What a Finite-Measure Cover Must Do
Sigma-finiteness is a property of a measure space, not merely a statement that some individual sets have finite measure. To test it, one must cover the whole space by countably many measurable sets, each of finite measure. Counting measure makes the distinction between countable and uncountable underlying sets especially visible; Dirac measure gives a contrasting case in which one set already suffices.
For a measure space \((X,\mathcal{M},\mu)\), a finite-measure cover is a sequence \((E_n)_{n\ge1}\) of sets in \(\mathcal{M}\) such that \(\bigcup_{n=1}^\infty E_n=X\) and \(\mu(E_n)<\infty\) for every \(n\). The measure space is sigma-finite if such a cover exists. A finite measure space, by contrast, has \(\mu(X)<\infty\). Thus sigma-finiteness permits \(\mu(X)=\infty\); it does not mean that the whole space has finite measure.
The Counting-Measure Test
Proof. First suppose \(X\) is countable. If \(X\) is finite, the sequence \(E_n=X\) for every \(n\) is a cover and \(\#(E_n)=|X|<\infty\). If \(X\) is empty, the same choice gives \(\#(E_n)=0\). If \(X\) is countably infinite, list its elements without repetition as \(x_1,x_2,\ldots\), and set \(E_n=\{x_1,\ldots,x_n\}\). Each \(E_n\) belongs to \(\mathcal{P}(X)\), has counting measure \(n<\infty\), and their union is \(X\). So the space is sigma-finite.
Conversely, suppose \((X,\mathcal{P}(X),\#)\) is sigma-finite, and let \((E_n)_{n\ge1}\) be a finite-measure cover. By the definition of counting measure, \(\#(E_n)<\infty\) implies that \(E_n\) is finite for every \(n\). A countable union of finite sets is countable: list the elements of each finite \(E_n\), and arrange these lists in a countable array indexed by pairs of positive integers. The pairs can be listed in diagonal order, so the union is at most countable. Since \(X=\bigcup_n E_n\), \(X\) is countable. This proves both implications. \(\square\)
The proposition identifies exactly what finite-measure sets can accomplish in this case: they are finite sets, so a countable collection of them can cover only a countable underlying set. The number of sets in the cover matters just as much as the measure of each set.
Worked Example: Counting Measure on the Positive Integers
Let \(X=\mathbb{N}=\{1,2,3,\ldots\}\), with \(\mathcal{P}(\mathbb{N})\) and counting measure \(\#\). We construct a finite-measure cover and check its measures.
For \(n\ge1\), let \(E_n=\{1,2,\ldots,n\}\). These are measurable because every subset of \(\mathbb{N}\) belongs to \(\mathcal{P}(\mathbb{N})\). They cover \(\mathbb{N}\): for each \(k\in\mathbb{N}\), \(k\in E_k\). Each set has exactly \(n\) elements, so the defining formula for counting measure gives \(\#(E_n)=n<\infty\). In particular, the first three sets have measures \(\#(E_1)=1\), \(\#(E_2)=2\), and \(\#(E_3)=3\). The countable sequence \((E_n)\) therefore verifies sigma-finiteness.
Check: the union is \(\bigcup_{n\ge1}\{1,\ldots,n\}=\mathbb{N}\), and every measure in the cover is finite by the explicit calculation \(\#(E_n)=n\). But \(\#(\mathbb{N})=\infty\), since \(\mathbb{N}\) is infinite. Thus this measure space is sigma-finite but not a finite measure space.
Worked Example: Why Counting Measure on the Reals Is Not Sigma-Finite
Take \(X=\mathbb{R}\), \(\mathcal{P}(\mathbb{R})\), and counting measure \(\#\). Suppose, for contradiction, that there were a finite-measure cover \((E_n)_{n\ge1}\) of \(\mathbb{R}\). Since \(\#(E_n)<\infty\), each \(E_n\) would have to be finite. A countable union of finite sets is countable, so \(\bigcup_{n\ge1}E_n\) would be countable. But the cover condition says that this union equals \(\mathbb{R}\), which is uncountable. The contradiction shows that no such cover exists.
Check: the obstruction is not that a proposed covering set has infinite measure and needs to be replaced. Every finite-counting-measure set is finite, and no sequence of finite sets covers \(\mathbb{R}\). The proposition therefore gives the same conclusion directly: the underlying set is uncountable, so the counting-measure space is not sigma-finite.
