# Well-definedness in measure theory

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

(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:

**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

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,

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

**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 \(\tilde{F}\in\mathcal{R}\),

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

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.