What Completeness Requires
A measurable set of measure zero is negligible for measuring sets, but that alone does not make all its subsets measurable. Completeness is the additional requirement that no subset of a measurable null set is left outside the sigma-algebra. It is a property of the measure space, not merely a statement that certain sets have measure zero.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space: \(\mathcal{M}\) is a sigma-algebra on \(X\), and \(\mu:\mathcal{M}\to[0,\infty]\) is a countably additive measure with \(\mu(\varnothing)=0\). Unless specified otherwise, no finiteness or sigma-finiteness assumption is imposed.
In a complete measure space, every such subset \(A\) also has measure zero. Indeed, once completeness gives \(A\in\mathcal{M}\), monotonicity of the measure gives \(0\le\mu(A)\le\mu(N)=0\). The definition does not say that every subset of every measurable set is measurable; it concerns subsets of measurable sets of measure zero only.
A Two-Point Test
Worked Example: A Null Whole Space That Is Not Complete
Let \(X=\{a,b\}\), let \(\mathcal{M}=\{\varnothing,X\}\), and define \(\mu(E)=0\) for every \(E\in\mathcal{M}\). Determine whether \((X,\mathcal{M},\mu)\) is complete.
First, \(\mathcal{M}\) is a sigma-algebra: it contains \(\varnothing\) and \(X\), is closed under complements in \(X\), and every countable union of its members is either \(\varnothing\) or \(X\). The set function is a measure. For any pairwise disjoint sequence of sets from \(\mathcal{M}\), either all sets are empty, or exactly one can be \(X\) and the rest are empty. In either case, the measure of the union and the sum of the measures are both zero. This is a finite measure, since \(\mu(X)=0\), and it is sigma-finite, since \(X\) itself has finite measure.
Now \(X\in\mathcal{M}\) and \(\mu(X)=0\). But \(\{a\}\subseteq X\) and \(\{a\}\notin\mathcal{M}\). Therefore the space is not complete. The same check works with \(\{b\}\). Check: the relevant subset is not a set of measure zero in this space; it is not measurable at all, so \(\mu(\{a\})\) is undefined.
This example isolates the issue: a measurable null set can have subsets that the sigma-algebra does not contain. Having a null whole space does not force the sigma-algebra to be the power set.
Examples of Complete Spaces
Worked Example: Counting Measure on the Natural Numbers
Take \(X=\mathbb{N}=\{1,2,\ldots\}\), \(\mathcal{M}=\mathcal{P}(\mathbb{N})\), and counting measure \(\mu(E)=\#E\) when \(E\) is finite and \(\mu(E)=\infty\) when \(E\) is infinite. Is this space complete?
It is a measure space: for pairwise disjoint subsets, the number of elements in their union is the sum of their numbers of elements, with an infinite value if the union is infinite. It is sigma-finite because \(\mathbb{N}=\bigcup_{n=1}^{\infty}\{n\}\) and \(\mu(\{n\})=1\) for each \(n\). It is not a finite measure space, since \(\mu(\mathbb{N})=\infty\).
The only measurable null set is \(\varnothing\). In fact, every nonempty subset contains some \(n\), and its counting measure is at least \(1\). The only subset of \(\varnothing\) is \(\varnothing\), which belongs to \(\mathcal{M}\). Thus the space is complete. Check: the completeness condition holds, although there are many measurable sets of positive measure and the total measure is infinite.
Worked Example: A Complete Space With a Smaller Sigma-Algebra
Let \(X=\{a,b,c\}\), \(\mathcal{M}=\{\varnothing,\{a\},\{b,c\},X\}\), and assign \(\mu(\{a\})=0\), \(\mu(\{b,c\})=5\), \(\mu(\varnothing)=0\), and \(\mu(X)=5\). Determine completeness.
The sigma-algebra consists of all unions of the two disjoint blocks \(\{a\}\) and \(\{b,c\}\). The assigned block weights are nonnegative, and the measure of a union of blocks is the sum of their weights. For a disjoint sequence of measurable sets, each of the two blocks can occur in at most one nonempty member of the sequence; therefore the measure of the union is the sum of the measures in the sequence. This verifies countable additivity. The measure is finite, with \(\mu(X)=5\), and hence sigma-finite.
