SYSTEMATIC MATHEMATICS

Measure & Lebesgue Integration

This course contains 40 visible knowledge chapters in six content-sized units of lengths 10, 6, 8, 6, 5, and 5. What should “size” and “area under a function” mean when intervals and Riemann sums are no longer enough? Begin by reading X, 𝓕, μ, set operations, and a loaded-die probability list one symbol and one number at a time; then build measurable sets, interval-cover outer measure, regular approximation, and the Carathéodory extension that turns a premeasure into a measure. The symbol σ is read before a σ-algebra is built, “almost everywhere” means “outside a null set,” and each integral begins as accumulated area or signed accumulation. π–λ and monotone-class theorems explain how verified identities extend from generators, while Lusin explains why measurable functions become continuous after a controlled small deletion. Layer-cake formulas, a theorem-selection record, Jensen’s inequality, reliable convergence theorems, Lᵖ spaces, repeated integration, changes of variables, densities, and the Hardy–Littlewood maximal estimate follow with an exact permission and failure example beside every reordered limit, sum, derivative, or integral.

Before this course: Completed Real Analysis and Proof, Logic & Set Theory; Single-Variable Calculus, Multivariable Calculus, Linear Algebra, and Probability & Statistics are used for examples and applications. Students must be comfortable with countable set operations, rigorous limits, sequences of functions, Riemann integration, norms, inner products, and proof by approximation.

COURSE FACTSLevel, chapters, units, prerequisite, and outcome
Chapter 1

Reading measure notation: set, collection, and size rule

Objective: Can the same set A have different measures?

Measure theory separates three things that ordinary length often hides: the possible objects, the collections we are allowed to measure, and the numerical rule used to measure them. Reading that three-layer structure first prevents the symbols from becoming empty code.

Start from the definition: In (X,𝓕,μ), X is the full set of possible points; 𝓕 (read “script F”) is the collection of subsets declared measurable; μ (read “mu”) is the size rule; and μ(A) is the number assigned to a measurable subset A. Also, ∅ is the empty set, A⊆X means every point of A lies in X, A^c is the part of X outside A, ∪ means union, ∩ means intersection, and Σ means add the listed terms. The result to establish is: On a finite set X={x₁,…,x_n}, choose nonnegative weights w₁,…,w_n and define μ(A) as the sum of w_i over points x_i in A. This is a measure on all subsets. If the weights add to one, μ is a probability measure and μ(A) is the probability that the outcome lies in A.

The exact worked question is: A loaded six-sided die is assigned (1,1,1,1,2,2)/8. State the probability of every face and check the total. Compare it with this failure boundary: The word “mass” in “probability mass function” is only an analogy for how the total probability one is distributed among exact outcomes. It is not physical mass, and it does not explain a tuple by itself. A list such as (1,1,1,1,2,2)/8 must be unpacked entry by entry.

On a finite set X={x₁,…,x_n}, choose nonnegative weights w₁,…,w_n and define μ(A) as the sum of w_i over points x_i in A. This is a measure on all subsets. If the weights add to one, μ is a probability measure and μ(A) is the probability that the outcome lies in A.

The empty set contains no points, so its weight sum is zero: μ(∅)=0.

If measurable sets A₁,A₂,… are pairwise disjoint, no point is counted twice when their weights are added.

Therefore the weight of their union equals the sum of their separate weights; on a finite X only finitely many disjoint sets can be nonempty, so countable additivity follows.

A loaded six-sided die is assigned (1,1,1,1,2,2)/8. State the probability of every face and check the total.

  1. The six positions correspond in order to faces 1,2,3,4,5,6; the common denominator 8 means divide every listed number by 8.
  2. Thus P(1)=P(2)=P(3)=P(4)=1/8, while P(5)=P(6)=2/8=1/4.
  3. Add all six probabilities: four copies of 1/8 plus two copies of 2/8 give (4+4)/8=1.

Result: Faces 1–4 each occur with probability 1/8; faces 5–6 each occur with probability 1/4. The final two faces are each twice as likely as any one of the first four faces.