Phase 6 - Lesson 6.2
Measures and Sigma-Algebras
How to assign a consistent notion of size to sets - and why we cannot measure everything.
Builds on: Compactness/topology (6.1) supply the Borel sets we will measure.
Leads to: A probability measure (Phase 7) is exactly a measure of total mass 1.
Leads to: A probability measure (Phase 7) is exactly a measure of total mass 1.
Learning Objectives
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- Define a sigma-algebra and give the Borel sigma-algebra as the key example.
- Define a measure and prove countable additivity implies monotonicity and continuity.
- Explain why Lebesgue measure cannot be defined on all subsets (non-measurable sets).
- Compute the measure of simple and limiting sets, including sets of measure zero.
- Interpret ‘almost everywhere’ and its role in later integration theory.
Key Vocabulary
- Sigma-algebra
- A collection \(\mathcal{F}\) containing \(X\) and closed under complement and countable unions.
- Measure
- A function \(\mu:\mathcal{F}\to[0,\infty]\) with \(\mu(\emptyset)=0\) and countable additivity on disjoint sets.
- Borel sigma-algebra
- The smallest sigma-algebra containing all open sets; the natural domain for measuring on \(\mathbb{R}\).
- Lebesgue measure
- The translation-invariant measure on \(\mathbb{R}^n\) assigning length/area/volume, extending intervals \(\mu([a,b])=b-a\).
- Null set
- A set of measure zero; e.g. any countable set has Lebesgue measure 0.
- Almost everywhere
- A property holding except on a null set; written ‘a.e.’
Intuition & Motivation
Intuition
We want to say how ‘big’ a set is - its length, area, or (in probability) its likelihood. The naive hope is to assign a size to every subset, but that hope is provably impossible: the Vitali construction builds a set whose translates tile the line, so any translation-invariant, countably-additive size would have to be both 0 and positive. The fix is to restrict attention to a rich but not universal collection of ‘nice’ sets - a sigma-algebra - on which a consistent measure lives. Everything you will ever integrate is measurable; the pathological sets are simply excluded.
Sigma-algebras
Definition - Sigma-algebra
A collection \(\mathcal{F}\) of subsets of \(X\) with (i) \(X\in\mathcal{F}\), (ii) \(A\in\mathcal{F}\Rightarrow A^c\in\mathcal{F}\), and (iii) closure under countable unions \(\bigcup_n A_n\in\mathcal{F}\). De Morgan then gives closure under countable intersections.
The Borel sigma-algebra \(\mathcal{B}(\mathbb{R})\) is the smallest sigma-algebra containing every open interval; it contains all open, closed, countable, \(F_\sigma,G_\delta\) sets - essentially every set arising in practice.
Definition - Measure
A function \(\mu:\mathcal{F}\to[0,\infty]\) with \(\mu(\emptyset)=0\) and countable additivity: for disjoint \(A_1,A_2,\dots\), \(\mu\big(\bigcup_n A_n\big)=\sum_n\mu(A_n)\).
Theorem - Consequences of countable additivity
Any measure is monotone (\(A\subseteq B\Rightarrow\mu(A)\le\mu(B)\)), countably subadditive, and continuous from below/above: \(A_n\uparrow A\Rightarrow\mu(A_n)\to\mu(A)\) (and from above when some \(\mu(A_n)\lt \infty\)).
Proof
For monotonicity write \(B=A\cup(B\setminus A)\) disjointly; additivity gives \(\mu(B)=\mu(A)+\mu(B\setminus A)\ge\mu(A)\). For continuity from below, telescope \(A=\bigsqcup_n(A_n\setminus A_{n-1})\) and apply countable additivity to the partial sums. ∎
Why not measure everything?
Non-measurable sets exist
Using the axiom of choice, Vitali builds a subset of \([0,1]\) with no consistent Lebesgue measure. Hence Lebesgue measure is defined only on the (very large) sigma-algebra of measurable sets, not on the full power set. This is a feature: measurability is exactly the closure property that makes limits behave.
