Two Values Encode a Set
An indicator function converts a set into a function by assigning value \(1\) on the set and value \(0\) off it. This simple construction is a basic way to represent sets as functions: questions about which points belong to the set become questions about where the function takes each of its two values. In particular, measurability of the function is exactly measurability of the set it records.
Let \((X,\mathcal{M})\) be a measurable space, meaning that \(\mathcal{M}\) is a sigma-algebra of subsets of \(X\). For a function \(f:X\to\mathbb{R}\), Borel measurability means that \(f^{-1}(B)\in\mathcal{M}\) for every \(B\in\mathcal{B}_{\mathbb{R}}\), the Borel sigma-algebra on \(\mathbb{R}\). This is a property relative to the chosen domain sigma-algebra \(\mathcal{M}\).
The indicator takes only the values \(0\) and \(1\). Its strict and non-strict level sets therefore change only when the threshold crosses one of these values. The endpoint cases matter: for example, \(\{\chi_E>1\}\) is empty, whereas \(\{\chi_E\ge 1\}=E\).
Level Sets of an Indicator
Proof. If \(x\in E\), then \(\chi_E(x)=1\); if \(x\in E^c\), then \(\chi_E(x)=0\). For the strict inequality, when \(a<0\), both possible values \(0\) and \(1\) exceed \(a\), so every \(x\in X\) belongs to the level set. When \(0\le a<1\), the value \(1\) exceeds \(a\), but \(0\) does not, so the set is exactly \(E\). When \(a\ge1\), neither value exceeds \(a\), and the set is empty. For the non-strict inequality, if \(a\le0\), both values are at least \(a\); if \(0<a\le1\), only the value \(1\) is at least \(a\); and if \(a>1\), neither value is at least \(a\). These give \(X\), \(E\), and \(\varnothing\), respectively. For equality, the value \(0\) occurs exactly on \(E^c\), the value \(1\) occurs exactly on \(E\), and no other value is taken. \(\square\)
Worked Example: An Interval with Both Endpoints Checked
Take \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{L}\), the Lebesgue sigma-algebra, and \(E=[-1,2)\). Determine the strict, non-strict, and equality level sets of \(\chi_E\) at \(a=0\), \(a=1\), and \(a=2\).
The interval includes \(-1\) and excludes \(2\). Thus \(\chi_E(-1)=1\), \(\chi_E(2)=0\), and the two possible function values everywhere are \(0\) and \(1\). At \(a=0\), the strict level set is \(\{\chi_E>0\}=E=[-1,2)\); the non-strict level set is \(\{\chi_E\ge0\}=\mathbb{R}\); and the equality level set is \(\{\chi_E=0\}=E^c=(-\infty,-1)\cup[2,\infty)\).
At \(a=1\), the strict level set is \(\{\chi_E>1\}=\varnothing\); the non-strict level set is \(\{\chi_E\ge1\}=E=[-1,2)\); and \(\{\chi_E=1\}=E\). At \(a=2\), both \(\{\chi_E>2\}\) and \(\{\chi_E\ge2\}\) are empty, and \(\{\chi_E=2\}\) is empty. Check: the endpoint \(2\) lies outside \(E\), so \(\chi_E(2)=0\); it belongs to the computed set \(\{\chi_E=0\}\), but not to \(\{\chi_E=1\}\). This agrees with the definition of the half-open interval.
The Measurability Characterization
Proof. First suppose \(E\in\mathcal{M}\). Fix any Borel set \(B\in\mathcal{B}_{\mathbb{R}}\). Since \(\chi_E\) takes only the values \(0\) and \(1\), its preimage of \(B\) has one of four forms: it is \(X\) if both \(0\) and \(1\) belong to \(B\); it is \(\varnothing\) if neither belongs to \(B\); it is \(E\) if \(1\in B\) and \(0\notin B\); and it is \(E^c\) if \(0\in B\) and \(1\notin B\). Each of \(X,\varnothing,E,E^c\) belongs to \(\mathcal{M}\), because \(\mathcal{M}\) is a sigma-algebra and \(E\in\mathcal{M}\). Thus every Borel preimage is measurable, so \(\chi_E\) is measurable.
Conversely, suppose \(\chi_E\) is measurable. The interval \((1/2,\infty)\) is Borel, and \(\chi_E^{-1}((1/2,\infty))=\{x:\chi_E(x)>1/2\}=E\), since \(1>1/2\) and \(0\) is not greater than \(1/2\). Measurability of \(\chi_E\) makes this preimage a member of \(\mathcal{M}\). Hence \(E\in\mathcal{M}\). Both implications hold, including \(E=\varnothing\) and \(E=X\), so the equivalence is proved. \(\square\)
The proof also shows why a threshold strictly between \(0\) and \(1\) is useful: it separates the two possible values. Thresholds outside that range can produce only \(X\) or \(\varnothing\), which do not reveal the defining set.
Worked Example: An Indicator Supported on a Null Set
On \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure, let \(N=\{1/n:n\in\mathbb{N},\,n\ge1\}\). Show that \(\chi_N\) is measurable, compute its level sets, and explain what “zero almost everywhere” means here.
