Finite Range Is Only Half the Definition
A function taking only finitely many real values can be described by listing the sets on which each value occurs. For it to be a simple function, those sets must also be measurable. This distinction matters: the range describes what values the function takes, while measurability controls the inverse images of sets of values.
Throughout, let \((X,\mathcal{M},\mu)\) be an arbitrary measure space; no finiteness, completeness, or sigma-finiteness assumption on \(\mu\) is needed here. The definitions and results below depend on \((X,\mathcal{M})\), not on the measure itself. Functions in this tutorial are real-valued, meaning \(f(x)\in\mathbb{R}\) for every \(x\in X\); they do not take the values \(+\infty\) or \(-\infty\). A real-valued function \(f\) is measurable if \(f^{-1}(B)=\{x\in X:f(x)\in B\}\) belongs to \(\mathcal{M}\) for every Borel set \(B\in\mathcal{B}_{\mathbb{R}}\).
The zero function on a nonempty \(X\) is therefore simple: it is measurable and has range \(\{0\}\). An indicator \(\chi_E\) of a set \(E\subseteq X\) takes values in \(\{0,1\}\), but it is simple only when \(E\) is measurable. Neither the measure of \(E\) nor whether it is a null set is needed for this definition.
Finite Measurable Fibers and Partitions
For a value \(c\in\mathbb{R}\), the fiber of \(f\) at \(c\) is the set \(\{x\in X:f(x)=c\}\). The following characterization makes the two requirements in the definition explicit.
Proof. Suppose first that \(f\) is measurable. For each \(c\in f(X)\), the singleton \(\{c\}\) is a Borel subset of \(\mathbb{R}\). Thus \(\{x:f(x)=c\}=f^{-1}(\{c\})\in\mathcal{M}\).
Conversely, suppose every fiber at an attained value is in \(\mathcal{M}\). Write the distinct values of \(f\) as \(c_1,\ldots,c_r\), where \(r\) is finite, and put \(E_i=\{x:f(x)=c_i\}\). For every Borel set \(B\subseteq\mathbb{R}\),
This is a finite union of measurable sets, including the empty union when no \(c_i\) belongs to \(B\). Hence \(f^{-1}(B)\in\mathcal{M}\) for every Borel \(B\), so \(f\) is measurable. If \(X=\varnothing\), the finite-range function has no fibers to check and each inverse image is empty, so the conclusion also holds. \(\square\)
A finite measurable partition of \(X\) means a finite collection \(E_1,\ldots,E_r\in\mathcal{M}\) of pairwise disjoint sets whose union is \(X\). Partition cells are taken to be nonempty; for \(X=\varnothing\), the empty collection is a finite partition. Each point of \(X\) then belongs to exactly one cell.
Proof. First suppose \(X\ne\varnothing\) and \(f\) is simple. List its distinct attained values as \(c_1,\ldots,c_r\), and set \(E_i=\{x:f(x)=c_i\}\). The theorem shows that every \(E_i\) is measurable. Distinct fibers are disjoint, and every \(x\in X\) has one of the listed values, so the fibers cover \(X\). At each \(x\), exactly one indicator \(\chi_{E_i}(x)\) equals \(1\); all the others equal \(0\). Consequently \(\sum_{i=1}^r c_i\chi_{E_i}(x)=f(x)\).
In the other direction, suppose \(f=\sum_{i=1}^{r}c_i\chi_{E_i}\) for a finite measurable partition. At every \(x\), exactly one cell contains \(x\), so \(f(x)\) equals the corresponding \(c_i\); in particular, its range is finite. For every Borel \(B\), its inverse image is the union of those cells \(E_i\) for which \(c_i\in B\). This finite union is measurable, so \(f\) is measurable and hence simple. When \(X=\varnothing\), the empty function is measurable and has empty finite range, and the empty sum gives it the stated representation. \(\square\)
The constants in a partition representation need not be distinct. If two cells are assigned the same value, their union is the fiber at that value. The representation by distinct attained values uses exactly the nonempty fibers; a partition representation may instead split one fiber into several measurable cells.
Worked Examples
Worked Example: An Indicator on the Real Line
Take \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{L}\), and \(\mu=m\), Lebesgue measure. Let \(E=[0,1)\cup\{2\}\) and \(f=\chi_E\). The set \(E\) is Borel, hence Lebesgue measurable, and \(E^c\) is measurable because \(\mathcal{L}\) is a sigma-algebra. The fibers are \(\{x:f(x)=1\}=E\) and \(\{x:f(x)=0\}=E^c\). Thus \(f\) is simple, with partition representation \(f=1\chi_E+0\chi_{E^c}\).
To check the inverse-image description directly, let \(B=(-\tfrac12,\tfrac12)\). This Borel set contains \(0\) but not \(1\), so \(f^{-1}(B)=E^c\). If \(B=(\tfrac12,\tfrac32)\), it contains \(1\) but not \(0\), so \(f^{-1}(B)=E\). If \(B=(2,3)\), it contains neither attained value, and \(f^{-1}(B)=\varnothing\). These cases follow from membership of the actual values \(0\) and \(1\), not from the size of \(B\). Check: \(f^{-1}(\{1\})=E\) and \(f^{-1}(\{0\})=E^c\), both measurable, which verifies the fiber criterion.