Worked Example - Sets of measure zero
1
A single point has measure 0: cover it by \([a-\varepsilon,a+\varepsilon]\), so \(\mu(\{a\})\le2\varepsilon\) for all \(\varepsilon\gt 0\).
2
By countable additivity any countable set is null: \(\mu(\mathbb{Q}\cap[0,1])=\sum_n0=0\).
3
The Cantor set is uncountable yet has measure 0: at step \(k\) a fraction \((2/3)^k\) remains, and \((2/3)^k\to0\).
4
Null sets are ‘invisible’ to integration - changing a function on one changes no integral, which is what ‘almost everywhere’ captures.
Interactive: the Cantor set has measure zero
Track the remaining Lebesgue measure of the Cantor construction as you remove middle thirds - an uncountable set shrinking to size 0.
Common Mistakes to Avoid
- Assuming every subset of \(\mathbb{R}\) has a length - non-measurable sets show it cannot.
- Confusing countable additivity with finite additivity; the ‘countable’ is what powers limit theorems.
- Believing measure zero means countable - the Cantor set is uncountable yet null.
- Forgetting a sigma-algebra must be closed under complements, not just unions.
Quant Practitioner Tips
- ‘Almost everywhere’ lets you ignore null sets; most integration theorems only hold a.e.
- To show a set is Borel, build it from open sets via countable unions/intersections and complements.
- Countable additivity + continuity from below is the standard route to compute measures of limits.
- In probability, replace ‘a.e.’ with ‘almost surely’ - same concept, mass-1 measure.
Knowledge Check
Q1 Medium
Which property must a sigma-algebra satisfy that a mere algebra of sets need not?
Q2 Medium
Countable additivity of a measure directly implies all of the following EXCEPT:
Q3 Medium
The Cantor set is an example of a set that is:
Practical Exercise
Prove that a countable set \(A=\{a_1,a_2,\dots\}\subset\mathbb{R}\) has Lebesgue measure zero, using countable subadditivity.
▶ Show full solution
Fix \(\varepsilon\gt 0\). Cover \(a_n\) by the interval \(I_n=(a_n-\varepsilon 2^{-n-1},\,a_n+\varepsilon 2^{-n-1})\), of length \(\varepsilon 2^{-n}\).
By countable subadditivity, \(\mu(A)\le\sum_{n=1}^\infty\mu(I_n)=\sum_{n=1}^\infty\varepsilon 2^{-n}=\varepsilon\). Since this holds for every \(\varepsilon\gt 0\), \(\mu(A)=0\). In particular \(\mu(\mathbb{Q})=0\).
After the reveal, answer for yourself: Why does the same trick fail for an uncountable set like \([0,1]\)?
Lesson Summary
A sigma-algebra is a collection of sets closed under complements and countable unions; a measure assigns them sizes with countable additivity, from which monotonicity and continuity follow. Not every set is measurable (Vitali), so we work on the Borel/Lebesgue sigma-algebra. Null sets - even uncountable ones like the Cantor set - are the ‘almost everywhere’ we may ignore.
Retrieval Practice
Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.
▶ Show retrieval prompts & answers
Q: What are the two closure axioms of a sigma-algebra beyond containing \(X\)?
A: Closure under complement and under countable unions (hence, by De Morgan, countable intersections).
A: Closure under complement and under countable unions (hence, by De Morgan, countable intersections).
Q: Give an uncountable set of measure zero and why it is null.
A: The Cantor set: after \(k\) middle-third removals a fraction \((2/3)^k\to0\) remains, so its Lebesgue measure is 0, yet it is uncountable.
A: The Cantor set: after \(k\) middle-third removals a fraction \((2/3)^k\to0\) remains, so its Lebesgue measure is 0, yet it is uncountable.
Completion Checklist
- I can explain the core ideas in my own words
- I worked the derivations/examples by hand
- I completed the interactive workbench(es)
- I passed the knowledge check
Confidence / mastery rating
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