5 — Cantor Set, Borel vs Lebesgue σ-Algebras, and Integration of Positive Functions
Lecture 5 completes the construction of the Lebesgue integral for non-negative measurable functions, using:
Simple functions as the foundation
The supremum definition from below
The infimum definition from above
A proof that the two definitions agree
Integration properties (positivity, homogeneity, additivity)
1. Cantor Set Review and Measurability
Let $K$ be the Cantor set constructed inside $[0,1]$.
Properties:
- $K$ is uncountable, with cardinality: \(\vert K\vert = \vert \mathbb{R}\vert = 2^{\aleph_0}.\)
- $K = \bigcap_{n=1}^\infty K_n$, with $K_n \downarrow K$, and each $K_n$ closed
⇒ $K$ is closed.
Question: Is the Cantor set Borel?
Yes. Every closed set is Borel, and the Borel σ-algebra is the smallest σ-algebra containing all closed sets.
However:
- The Lebesgue σ-algebra $ \mathcal{F}^* $ is strictly larger than the Borel σ-algebra $ \mathcal{F} $.
- Why? Because $ \mathcal{F}^* $ contains all subsets of null sets.
If $A \subset N$ where $N$ is Borel and $\mu(N)=0$, then: \(A \in \mathcal{F}^*.\)
Since the Cantor set is uncountable and has measure zero, it has $2^{\vert K\vert } = 2^{\vert \mathbb{R}\vert }$ subsets, most of which are not Borel — but all are Lebesgue measurable.
Thus: \(\vert \mathcal{F}\vert = \vert \mathbb{R}\vert < \vert \mathcal{F}^*\vert = 2^{\vert \mathbb{R}\vert }.\)
2. Integration of Simple Functions (Review)
Given a σ-finite measure space $(\Omega, \mathcal{F}, \mu)$, where: \(\exists E_n \uparrow \Omega, \quad \mu(E_n) < \infty,\) we define the integral of a simple function: \(\varphi = \sum_{i=1}^n a_i \mathbf{1}_{A_i}, \quad a_i \in \mathbb{R}, \quad A_i \in \mathcal{F},\; \mu(A_i) < \infty.\)
Definition: \(I(\varphi) = \int \varphi \, d\mu = \sum_{i=1}^n a_i\, \mu(A_i).\)
Properties for simple functions
-
Positivity:
\(\varphi \ge 0 \implies I(\varphi) \ge 0.\) -
Homogeneity:
\(I(a\varphi) = a I(\varphi).\) -
Additivity:
\(I(\varphi + \psi) = I(\varphi) + I(\psi).\)
The diagram on page 1 shows decomposing
$\varphi = \sum a_i 1{A_i}$
and
$\psi = \sum b_j 1{B_j}$
into rectangles $A_i \cap B_j$.
3. Extending Integration to a Non-Negative Measurable Function
Now assume:
- $\mu(\Omega) < \infty$ or more generally $\mu$ is σ-finite,
- $f : \Omega \to [0,\infty)$ is measurable.
We want to define: \(I(f) = \int_\Omega f\, d\mu.\)
Step 1: Define lower integral
\(I(f) := \sup\{\, I(\varphi): 0 \le \varphi \le f, \; \varphi \text{ simple} \}.\)
Step 2: Define upper integral
\(\tilde{I}(f) := \inf\{\, I(\psi) : \psi \ge f, \; \psi \text{ simple} \}.\)
Goal
Show: \(I(f) = \tilde{I}(f),\) so the integral is well-defined.
4. Proof That Lower and Upper Integrals Agree
Let $f \ge 0$ and bounded: $0 \le f \le M$.
Partition the range into $n$ equal pieces:
Define measurable sets: \(E_k^{(n)} = \left\{ x : \frac{(k-1)M}{n} < f(x) \le \frac{kM}{n} \right\}, \quad 1 \le k \le n.\)
Construct simple functions:
-
Upper step function
\(\psi_n(x) = \sum_{k=1}^n \frac{kM}{n}\, 1_{E_k^{(n)}} \ge f.\) -
Lower step function
\(\varphi_n(x) = \sum_{k=1}^n \frac{(k-1)M}{n}\, 1_{E_k^{(n)}} \le f.\)
Then: \(\psi_n(x) - \varphi_n(x) = \sum_{k=1}^n \frac{M}{n}\, 1_{E_k^{(n)}}.\)
Integrating: \(I(\psi_n) - I(\varphi_n) = \frac{M}{n} \sum_{k=1}^n \mu(E_k^{(n)}) = \frac{M}{n} \mu(\Omega) \xrightarrow[n\to\infty]{} 0.\)
Thus: \(0 \le \tilde{I}(f) - I(f) \le I(\psi_n) - I(\varphi_n) \to 0.\)
Therefore: \(\boxed{I(f) = \tilde{I}(f)}.\)
This completes the definition of $\int f\, d\mu$ for non-negative measurable $f$.
5. Properties of the Integral for Non-Negative $f,g$
For measurable $f,g \ge 0$ and $a \ge 0$:
-
Positivity: \(I(f) \ge 0.\)
-
Homogeneity: \(I(af) = a \, I(f).\)
-
Additivity: \(I(f+g) = I(f) + I(g).\)
The handwritten proof on pages 2–3 shows both the “inf over upper simple functions” and “sup over lower simple functions” arguments, concluding equality from both sides.
6. Next Topic (Preview)
The next lecture will move to:
- Defining the integral of general (not necessarily non-negative) measurable functions,
- Using positive/negative part:
\(f = f^+ - f^-, \quad f^\pm = \max(\pm f,0).\)
This will complete the construction of the Lebesgue integral.
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