Tutorials › Measure Theory › Indicators of Lebesgue-Measurable Non-Borel Sets

Home  /  Tutorials  /  Measurable Functions

Measurable Functions · Tutorial 199 of 798

Indicators of Lebesgue-Measurable Non-Borel Sets

See how a null set can contain a Lebesgue-measurable non-Borel subset, and exactly how that set determines the measurability of its indicator.

Intermediate 9 min read

What You'll Learn

  • Use the Cantor set and completeness of Lebesgue measure to obtain a measurable non-Borel set.
  • Compute the preimage of any Borel target set under an indicator function.
  • Prove the exact criteria for an indicator to be Lebesgue-measurable or Borel-measurable.
  • Distinguish a function's range from the measurability of its preimages.
  • Identify a common error when checking measurability of an indicator.

Why Indicators Separate Two Notions of Measurability

A real-valued function can take only two values and still distinguish Lebesgue measurability from Borel measurability. The key is that measurability concerns preimages: for an indicator, one of those preimages is precisely the set whose measurability is in question. We will construct a Lebesgue-measurable set that is not Borel, and then calculate every Borel preimage of its indicator.

Throughout the first part, the measure space is \((\mathbb{R},\mathcal{L},m)\), with Lebesgue measure \(m\). It is complete and sigma-finite, but not finite. Completeness means that every subset of an \(m\)-null set belongs to \(\mathcal{L}\). The Borel sigma-algebra \(\mathcal{B}_{\mathbb{R}}\) is contained in \(\mathcal{L}\), but the two sigma-algebras are not equal.

Definition: If \(E\subseteq\mathbb{R}\), its indicator is the real-valued function \(\chi_E:\mathbb{R}\to\mathbb{R}\) defined by \(\chi_E(x)=1\) for \(x\in E\) and \(\chi_E(x)=0\) for \(x\notin E\). A function \(f:\mathbb{R}\to\mathbb{R}\) is Borel-measurable if \(f^{-1}(B)\in\mathcal{B}_{\mathbb{R}}\) for every \(B\in\mathcal{B}_{\mathbb{R}}\); it is Lebesgue-measurable if \(f^{-1}(B)\in\mathcal{L}\) for every such \(B\).

Finding a Lebesgue-Measurable Non-Borel Set

Let \(C\subseteq[0,1]\) be the middle-thirds Cantor set. At stage \(n\), the set remaining in its construction is covered by \(2^n\) closed intervals, each of length \(3^{-n}\). Their total length is \(2^n3^{-n}=(2/3)^n\). Since \(C\) is contained in the stage-\(n\) set for every \(n\), its Lebesgue outer measure satisfies \(m^*(C)\le(2/3)^n\) for every \(n\). As \((2/3)^n\to0\), \(m^*(C)=0\); in particular, \(C\) is a Borel null set.

There are strictly more subsets of \(C\) than there are Borel subsets of \(\mathbb{R}\). Here are the cardinality facts behind that assertion. The Cantor set has cardinality \(\mathfrak{c}\), the cardinality of \(\mathbb{R}\), so its collection of subsets has cardinality \(2^{\mathfrak{c}}>\mathfrak{c}\), by Cantor's theorem. On the other hand, there are at most \(\mathfrak{c}\) Borel subsets of \(\mathbb{R}\). To see the upper bound, start with the countable basis of open intervals with rational endpoints. Borel sets are obtained from this basis using complements and countable unions. Each such construction can be encoded by a countable, well-founded tree: leaves identify basis elements, and operation nodes record complements or countable unions. Each tree is a subset of the countable set of finite sequences of natural numbers, with countably many possible labels. There are at most \(\mathfrak{c}\) such codes. The recursive description covers every Borel set: at each countable stage of the Borel construction, a countable union combines countably many countable codes into another countable code, and complementation adds one operation node.

Worked Example: Choosing a Measurable Set That Is Not Borel

Choose a subset \(A\subseteq C\) that is not Borel. Such a subset exists because \(C\) has \(2^{\mathfrak{c}}\) subsets while \(\mathbb{R}\) has at most \(\mathfrak{c}\) Borel sets, and \(2^{\mathfrak{c}}>\mathfrak{c}\).

Because \(C\) is an \(m\)-null measurable set and \((\mathbb{R},\mathcal{L},m)\) is complete, every subset of \(C\) is Lebesgue measurable. Thus \(A\in\mathcal{L}\), although \(A\notin\mathcal{B}_{\mathbb{R}}\). Also \(m(A)=0\), since \(A\subseteq C\) and \(m(C)=0\). Check: if \(A\) were Borel, then it would be one of the at most \(\mathfrak{c}\) Borel sets, but the cardinality argument ensures a subset of \(C\) outside that collection. The conclusion is existence; this construction does not specify a particular named subset \(A\).

This use of completeness matters. Knowing only that \(C\) is Borel and null would not, without completeness, imply that every subset of \(C\) is measurable in an arbitrary measure space. The assertion here uses the completed Lebesgue measure on \(\mathbb{R}\).

Preimages of an Indicator

For any set \(E\subseteq\mathbb{R}\) and any set \(B\subseteq\mathbb{R}\), the values of \(\chi_E\) show that its preimage has only four possibilities. If \(0,1\in B\), every point maps into \(B\); if \(1\in B\) but \(0\notin B\), exactly the points of \(E\) map into \(B\); if \(0\in B\) but \(1\notin B\), exactly the points outside \(E\) do; and if neither value belongs to \(B\), no point does.

$$ \chi_E^{-1}(B)= \begin{cases} \mathbb{R}, & 0\in B\text{ and }1\in B,\\ E, & 1\in B\text{ and }0\notin B,\\ E^c, & 0\in B\text{ and }1\notin B,\\ \varnothing, & 0\notin B\text{ and }1\notin B. \end{cases} $$

This calculation proves the exact measurability criterion, with no restriction that \(E\) be null or have positive measure.