The measurable sets and their measures are exactly \(\varnothing\) with measure \(0\), \(\{a\}\) with measure \(0\), \(\{b,c\}\) with measure \(5\), and \(X\) with measure \(5\). Thus the measurable null sets are \(\varnothing\) and \(\{a\}\). Each subset of either set is measurable: the subsets of \(\{a\}\) are \(\varnothing\) and \(\{a\}\). The space is complete, even though \(\mathcal{M}\neq\mathcal{P}(X)\). Check: the nonmeasurable subsets \(\{b\}\) and \(\{c\}\) are not subsets of any measurable null set, so they do not violate completeness.
A Criterion for Finite Partition Spaces
The finite examples can be organized by examining the blocks that generate the sigma-algebra. Let \(X\) be a finite set partitioned into nonempty, pairwise disjoint blocks \(P_1,\ldots,P_r\), and let \(\mathcal{M}\) consist of all unions of these blocks. Suppose \(\mu\) is a measure on \((X,\mathcal{M})\), and write \(w_i=\mu(P_i)\). Since there are finitely many blocks and each \(w_i\in[0,\infty]\), this measure space has finite measure if and only if every \(w_i\) is finite.
Proof. Suppose first that the space is complete. If a block \(P_i\) has measure zero and contains two distinct points \(x\) and \(y\), then \(\{x\}\subseteq P_i\). Completeness would imply \(\{x\}\in\mathcal{M}\). But every member of \(\mathcal{M}\) is a union of whole partition blocks, and the block containing \(x\) is \(P_i\), which also contains \(y\). Thus no union of whole blocks can equal \(\{x\}\), a contradiction. So every zero-measure block is a singleton.
Conversely, suppose every zero-measure block is a singleton. Let \(N\in\mathcal{M}\) satisfy \(\mu(N)=0\). The set \(N\) is a union of some of the blocks. Since the measure is nonnegative and additive over this finite union, every block in that union has measure zero: if any had positive measure, the measure of \(N\) would be positive. By hypothesis, each block in \(N\) is a singleton. Every subset \(A\subseteq N\) is therefore a union of some of those singleton blocks, so \(A\in\mathcal{M}\). This verifies completeness, including the case \(N=\varnothing\), whose only subset is itself. \(\square\)
Worked Example: Applying the Finite Partition Criterion
Let \(X=\{a,b,c,d\}\), partitioned into \(P_1=\{a,b\}\) and \(P_2=\{c,d\}\). Let \(\mathcal{M}\) be all unions of these blocks, with \(\mu(P_1)=2\) and \(\mu(P_2)=3\). Decide completeness and verify the conclusion directly.
The measurable sets are \(\varnothing\), \(P_1\), \(P_2\), and \(X\), with measures \(0,2,3,5\), respectively. These values follow by finite additivity on the disjoint blocks. The only measurable null set is \(\varnothing\); hence its only subset is measurable, and the space is complete. The criterion gives the same answer: neither block has measure zero, so the condition on zero-measure blocks holds. Check: \(\mu(X)=\mu(P_1)+\mu(P_2)=2+3=5\), and no nonempty measurable set has measure zero.
A Common Mistake and Its Correction
A frequent error is to infer measurability from containment in a null set. In the two-point example, one might write, “Since \(\{a\}\subseteq X\) and \(\mu(X)=0\), \(\mu(\{a\})=0\).” This calculation is invalid: \(\{a\}\notin\mathcal{M}\), so \(\mu(\{a\})\) is not defined. Monotonicity says \(\mu(A)\le\mu(X)\) when \(A\in\mathcal{M}\) and \(A\subseteq X\); it does not establish \(A\in\mathcal{M}\). The correct test is to check membership first. Since \(\{a\}\notin\mathcal{M}\) while it is a subset of the measurable null set \(X\), the space is incomplete.
Completeness matters whenever arguments discard or modify subsets of null sets. Its role is specific: it guarantees measurability of those subsets. It should not be confused with sigma-finiteness, which asks whether the space is a countable union of measurable sets of finite measure, or with finiteness of the total measure. The counting-measure example is complete and sigma-finite but has infinite total measure; the two-point example is finite and sigma-finite but incomplete.
Check Your Understanding
For each question, check both sigma-algebra membership and measure values; a measure is defined only on measurable sets.
- In the two-point example, why does \(\mu(X)=0\) not give a value for \(\mu(\{a\})\)?
- Show that counting measure on a finite set \(X\), with \(\mathcal{M}=\mathcal{P}(X)\), is complete.
- For a partition sigma-algebra, what condition must each zero-measure block satisfy for completeness?
- In the three-point example, list all measurable null sets and all subsets of each one.
- Can a measure space be complete without its sigma-algebra being the power set? Verify your answer using one worked example.