Worked Example: Three Values on a Measurable Partition
Let \(X=[0,3]\), let \(\mathcal{M}\) be the trace of \(\mathcal{L}\) on \(X\), and let \(\mu\) be Lebesgue measure restricted to \(X\). Partition \(X\) into \(E_1=[0,1)\), \(E_2=[1,2]\), and \(E_3=(2,3]\). These sets are measurable, are pairwise disjoint, and their union is \(X\). Define \(f=4\chi_{E_1}-2\chi_{E_2}+1\chi_{E_3}\). Point by point, \(f=4\) on \([0,1)\), \(f=-2\) on \([1,2]\), and \(f=1\) on \((2,3]\). Thus \(f(X)=\{-2,1,4\}\), and its three fibers are exactly the three partition cells.
For the Borel set \(B=(-3,2)\), the values \(-2\) and \(1\) belong to \(B\), while \(4\) does not. Therefore \(f^{-1}(B)=E_2\cup E_3=[1,3]\). For \(C=(0,5)\), the values \(1\) and \(4\) belong to \(C\), while \(-2\) does not, giving \(f^{-1}(C)=E_1\cup E_3=[0,1)\cup(2,3]\). The endpoint \(2\) is included in \(E_2\), not \(E_3\), so it is absent from the latter inverse image. Check: \([1,3]\) contains exactly the points where the assigned value is \(-2\) or \(1\); \([0,1)\cup(2,3]\) contains exactly those where it is \(4\) or \(1\).
Worked Example: A Finite-Range Map That Is Not Measurable
Let \(X=\{a,b\}\) and \(\mathcal{M}=\{\varnothing,X\}\), with \(\mu\) the counting measure restricted to \(\mathcal{M}\). Define \(f(a)=0\) and \(f(b)=1\). Its range is the finite set \(\{0,1\}\), but the fiber at \(1\) is \(\{b\}\), which is not in \(\mathcal{M}\). In particular, \(\{1\}\) is Borel and \(f^{-1}(\{1\})=\{b\}\notin\mathcal{M}\), so \(f\) is not measurable and is not simple.
The distinction can also be checked directly: \(f^{-1}(\{0\})=\{a\}\notin\mathcal{M}\), while \(f^{-1}(\{0,1\})=X\in\mathcal{M}\). Measurability requires the inverse image of every Borel set to be measurable, not merely some of them. Check: the range has two values, but its fibers are not measurable; this is exactly the missing condition in the definition.
Indicators, Uses, and a Common Pitfall
The indicator case gives a useful special criterion on any measure space \((X,\mathcal{M},\mu)\). For \(E\subseteq X\), \(\chi_E\) is real-valued and has range contained in \(\{0,1\}\). If \(E\in\mathcal{M}\), then for any Borel \(B\subseteq\mathbb{R}\), the inverse image \(\chi_E^{-1}(B)\) is one of \(\varnothing\), \(E\), \(E^c\), or \(X\), all measurable. Conversely, if \(\chi_E\) is measurable, then \(E=\chi_E^{-1}(\{1\})\in\mathcal{M}\). Thus \(\chi_E\) is simple if and only if \(E\) is measurable.
A common mistake is to treat “finite range” and “simple” as synonyms without the measurability condition. In the two-point example above, the incorrect argument is: “\(f(X)=\{0,1\}\) is finite, therefore \(f\) is simple.” The conclusion does not follow: \(f^{-1}(\{1\})=\{b\}\notin\mathcal{M}\), so \(f\) is not measurable. The correction is to check the fibers, or equivalently to verify the inverse images of all Borel sets. On a finite measurable partition, the check succeeds because every fiber is a finite union of measurable partition cells.
Finite measurable partitions let us describe a simple function using finitely many values and measurable regions. This is the structural reason such functions are convenient in measure theory: the function’s behavior is reduced to finitely many measurable sets. The definition alone does not assert that a function is integrable, nor does this discussion assign an integral to it; those are separate questions requiring the relevant integral definitions and hypotheses.
Check Your Understanding
For each question, distinguish the range condition from the measurability condition.
- On an arbitrary measure space, prove that every constant real-valued function is simple, including the case \(X=\varnothing\).
- If \(f\) has range \(\{c_1,c_2\}\), with \(c_1\ne c_2\), and both fibers are measurable, show that \(f\) is measurable by computing \(f^{-1}(B)\) for a Borel set \(B\).
- For \(E\subseteq X\), explain why measurability of \(\chi_E\) implies \(E\in\mathcal{M}\).
- On \(X=\{a,b\}\) with \(\mathcal{M}=\{\varnothing,X\}\), which functions \(f:X\to\mathbb{R}\) are measurable? Use their fibers to justify your answer.
- In the three-value example, compute \(f^{-1}((-3,-1))\) and \(f^{-1}((1,4])\), paying attention to which values belong to each open interval.