Why Exhaustion Gives Uniqueness
An extension theorem answers whether a premeasure can be extended to a measure. A separate question is whether that extension is the only one. If the premeasure is finite on the whole underlying space, the finite-measure uniqueness argument applies directly. When the total measure is infinite, that argument cannot be applied globally. Sigma-finiteness resolves the problem locally: cover the space by sets of finite premeasure, compare the extensions on each such piece, and then let the pieces increase to the whole space.
Throughout, let \(\mathcal{R}\) be a ring of subsets of a set \(X\). A ring is closed under finite unions and set differences; in particular, it is closed under intersections and contains \(\varnothing\). Let \(\sigma(\mathcal{R})\) denote the sigma-algebra generated by \(\mathcal{R}\). A premeasure \(\mu_0:\mathcal{R}\to[0,\infty]\) is countably additive whenever a disjoint sequence in \(\mathcal{R}\) has its union in \(\mathcal{R}\). It is sigma-finite if \(X\) is covered by countably many sets in \(\mathcal{R}\) on which \(\mu_0\) is finite.
Here “extension” means agreement with \(\mu_0\) on every set in the ring. The conclusion concerns the generated sigma-algebra. If measures are instead defined on a larger sigma-algebra, this argument establishes their agreement on \(\sigma(\mathcal{R})\); agreement outside that sigma-algebra is not asserted.
Proof by Finite-Measure Restrictions
We use the pi-lambda theorem in its standard form: if a pi-system \(\mathcal{P}\) is contained in a lambda-system \(\mathcal{D}\), then \(\sigma(\mathcal{P})\subseteq\mathcal{D}\). A pi-system is a nonempty family closed under finite intersections. A lambda-system contains the whole space, is closed under complements relative to that space, and is closed under countable disjoint unions. The pi-lambda theorem is established in the earlier material on the pi-lambda method.
Proof. By sigma-finiteness, choose \(F_1,F_2,\ldots\in\mathcal{R}\) such that \(X=\bigcup_{j=1}^{\infty}F_j\) and \(\mu_0(F_j)<\infty\) for every \(j\). Set \(E_n=\bigcup_{j=1}^{n}F_j\). Since \(\mathcal{R}\) is a ring, \(E_n\in\mathcal{R}\), the sets increase, and \(\bigcup_n E_n=X\). Finite subadditivity of a premeasure gives \(\mu_0(E_n)\le\sum_{j=1}^n\mu_0(F_j)<\infty\).
Fix \(n\). Define measures on \(\sigma(\mathcal{R})\) by \(\mu_n(B)=\mu(B\cap E_n)\) and \(\nu_n(B)=\nu(B\cap E_n)\). These are measures: for disjoint measurable \(B_k\), the sets \(B_k\cap E_n\) are disjoint, and their union is \((\bigcup_k B_k)\cap E_n\), so countable additivity follows from that of \(\mu\) and \(\nu\). They are finite because \(\mu_n(X)=\mu(E_n)=\mu_0(E_n)<\infty\) and \(\nu_n(X)=\nu(E_n)=\mu_0(E_n)<\infty\).
For every \(A\in\mathcal{R}\), the intersection \(A\cap E_n\) belongs to \(\mathcal{R}\). Therefore \(\mu_n(A)=\mu_0(A\cap E_n)=\nu_n(A)\). The ring \(\mathcal{R}\) is a pi-system, since it is nonempty and closed under intersections, and it generates \(\sigma(\mathcal{R})\). Consider the agreement class \(\mathcal{D}_n=\{B\in\sigma(\mathcal{R}):\mu_n(B)=\nu_n(B)\}\). It contains \(X\), because the two finite total measures are equal. If \(B\in\mathcal{D}_n\), then finite additivity and equality of the totals give \(\mu_n(B^c)=\mu_n(X)-\mu_n(B)=\nu_n(X)-\nu_n(B)=\nu_n(B^c)\). For pairwise disjoint \(B_k\in\mathcal{D}_n\), countable additivity gives \(\mu_n(\bigcup_kB_k)=\sum_k\mu_n(B_k)=\sum_k\nu_n(B_k)=\nu_n(\bigcup_kB_k)\). Thus \(\mathcal{D}_n\) is a lambda-system. It contains \(\mathcal{R}\), so the pi-lambda theorem gives \(\mathcal{D}_n=\sigma(\mathcal{R})\).
Now fix any \(B\in\sigma(\mathcal{R})\). Since \(E_n\) increases to \(X\), the sets \(B\cap E_n\) increase to \(B\). Continuity from below for measures, applied separately to \(\mu\) and \(\nu\), yields \(\mu(B)=\lim_n\mu(B\cap E_n)\) and \(\nu(B)=\lim_n\nu(B\cap E_n)\), with infinite limits allowed. The restricted measures agree on every measurable set, so \(\mu(B\cap E_n)=\nu(B\cap E_n)\) for every \(n\). Their limits are equal, and hence \(\mu(B)=\nu(B)\). This holds for every \(B\in\sigma(\mathcal{R})\). \(\square\)
Worked Applications of Exhaustion
Worked Example: Counting Measure on the Natural Numbers
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), and let \(\mathcal{R}\) be the ring of finite subsets of \(\mathbb{N}\). Set \(\mu_0(F)=|F|\) for \(F\in\mathcal{R}\). This is a premeasure: whenever disjoint finite sets have a finite union, the cardinality of that union is the sum of their cardinalities. It is sigma-finite because \(E_n=\{1,\ldots,n\}\) increases to \(\mathbb{N}\) and \(\mu_0(E_n)=n<\infty\). Also, \(\sigma(\mathcal{R})\) is the power set of \(\mathbb{N}\), since every subset is a countable union of its singleton sets.