Each singleton \(\{1/n\}\) has Lebesgue measure zero, and countable subadditivity gives \(m(N)\le\sum_{n=1}^{\infty}m(\{1/n\})=0\). Since measures are nonnegative, \(m(N)=0\). In particular \(N\in\mathcal{L}\), so the theorem implies that \(\chi_N\) is measurable.
For \(0\le a<1\), the strict level set \(\{\chi_N>a\}\) is \(N\); for \(a<0\), it is \(\mathbb{R}\); and for \(a\ge1\), it is empty. For example, \(\{\chi_N>1/2\}=N\), while \(\{\chi_N>1\}=\varnothing\). The equality sets are \(\{\chi_N=1\}=N\) and \(\{\chi_N=0\}=N^c\). Check: \(\chi_N\) is nonzero at every point \(1/n\), so it is not zero everywhere. It is zero outside \(N\), and \(m(N)=0\), so it is zero \(m\)-almost everywhere. No integral or convergence claim is needed for this distinction.
A Lebesgue Measurable Set That Is Not Borel
Measurability depends on the sigma-algebra on the domain. On \(\mathbb{R}\), every Borel set is Lebesgue measurable, but not every Lebesgue measurable set is Borel. An indicator of a set of the latter kind is Lebesgue measurable when the domain sigma-algebra is \(\mathcal{L}\), but it is not measurable when the domain sigma-algebra is only \(\mathcal{B}_{\mathbb{R}}\).
Worked Example: A Non-Borel Subset of the Cantor Set
Let \(C\subseteq[0,1]\) be the middle-thirds Cantor set. There exists a subset \(S\subseteq C\) that is not Borel. Show that \(S\) is nevertheless Lebesgue measurable, and compute the level sets of its indicator.
At stage \(n\) of the Cantor construction, \(C\) is covered by \(2^n\) closed intervals, each of length \(3^{-n}\). Their total length is \(2^n3^{-n}=(2/3)^n\), which tends to \(0\). Therefore \(m(C)=0\). The Lebesgue sigma-algebra is complete: every subset of a Lebesgue null set is Lebesgue measurable. Thus every \(S\subseteq C\) belongs to \(\mathcal{L}\).
To see why a non-Borel choice of \(S\) exists, \(C\) has cardinality equal to that of \(\mathbb{R}\), so its collection of subsets has strictly larger cardinality than \(\mathbb{R}\), by Cantor’s theorem. There are at most as many Borel subsets of \(\mathbb{R}\) as real numbers: Borel sets can be described by countable codes built from a countable base using countable set operations. Hence not every subset of \(C\) can be Borel. Choose one such non-Borel subset and call it \(S\).
For the real-valued function \(\chi_S\), the strict level sets are \(\{\chi_S>a\}=\mathbb{R}\) when \(a<0\), \(S\) when \(0\le a<1\), and \(\varnothing\) when \(a\ge1\). In particular, \(\{\chi_S>1/2\}=S\). The theorem shows that \(\chi_S\) is measurable on \((\mathbb{R},\mathcal{L})\), because \(S\in\mathcal{L}\). It is not measurable on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), because the Borel preimage \(\{\chi_S>1/2\}=S\) is not a Borel set. Check: the same function has different measurability status only because the domain sigma-algebra changed; its values and its level sets did not change.
A Common Threshold Mistake
A tempting but incorrect argument is: “If \(\chi_E\) is measurable, then \(E=\{\chi_E>1\}\), so \(E\) is measurable.” The set identity is wrong. Because the only values of \(\chi_E\) are \(0\) and \(1\), neither value is strictly greater than \(1\), and \(\{\chi_E>1\}=\varnothing\) for every \(E\). The correction is to choose a threshold between the two values: \(\{\chi_E>1/2\}=E\). This is the preimage of a Borel set, so it gives the required implication.
The characterization is useful whenever sets are represented by functions: it allows a set-membership question to be tested through a function’s level set, and it prevents confusion between different notions of measurability. In particular, a null set is measurable in the Lebesgue sigma-algebra, but a null set need not be Borel; these facts do not conflict with the equivalence, which always uses the same domain sigma-algebra for the set and the function.
Check Your Understanding
For each question, keep track of the strict or non-strict inequality and of the sigma-algebra on the domain.
- For \(E=(0,3]\subseteq\mathbb{R}\), compute \(\{\chi_E>0\}\), \(\{\chi_E\ge1\}\), and \(\{\chi_E=0\}\), including the endpoint \(3\).
- Why does the equality \(\{\chi_E>1/2\}=E\) prove one direction of the measurability characterization?
- If \(N\) is a Lebesgue null set, what are \(\{\chi_N>1/2\}\) and \(\{\chi_N=0\}\)? Is \(\chi_N\) necessarily zero everywhere?
- How can \(\chi_S\) be Lebesgue measurable but not Borel measurable when \(S\) is Lebesgue measurable and not Borel?
- For which real numbers \(a\) is \(\{\chi_E\ge a\}=E\)? Include both endpoint cases for \(a\).