What Completeness Adds
Lebesgue measure regards every subset of a null set as measurable, even when that subset is not Borel. This is stronger than merely knowing that the null set itself is measurable: it enlarges the collection of measurable sets while assigning measure zero to all the newly included subsets. The key step is to verify the Caratheodory criterion directly using outer measure.
Throughout, \(m^*\) denotes Lebesgue outer measure on \(\mathbb{R}\), and \(m\) denotes Lebesgue measure on \((\mathbb{R},\mathcal{L})\). This measure space is complete and sigma-finite, but not finite, since \(m(\mathbb{R})=\infty\). The Borel restriction is the measure \(m_{\mathcal{B}}\) on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\) defined by \(m_{\mathcal{B}}(B)=m(B)\) for Borel \(B\); it is sigma-finite, but it is not complete. Here, sigma-finite means that the space is a countable union of measurable sets of finite measure. Completeness is a different property.
Recall the Caratheodory criterion: a set \(E\subseteq\mathbb{R}\) is Lebesgue measurable exactly when, for every \(A\subseteq\mathbb{R}\), \(m^*(A)=m^*(A\cap E)+m^*(A\setminus E)\). We use the standard properties of outer measure: monotonicity and countable subadditivity. In particular, if \(N\) is a Lebesgue null set, then \(m^*(N)=0\).
The Completeness Theorem
Proof. Fix \(S\subseteq N\). By monotonicity of outer measure, \(0\le m^*(S)\le m^*(N)=0\), so \(m^*(S)=0\). Now let \(A\subseteq\mathbb{R}\) be arbitrary. Since \(A\cap S\subseteq S\), monotonicity gives \(m^*(A\cap S)=0\). Also \(A\setminus S\subseteq A\), so \(m^*(A\setminus S)\le m^*(A)\). In the other direction, \(A=(A\setminus S)\cup(A\cap S)\), and subadditivity gives \(m^*(A)\le m^*(A\setminus S)+m^*(A\cap S)=m^*(A\setminus S)\). Thus \(m^*(A)=m^*(A\setminus S)\), including when this value is infinite. Therefore \(m^*(A)=m^*(A\setminus S)+m^*(A\cap S)\), which is the Caratheodory criterion for \(S\). Hence \(S\in\mathcal{L}\), and \(m(S)=m^*(S)=0\). This holds for every subset of every Lebesgue null set, so the Lebesgue measure space is complete. \(\square\)
The proof does not assume that \(A\) has finite measure. In particular, it avoids any subtraction of infinite quantities: it establishes equality by comparing the outer measures of \(A\) and \(A\setminus S\) directly.
Worked Example: Every Subset of a Singleton Is Null
Take \(N=\{0\}\subset\mathbb{R}\). The set \(N\) has Lebesgue measure zero: for every \(\varepsilon>0\), it is covered by the interval \((-\varepsilon/2,\varepsilon/2)\), whose length is \(\varepsilon\). Thus \(m^*(N)\le\varepsilon\) for every \(\varepsilon>0\), so \(m^*(N)=0\). Its only subsets are \(\varnothing\) and \(\{0\}\). For either subset \(S\), monotonicity gives \(0\le m^*(S)\le m^*(N)=0\), hence \(m^*(S)=0\).
For example, for any \(A\subseteq\mathbb{R}\), the pieces \(A\cap\{0\}\) and \(A\setminus\{0\}\) split \(A\), and the first piece has outer measure zero. Subadditivity gives \(m^*(A)\le m^*(A\setminus\{0\})\), while monotonicity gives the reverse inequality. The criterion therefore holds for \(\{0\}\). Check: applying the interval cover with \(\varepsilon=1/10\) gives length \((1/10)/2-(-(1/10)/2)=1/10\); since this works for every positive \(\varepsilon\), the outer measure is zero, not merely at most \(1/10\).
Why the Borel Restriction Is Not Complete
Every Borel set is Lebesgue measurable, but not every Lebesgue measurable set is Borel. Completeness explains one source of this difference: a subset of a Borel null set is Lebesgue measurable whether or not it is Borel. The following construction exhibits a Borel null set with a non-Borel subset. It is used here to distinguish the two sigma-algebras.
Worked Example: A Null Borel Set with a Non-Borel Subset
For each \(n\ge1\), form \(U_n\) as the union of the \(2^n\) closed intervals \[ \left[\sum_{j=1}^{n}\frac{2a_j}{3^j},\ \sum_{j=1}^{n}\frac{2a_j}{3^j}+\frac{1}{3^n}\right], \qquad (a_1,\ldots,a_n)\in\{0,1\}^n. \] Each interval has length \(3^{-n}\), so the sum of the lengths at stage \(n\) is \(2^n3^{-n}=(2/3)^n\). The intervals are nested across stages: each interval at stage \(n\) contains the two intervals obtained by choosing \(a_{n+1}=0\) and \(a_{n+1}=1\). Thus \(N=\bigcap_{n=1}^{\infty}U_n\) is Borel, as a countable intersection of finite unions of closed intervals.