The counting measure \(\mu(A)=|A|\), allowing \(+\infty\), is an extension. The theorem says every measure extension to the power set is this counting measure. For \(B=\{2,4,6,\ldots\}\), its restriction to \(E_n\) counts the even integers from \(1\) through \(n\). If \(r=\lfloor n/2\rfloor\), then \(\mu(B\cap E_n)=r\), so \(\lim_{n\to\infty}\mu(B\cap E_n)=+\infty=\mu(B)\). Check: at \(n=7\), \(r=\lfloor7/2\rfloor=3\), and \(B\cap E_7=\{2,4,6\}\) has measure \(3\), as the formula gives.
Worked Example: A Weighted Premeasure with Infinite Total Measure
Again take \(X=\mathbb{N}\) and the ring of finite subsets, but define \(\mu_0(F)=\sum_{k\in F}k\). This is a premeasure because finite additivity holds on disjoint sets, and a disjoint sequence of nonempty subsets of a finite union can have only finitely many nonempty terms. For \(E_n=\{1,\ldots,n\}\), the premeasure is finite and equals \(\mu_0(E_n)=1+2+\cdots+n=n(n+1)/2\). Thus the premeasure is sigma-finite, although its extension has infinite total mass.
Define \(\mu(A)=\sum_{k\in A}k\), where a sum of nonnegative terms is allowed to equal \(+\infty\). This is a measure: for disjoint subsets \(A_j\), each integer \(k\) occurs in at most one \(A_j\), so summing the nonnegative weights over their union gives the sum of the separate sums. Consider the set \(B=\{2,4,6,\ldots\}\). If \(r=\lfloor n/2\rfloor\), then \(\mu(B\cap E_n)=2+4+\cdots+2r=2(1+\cdots+r)=r(r+1)\). As \(n\) increases, \(r\) increases without bound, so the restricted values tend to \(+\infty\), agreeing with \(\mu(B)=+\infty\). Check: for \(n=6\), \(r=3\), and the direct sum \(2+4+6=12\) equals \(3(3+1)=12\).
Worked Example: Finite Total Mass Recovered by Exhaustion
On the same ring of finite subsets of \(\mathbb{N}\), set \(\mu_0(F)=\sum_{k\in F}2^{-k}\). The finite-set additivity makes this a premeasure, and \(E_n=\{1,\ldots,n\}\) is a finite-measure exhaustion. The extension to the power set is \(\mu(A)=\sum_{k\in A}2^{-k}\), which is a measure by additivity of nonnegative series. Its total mass is \(\sum_{k=1}^{\infty}2^{-k}=1\).
For the even integers \(B=\{2,4,6,\ldots\}\), let \(r=\lfloor n/2\rfloor\). The restricted sum is \(\mu(B\cap E_n)=\sum_{j=1}^{r}2^{-2j}=\sum_{j=1}^{r}4^{-j}=\frac13(1-4^{-r})\), with value \(0\) when \(r=0\). Therefore continuity from below gives \(\mu(B)=\lim_{r\to\infty}\frac13(1-4^{-r})=\frac13\). Independently, the geometric series for the even-indexed terms is \(\frac{1/4}{1-1/4}=\frac13\). Check: for \(n=4\), the restricted sum is \(2^{-2}+2^{-4}=1/4+1/16=5/16\); the formula gives \(\frac13(1-4^{-2})=\frac13\cdot\frac{15}{16}=\frac5{16}\).
The Pitfall: A Piece Is Not the Whole Space
A common error is to prove agreement on \(B\cap E_n\) for each \(n\), then assert that the measures agree on \(B\) for one fixed \(n\). The valid conclusion uses the limit over the exhaustion. For example, with counting measure on \(\mathbb{N}\), take \(B=\mathbb{N}\) and \(E_n=\{1,\ldots,n\}\). Then \(\mu(B\cap E_n)=n\), while \(\mu(B)=+\infty\). No fixed finite \(n\) gives the value of \(\mu(B)\); only \(\lim_n\mu(B\cap E_n)=+\infty\) does.
The finite restriction is essential to the pi-lambda step: the agreement class must contain \(X\), and that requires equal finite total masses for the restricted measures. The global measures can have infinite total mass because the restriction to each \(E_n\) is finite. Exhaustion then reconnects these local comparisons to every set in the generated sigma-algebra.
Check Your Understanding
Answer using the distinction between agreement on a finite-measure piece and agreement on the full generated sigma-algebra.
- Why does replacing a sigma-finite cover by its successive finite unions produce an increasing exhaustion in the ring?
- For fixed \(n\), explain why \(B\mapsto\mu(B\cap E_n)\) is finite even if \(\mu(X)=+\infty\).
- Which property of a ring ensures that \(A\cap E_n\) is an admissible set on which the original extensions agree?
- In the counting-measure example, compute \(\mu(\{2,4,6,\ldots\}\cap E_{11})\), and state its limit as \(n\) tends to infinity.
- Why does the proof establish uniqueness on \(\sigma(\mathcal{R})\), but not automatically on a larger sigma-algebra?