Dirac Measures: A One-Set Cover
Proof. The whole space \(Y\) belongs to \(\mathcal{A}\), and \(x\in Y\). The definition of the Dirac measure therefore gives \(\delta_x(Y)=1<\infty\). The sequence \(E_n=Y\) for every \(n\ge1\) is a countable measurable cover of \(Y\) by finite-measure sets. Hence the space is sigma-finite. \(\square\)
This argument needs no assumption that the singleton \(\{x\}\) belongs to \(\mathcal{A}\). The measurable set used in the cover is \(Y\) itself, which is always measurable. In fact, every measurable set has Dirac measure either \(0\) or \(1\), so all measurable sets have finite Dirac measure; the single-set cover is already enough to establish sigma-finiteness.
Worked Example: A Dirac Measure on the Borel Real Line
Let \(Y=\mathbb{R}\), \(\mathcal{A}=\mathcal{B}_{\mathbb{R}}\), and take the Dirac measure \(\delta_3\). The sequence \(E_n=\mathbb{R}\) is a measurable cover. Since \(3\in\mathbb{R}\), the definition gives \(\delta_3(E_n)=\delta_3(\mathbb{R})=1\) for every \(n\). Thus each covering set has finite measure and the cover proves sigma-finiteness.
One can also see how membership determines the measure using the Borel sets \(A=(-\infty,3)\) and \(B=[3,\infty)\). Since \(3\notin A\), \(\delta_3(A)=0\); since \(3\in B\), \(\delta_3(B)=1\). Both sets are measurable, \(A\cup B=\mathbb{R}\), and both measures are finite. Check: the point \(3\) belongs to exactly one of these two sets, so the two values add to \(0+1=1=\delta_3(\mathbb{R})\).
Worked Example: A Dirac Measure When Singletons Are Not Measurable
Let \(Y=\{a,b\}\) and let \(\mathcal{A}=\{\varnothing,Y\}\), the sigma-algebra containing only the empty set and the whole space. Choose \(x=a\). The singleton \(\{a\}\) is not in \(\mathcal{A}\), so it cannot be used as a measurable member of a cover. Nevertheless, the Dirac measure on this measurable space is well-defined by membership: \(\delta_a(\varnothing)=0\) and \(\delta_a(Y)=1\).
Take \(E_n=Y\) for every \(n\). Then \(\bigcup_{n\ge1}E_n=Y\), each \(E_n\) is in \(\mathcal{A}\), and \(\delta_a(E_n)=1<\infty\). This explicit cover verifies sigma-finiteness without using the nonmeasurable singleton. Check: the only two measurable sets have measures \(0\) and \(1\), so the whole-space covering set has finite measure as required.
A Common Mistake and a Reliable Test
Common mistake: “Counting measure on \(\mathbb{R}\) is sigma-finite because \(\mathbb{R}\) is the union of its singletons, and each singleton has measure \(1\).” The calculation for an individual singleton is correct: \(\#(\{x\})=1\) for each \(x\in\mathbb{R}\). The error is in the cover: sigma-finiteness requires a countable family of finite-measure sets, whereas the family \(\{\{x\}:x\in\mathbb{R}\}\) is uncountable. It is not a permissible sigma-finite cover. The correction is to use the proposition: a countable family of finite-counting-measure sets consists of finite sets, whose union is countable, so it cannot cover \(\mathbb{R}\).
For these measures, a useful test is to write down the finite-measure sets before trying to build a cover. Under counting measure they are exactly the finite sets, so sigma-finiteness is equivalent to countability of the underlying set. Under a Dirac measure every measurable set has measure at most \(1\), and the whole space itself is a finite-measure cover. These conclusions depend on the definitions of the two measures; they should not be transferred to other measures without checking their finite-measure sets.
Check Your Understanding
For each question, identify the measurable space, specify the proposed covering sets, and verify the measure condition rather than relying only on the name of the measure.
- For counting measure on a finite set \(X=\{u,v,w\}\), give a finite-measure cover and compute the measure of each covering set.
- Explain why a countable family of finite-measure sets cannot cover an uncountable set under counting measure.
- For a Dirac measure \(\delta_x\) on an arbitrary measurable space \((Y,\mathcal{A})\), which one-set cover proves sigma-finiteness, and what is the measure of its covering set?
- In the measurable space \((\{a,b\},\{\varnothing,\{a,b\}\})\), explain why \(\{a\}\) cannot be used in a measurable cover but the Dirac measure at \(a\) is still sigma-finite.
- Does sigma-finiteness of counting measure on \(\mathbb{N}\) imply that the total measure \(\#(\mathbb{N})\) is finite? Justify your answer using the definitions.