This blog covers very niche topics. Today: How to easily show that the Lebesgue measure and the Lebesgue integral is well-defined.

I recently started reading reading the “Bandit Algorithms” book by Tor Lattimore and Csaba Szepesvari. I like it a lot. Chapter 2 there got me back to one of my favorite topics in math: Measure Theory. While measure theory is often held as being boring or dry, I disagree. It’s just often presented in a suboptimal way. In particular, many authors get so carried away by a progression of simple functions – positive functions – real-valued functions that they forget that linearity is a great thing (one symptom of this: Trying to find a “common partition” for two simple functions).

Here we present a simple lemma and its corollary that takes care of most well-definedness questions in measure theory, in particular: Why the extension of a pre-measure from a semiring to a ring is well-defined (and therefore unique), and why the Lebesgue integral of simple functions is well-defined.

Lemma 1. Let $$\H$$ be a family of sets with the following partition property: For each finite family $$(A_1, \ldots, A_n)$$ in $$\H$$ there is a finite disjoint family $$(B_k\st k \in K_0)$$ in $$\H$$ such that

$\forall j \in\set{1,\ldots,n}\, \exists K_j \subseteq K_0: A_j=\mathop{\dot{\bigcup}}_{k \in K_j} B_k.$

(In particular, either $$A_j \supseteq B_k$$ or $$A_j\cap B_k=\emptyset$$ is true at all times, with $$A_j\supseteq B_k \Leftrightarrow k \in K_j.$$)

Let $$\mu\from \H\to[0, \infty)$$ be an additive set function (i.e., if $$A, B\in\H$$ are disjoint with $$A\mathbin{\dot{\cup}} B\in\H$$, then $$\mu(A\mathbin{\dot{\cup}} B) = \mu(A) + \mu(B)$$). Let $$A_1, \ldots, A_n \in \H$$ and $$a_1, \ldots, a_n \in \F \in\set{\R, \mathbb{C}}$$. Define

$\qquad b_k:=\sum_{\substack{1\le j\le n\\ A_j\supseteq B_k}} a_j \where{k\in K_0}.$

Then $$\sum_{j=1}^n a_j \1_{A_j}=\sum_{k\in K_0} b_k\1_{B_k}$$ and

$\sum_{j=1}^n a_j \mu(A_j)=\sum_{k \in K_0} b_k \mu(B_k)$

Proof. The first statement follows directly from the definition of $$b_k$$. The second:

\begin{aligned} \sum_{j=1}^n a_j \mu(A_j) &=\sum_{j=1}^n a_j \sum_{k \in K_j} \mu(B_k) \\ &=\sum_{k \in K_0} \underbrace{\sum_{j=1}^n \1_{K_j}(k) a_j}_{=b_k} \mu(B_k)=\sum_{k \in K_0} b_k \mu(B_k). \qquad\text{//}\end{aligned}

Corollary 2. Let $$\H, \mu$$ as in Lemma 1. Let $$X:=\lin\set{\1_A\st A\in\H}$$. Then $$J\from X\to\F$$, defined via

$X\ni f=\sum_{j=1}^n a_j \1_{A_j} \mapsto \sum_{j=1}^n a_j \mu(A_j),$

is well-defined and linear.

Proof. It suffices to treat the case $$f=0$$. Let $$(B_k\st k \in K_0)$$ be the disjoint representation and $$(b_k\st b \in K_0)$$ as in Lemma 1. Then $$b_k=0$$ for all $$k \in k_0,$$ as $$f=\sum_{j=1}^n a_j \1_{A_j}=\sum_{k \in K_0} b_k \1_{B_k}$$ and the $$B_k$$ are pairwise disjoint. By Lemma 1,

$\sum_{j=1}^n a_j \mu(A_j) % =\sum_{j=1}^n a_j \mu(\mathop{\dot{\bigcup}}_{k \in k_j}B_k) = \sum_{k\in K_0} b_k \mu(B_k)=0.$

This shows $$J$$ is well-defined; it’s also clearly linear. //

With this simple tool, we can look at our first application: The Lebesgue-Stieltjes pre-measure.