Every point of \(N\) corresponds to an infinite sequence \(a_j\in\{0,1\}\) and has the form \(\sum_{j=1}^{\infty}2a_j3^{-j}\). To see the correspondence, the nested intervals containing a point determine successive digits; their lengths \(3^{-n}\) tend to zero, and their left endpoints converge to the displayed series. Conversely, each such series lies in the interval for its first \(n\) digits at every stage. Distinct digit sequences give distinct points: if the first differing digit is at \(k\), the leading difference is \(2\cdot3^{-k}\), while the sum of all possible later differences is at most \(\sum_{j>k}2\cdot3^{-j}=3^{-k}\). The two series therefore differ by at least \(3^{-k}>0\). There are continuum many infinite binary sequences, so \(N\) has cardinality \(\mathfrak{c}\), the cardinality of \(\mathbb{R}\).
The intervals of \(U_n\) cover \(N\), and their total length is \((2/3)^n\), which tends to zero. Hence \(0\le m^*(N)\le(2/3)^n\) for every \(n\), so \(m^*(N)=0\) and \(m(N)=0\). Since \(N\) has \(\mathfrak{c}\) points, its power set has cardinality strictly greater than \(\mathfrak{c}\), by Cantor's theorem. There are only \(\mathfrak{c}\) Borel subsets of \(\mathbb{R}\): Borel sets can be coded by countable constructions from a countable base, and the collection of such countable codes has cardinality at most \(\mathfrak{c}\). Therefore at least one subset \(S\subseteq N\) is not Borel.
Completeness now gives \(S\in\mathcal{L}\) and \(m(S)=0\), although \(S\notin\mathcal{B}_{\mathbb{R}}\). Thus \(N\) is a null set in the Borel restriction \(m_{\mathcal{B}}\), but that restriction does not contain every subset of \(N\) in its sigma-algebra. It is not complete. Check: the stage-\(n\) total cover length is \(2^n/3^n=(2/3)^n\); since \(0<2/3<1\), these bounds tend to zero, as required for the null-set conclusion.
The cardinality argument is only used to guarantee the existence of a non-Borel subset; it does not identify a particular such subset. Completeness itself is more direct: once a set is contained in a measurable null set, the outer-measure criterion proves its measurability without needing to describe it.
Changing a Function on a Null Subset
Worked Example: A Lebesgue-Measurable Indicator That Is Not Borel-Measurable
Let \(N\) and \(S\subseteq N\) be as in the preceding example, with \(S\) non-Borel. Define the real-valued function \(f=\chi_S\) on the Lebesgue measure space \((\mathbb{R},\mathcal{L},m)\). Since \(S\in\mathcal{L}\), \(f\) is Lebesgue measurable: for any real \(t\), the set \(\{x:f(x)>t\}\) is one of \(\varnothing\), \(S\), or \(\mathbb{R}\), according to the value of \(t\).
The function is a nonnegative simple function, and its integral is \[ \int_{\mathbb{R}}f\,dm=1\cdot m(S)=1\cdot0=0. \] It is not Borel-measurable, because \(\{x:f(x)>1/2\}=S\) is not Borel. Check: the only nonzero value of \(f\) is \(1\), and it occurs on a set of measure zero; the simple-function integral is therefore exactly zero. The strict inequality \(f(x)>1/2\) selects precisely \(S\).
This example separates three notions that should not be conflated. The set \(S\) is Lebesgue measurable, it is not Borel, and it has measure zero. The function \(\chi_S\) is Lebesgue measurable and integrable, but not Borel-measurable. Its integral is unchanged if it is replaced by the zero function, because the two functions differ only on the null set \(S\).
A Common Pitfall
A tempting but incorrect argument is: “\(m^*(S)=0\), so \(m(S)=0\); therefore \(S\) is measurable.” The expression \(m(S)\) is not available until \(S\) has been shown measurable, so this reasoning assumes its conclusion. The correction is to use \(m^*\) throughout and prove the Caratheodory identity for every \(A\subseteq\mathbb{R}\). When \(S\subseteq N\) and \(m^*(N)=0\), monotonicity gives \(m^*(A\cap S)=0\), and the two outer-measure inequalities give \(m^*(A)=m^*(A\setminus S)\). Only after this criterion proves \(S\in\mathcal{L}\) may we write \(m(S)=m^*(S)=0\).
Completeness is thus a property of both a measure and its chosen sigma-algebra. The Borel restriction and Lebesgue measure agree on Borel sets, but the Borel sigma-algebra omits some subsets of Borel null sets. Lebesgue measure includes those subsets by completing the Borel measure.
Check Your Understanding
Use the Caratheodory criterion and keep outer measure distinct from measure until measurability has been established.
- Let \(N\in\mathcal{L}\) be null and \(S\subseteq N\). For arbitrary \(A\subseteq\mathbb{R}\), show separately that \(m^*(A\cap S)=0\) and \(m^*(A\setminus S)=m^*(A)\).
- Explain why completeness is not the same property as sigma-finiteness.
- Why does a non-Borel subset of a Borel null set show that \(m_{\mathcal{B}}\) is not complete?
- For the indicator \(f=\chi_S\) in the worked example, compute \(\{x:f(x)>1/2\}\) and \(\int f\,dm\).
- In the pitfall argument, at what point is it legitimate to write \(m(S)\) rather than \(m^*(S)\)?