of rational numbers in [0,1] behaves differently under two different measures: it has Lebesgue measure zero but does NOT have Jordan measure zero. This explores a fundamental difference between these two ways of measuring sets!

The Approach:

This is a problem that shows how different mathematical tools can give different answers to seemingly similar questions. We'll compare two measurement systems and see how the concept of "countable vs. finite" creates this fascinating distinction.

Step-by-Step Solution:

Step 1: Understanding What We're Measuring

Let's call our set Q = {rational numbers in [0,1]}. We know:

• Q is countable (we can list all rationals: 0, 1, 1/2, 1/3, 2/3, 1/4, 3/4, ...)
• Q is dense in [0,1] (between any two real numbers, there's always a rational)

Step 2: Why Q Has Lebesgue Measure Zero

The key insight: Lebesgue measure can handle countable unions!

• Since Q is countable, we can write Q = {r₁, r₂, r₃, ...}
• For any ε > 0, we can cover each rational rᵢ with an interval of length ε/2ⁱ
• Total length of all intervals = ε/2¹ + ε/2² + ε/2³ + ... = ε
• Since ε can be arbitrarily small, the Lebesgue measure is 0

Why this works: Lebesgue measure is "countably additive" - it can handle infinitely many pieces as long as they're countable.

Step 3: Why Q Does NOT Have Jordan Measure Zero

The key insight: Jordan measure can only handle finite unions of intervals!

For Jordan measure zero, we need:

• For any ε > 0, cover Q with finitely many intervals whose total length < ε

But here's the problem:

• Q is dense in [0,1], so any finite collection of intervals that covers Q must have gaps
• But between any two intervals, there are more rationals!
• To truly cover all rationals with finite intervals, the total length must be at least 1

Therefore: The greatest lower bound of lengths needed is 1, not 0, so Q doesn't have Jordan measure zero.

Step 4: The Fundamental Difference

• Lebesgue: "I can use infinitely many intervals (countable), so I can make each one tiny!"
• Jordan: "I can only use finitely many intervals, and the rationals are too spread out!"

The Answer:

The set of rationals in [0,1] has:

• Lebesgue measure 0 because countable additivity allows us to cover each rational with arbitrarily small intervals
• Jordan measure ≠ 0 because finite covers of dense sets require substantial total length

This beautifully illustrates why Lebesgue measure is more powerful than Jordan measure - it can handle the "infinite but countable" nature of sets like the rationals.

Memory Tip:

Think of it this way: "Lebesgue lets you use infinitely many Band-Aids (countable), so you can make each one tiny. Jordan only gives you finitely many Band-Aids, so you need bigger ones to cover everything!"

The rationals are like scattered dots everywhere - infinite tiny Band-Aids work great, but a finite number of Band-Aids means some have to be big!