Theorem: On \((\mathbb{R},\mathcal{L},m)\), the real-valued indicator \(\chi_E\) is Lebesgue-measurable if and only if \(E\in\mathcal{L}\). It is Borel-measurable if and only if \(E\in\mathcal{B}_{\mathbb{R}}\).

Proof. First consider Lebesgue measurability. If \(E\in\mathcal{L}\), then \(E^c\in\mathcal{L}\), and \(\mathbb{R}\) and \(\varnothing\) belong to \(\mathcal{L}\). The displayed preimage classification therefore gives \(\chi_E^{-1}(B)\in\mathcal{L}\) for every Borel set \(B\), so \(\chi_E\) is Lebesgue-measurable. Conversely, if \(\chi_E\) is Lebesgue-measurable, take the Borel set \(B=(1/2,3/2)\). It contains \(1\) and not \(0\), so \(\chi_E^{-1}(B)=E\). Measurability of the function implies \(E\in\mathcal{L}\).

For Borel measurability, if \(E\in\mathcal{B}_{\mathbb{R}}\), then its complement and the two trivial sets are Borel, so the same classification gives Borel preimages for every Borel \(B\). Conversely, if \(\chi_E\) is Borel-measurable, the same test set \(B=(1/2,3/2)\) gives \(E=\chi_E^{-1}(B)\in\mathcal{B}_{\mathbb{R}}\). Both equivalences follow. \(\square\)

Worked Example: Every Borel Preimage of the Non-Borel Indicator

Use the set \(A\subseteq C\) chosen above and let \(f=\chi_A:\mathbb{R}\to\mathbb{R}\). For any Borel target set \(B\), apply the four cases with \(E=A\). In particular, for the Borel sets \((1/2,3/2)\), \((-1/2,1/2)\), \(\mathbb{R}\), and \(\mathbb{R}\setminus\{0,1\}\), the preimages are respectively \(A\), \(A^c\), \(\mathbb{R}\), and \(\varnothing\).

For the first target, \(1\in(1/2,3/2)\) and \(0\notin(1/2,3/2)\), so the preimage is \(A\). For the second, \(0\in(-1/2,1/2)\) and \(1\notin(-1/2,1/2)\), so the preimage is \(A^c\). The whole line contains both values, and \(\mathbb{R}\setminus\{0,1\}\) contains neither, giving the last two preimages. Since \(A\in\mathcal{L}\), all four preimages are Lebesgue measurable. But \(A\) is not Borel, so the first preimage also proves that \(f\) is not Borel-measurable. Check: \(f(x)\) is always either \(0\) or \(1\); hence it belongs to \((1/2,3/2)\) exactly when \(x\in A\), as calculated.

Worked Example: The Complement Indicator

Let \(g=\chi_{A^c}\) on the same measure space \((\mathbb{R},\mathcal{L},m)\). The set \(A^c\) is Lebesgue measurable because \(\mathcal{L}\) is a sigma-algebra. It is not Borel: if \(A^c\) were Borel, its complement \(A\) would be Borel as well, contradicting the choice of \(A\).

For the target \(B=(1/2,3/2)\), we have \(g^{-1}(B)=A^c\), which is Lebesgue measurable but not Borel. For \(B=(-1/2,1/2)\), we have \(g^{-1}(B)=(A^c)^c=A\), also Lebesgue measurable and not Borel. In general, the four-case formula applies with \(E=A^c\), so every Borel target has a Lebesgue-measurable preimage. Check: on \(A^c\), \(g=1\), and on \(A\), \(g=0\); these values give the two stated preimages directly.

A Common Mistake: Checking Only the Range

A tempting but incorrect argument is: “The indicator takes only the values \(0\) and \(1\), so it is Borel-measurable.” Having a finite range does not make a function Borel-measurable. For the specific function \(f=\chi_A\) above, the wrong calculation would assert that \(f^{-1}((1/2,3/2))\) is Borel because \((1/2,3/2)\) is Borel and \(f\) has two values. The actual calculation is

$$ f^{-1}((1/2,3/2))=\{x:f(x)=1\}=A, $$

and \(A\) is not Borel. The correction is to test preimages, not just values: a function with range \(\{0,1\}\) is Borel-measurable exactly when the set where it equals \(1\) is Borel. In contrast, the same indicator is Lebesgue-measurable because \(A\) is Lebesgue measurable.

The example also shows why the distinction between Borel and Lebesgue measurability is substantive. The domain's sigma-algebra determines which preimages are permitted. For an indicator, all the nontrivial preimages reduce to the defining set or its complement, making the distinction especially transparent.

Check Your Understanding

For each question, use the exact preimage classification and state which sigma-algebra is being tested.

  1. If \(E\subseteq\mathbb{R}\), compute \(\chi_E^{-1}((-\infty,1/2))\) and \(\chi_E^{-1}([1/2,\infty))\).
  2. Suppose \(E\in\mathcal{L}\setminus\mathcal{B}_{\mathbb{R}}\). Is \(\chi_E\) Lebesgue-measurable, Borel-measurable, both, or neither? Give a preimage that verifies your answer.
  3. Explain where completeness is used to show that every subset of the Cantor set is Lebesgue measurable.
  4. If \(\chi_E\) is Borel-measurable, use a Borel target set to prove that \(E\) is Borel.
  5. Why does a function's finite range alone not establish its Borel measurability?

Previous

Next

Back to Tutorials