Lemma 3. Let $$\H$$ be the set of half-open intervals and let $$\mathcal{R}$$ be the ring of finite unions of half-open intervals. Then each $$F \in \mathcal{R}$$ is the disjoint union of finitely many intervals.

Proof (Adapted from J. Voigt, “Einführung in die Integration”, Satz 1.1.1). Let $$F=\bigcup_{j=1}^n I_j$$ with $$I_j=[a_j, b_j)$$. The set $$\set{a_j, b_j\st j=1,\ldots,n}$$ can be written as $$\set{x_k\st k=0,\ldots,m}$$ with $$x_0 <x_2 < \cdots < x_m$$. A set of disjoint intervals with union $$F$$ is

$\{[x_{k-1}, x_k)\st 1\le k\le m \text{ und } \exists j\in\set{1,\ldots,n} : [x_{k-1}, x_k) \subseteq[a_j,b_j)\}. \qquad\text{//}$

Corollary 4. Let $$G\from\R\to\R$$ be monotone. The mapping $$\mu\from\H\to[0,\infty)$$, defined via $$\mu([a,b)) := G(b)-G(a)$$, can be uniquely extended to an additive set function $$\mu\from\mathcal{R}\to[0, \infty)$$.

Proof. $$\mu$$ is additive and $$\H$$ has the partition property from Lemma 1 by Lemma 3. Hence $$J\from X\to\F$$ from Corollary 2 is well-defined. Let $$F\in\mathcal{R}$$. Then $$\1_F\in X$$ and thus $$\mu(F):=J(\1_F)$$ is well-defined and $$\geq 0$$. This extension $$\mu$$ is additive, since for $$\tilde{F}\in\mathcal{R}$$,

$\mu(F \mathbin{\dot{\cup}} \tilde{F}) = J(\1_F+\1_{\tilde{F}}) = J(\1_F) + J(\1_{\tilde{F}}) = \mu(F)+\mu(\tilde{F}).$

For the uniqueness: Let $$\tilde{\mu}\from\mathcal{R}\to[0,\infty)$$ be another additive extension and let $$F = \mathop{\dot{\bigcup}}_{k=1}^m B_k \in\mathcal{R}$$ with $$B_1, \ldots, B_m\in\H$$. Then

$\tilde{\mu}(F)=\tilde{\mu}(\mathop{\dot{\bigcup}} B_k)=\sum \tilde{\mu}(B_k)=\sum \mu(B_k)=\mu(F). \qquad//$

If $$G$$ left-continuous, $$\mu$$ can be shown to be $$\sigma$$-additive (and hence a pre-measure). One can then use Carathéodory’s extension theorem to extend $$\mu$$ to a measure on the $$\sigma$$-algebra generated by $$\mathcal{R}$$, which is the Borel $$\sigma$$-algebra $$\mathcal{B}(\R)$$.

Later in the theory, Lemma 1 proves the well-definedness of the Lebesgue integral for simple functions:

Lemma 5. Let $$(\Omega, \mathcal{A}, \mu)$$ be a measure space and let $$f=\sum_{j=1}^n a_j \1_{A_j}$$ be a simple function. Then $$\int_\Omega f {~d}\mu:=\sum_{j=1}^n a_j \mu(A_j)$$ is well-defined.

Proof. For $$K\subseteq K_0:=\{1, \ldots, n\}$$, define

$B_K:=\bigcap_{k\in K} A_k \cap \bigcap_{k \in K_0 \setminus K}(\Omega \setminus A_k)$

Then $$(B_K\st \emptyset \neq K \subseteq K_0)$$ is a partition as in Corollary 2, and hence $$\int f {~d}\mu=J(f)$$ is well-defined. //

This defines the Lebesgue integral for simple functions. Since simple functions are dense in $$L_1(\mu)$$, one is tempted to just use the BLT theorem now. That doesn’t quite work however, since the norm of $$L_1(\mu)$$ is defined via that integral in the first place. Instead, one can now define the integral for positive functions via pointwise convergence almost everywhere from below, then extend to real-valued (and complex-valued) functions.