One Partition for Several Functions
A simple function has only finitely many values, but two simple functions need not take those values on the same measurable sets. To perform a pointwise operation, such as adding the functions or taking their maximum, we need to know their values at the same time. The useful device is a common finite measurable partition: intersect the level sets of each function, then compute the operation separately on each resulting cell.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space. No assumption that \(\mu\) is finite, sigma-finite, or complete is needed. A function called simple here is real-valued and measurable, with finite range. A finite measurable partition of \(X\) is a finite collection of pairwise disjoint measurable sets whose union is \(X\); empty cells may be discarded. If \(X=\varnothing\), its unique real-valued function is simple and all the closure statements below hold vacuously, so the partition arguments can be read for nonempty \(X\).
For two simple functions \(f\) and \(g\), let \(\{A_i\}_{i=1}^r\) and \(\{B_j\}_{j=1}^s\) be their respective level-set partitions. Their common refinement consists of the sets \(A_i\cap B_j\), with empty intersections removed. On each such cell, both functions have a fixed value. The same construction works for any finite family of simple functions.
These identities follow because each point belongs to exactly one \(A_i\) and exactly one \(B_j\). Every cell is measurable since \(\mathcal{M}\) is closed under finite intersections. The cells therefore give a finite measurable partition on which both functions are constant.
Closure Under the Basic Operations
Proof. For each \(k\in\{1,\ldots,n\}\), write the distinct values of \(f_k\) as \(a_{k,1},\ldots,a_{k,r_k}\), and set \(A_{k,i}=\{x:f_k(x)=a_{k,i}\}\). Each family \(\{A_{k,i}\}_{i=1}^{r_k}\) is a finite measurable partition of \(X\). Form the cells \(C_{i_1,\ldots,i_n}=\bigcap_{k=1}^n A_{k,i_k}\), retaining only nonempty cells. There are at most \(\prod_{k=1}^n r_k\) such cells. They are measurable because each is a finite intersection of measurable sets. They are pairwise disjoint: if two index tuples differ at \(k\), the corresponding cells lie in distinct, disjoint level sets of \(f_k\). They cover \(X\): for any \(x\in X\), each \(f_k(x)\) belongs to exactly one of its listed values, so \(x\) belongs to the cell indexed by those values.
On a cell \(C_{i_1,\ldots,i_n}\), each \(f_k\) is constantly \(a_{k,i_k}\). Hence \(\sum_{k=1}^n f_k\) is constantly \(\sum_{k=1}^n a_{k,i_k}\) there, and \(\prod_{k=1}^n f_k\) is constantly \(\prod_{k=1}^n a_{k,i_k}\) there. The maximum and minimum are constant there as well, with respective values \(\max_{1\le k\le n}a_{k,i_k}\) and \(\min_{1\le k\le n}a_{k,i_k}\). Each operation therefore has at most as many values as there are cells, hence finitely many. Each resulting function is measurable: for any subset \(D\subseteq\mathbb{R}\), its inverse image is the union of the cells whose assigned value lies in \(D\), a finite union of measurable sets. This proves simplicity for the sum, product, maximum, and minimum. Finally, if \(c\ne0\), \(cf_1\) is constant with value \(ca_{1,i}\) on each \(A_{1,i}\), so it is simple; if \(c=0\), it is the constant zero function, which is measurable and has a one-point range. All claims are pointwise, and no property of \(\mu\) beyond the measure-space setting was used. \(\square\)
The proof also records an important bound: if \(f_k\) has \(r_k\) distinct values, then the common refinement has at most \(\prod_{k=1}^n r_k\) nonempty cells. This is an upper bound, not necessarily the number of distinct values of the resulting function. Different cells can produce the same sum, product, maximum, or minimum.
Worked Tables on a Common Refinement
Worked Example: Computing Several Operations on a Finite Space
Let \(X=\{1,2,3,4\}\), let \(\mathcal{M}\) be its power set, and let \(\mu\) be counting measure. Define \(f\) and \(g\) by the following table. Since the level sets of \(f\) and \(g\) are unions of singletons, the singletons form a common measurable partition.
| \(x\) | \(f(x)\) | \(g(x)\) | \((f+g)(x)\) | \((fg)(x)\) | \(\max(f,g)(x)\) | \(\min(f,g)(x)\) |
|---|---|---|---|---|---|---|
| 1 | 2 | -3 | -1 | -6 | 2 | -3 |
| 2 | -1 | 4 | 3 | -4 | 4 | -1 |
| 3 | 2 | 1 | 3 | 2 | 2 | 1 |
| 4 | 0 | 4 | 4 | 0 | 4 | 0 |
For example, at \(x=2\), the input values are \(f(2)=-1\) and \(g(2)=4\). Thus the sum is \(-1+4=3\), the product is \((-1)4=-4\), the maximum is \(4\), and the minimum is \(-1\). Applying the same four operations to the other three rows gives the displayed values. The output ranges are respectively \(\{-1,3,4\}\), \(\{-6,-4,2,0\}\), \(\{2,4\}\), and \(\{-3,-1,0,1\}\), all finite. As a check, at each point the maximum is at least each input and the minimum is at most each input; the four rows satisfy these inequalities.
Worked Example: Intersecting Overlapping Indicator Sets
Let \(X=[0,4]\) with the restricted Lebesgue sigma-algebra and restricted Lebesgue measure. Put \(A=[0,2)\), \(B=[1,3)\), \(f=5\chi_A-\chi_{A^c}\), and \(g=2\chi_B+3\chi_{B^c}\). Both functions are simple: \(f\) takes values \(5,-1\), while \(g\) takes values \(2,3\). The common partition cells are \([0,1)\), \([1,2)\), \([2,3)\), and \([3,4]\); these are precisely the nonempty intersections of \(A,A^c\) with \(B,B^c\).
On \([0,1)\), the point is in \(A\) but not \(B\), so \(f=5\) and \(g=3\). The sum, product, maximum, and minimum are \(8,15,5,3\). On \([1,2)\), the point is in both sets, so \(f=5\) and \(g=2\); the four results are \(7,10,5,2\). On \([2,3)\), the point is in \(B\) but not \(A\), so \(f=-1\) and \(g=2\); the results are \(1,-2,2,-1\). On \([3,4]\), the point is in neither set, so \(f=-1\) and \(g=3\); the results are \(2,-3,3,-1\).
Every operation is constant on each of the four measurable cells, so each result is simple. As a check on endpoints, \(2\in B\) but \(2\notin A\), placing it in \([2,3)\); \(3\notin B\), placing it in \([3,4]\). The cells include every point of \(X\) exactly once, including \(4\).
Worked Example: A Scalar Multiple and Three-Function Extrema
Let \(X=\{a,b,c,d\}\) with the power-set sigma-algebra and counting measure. Define \(f_1,f_2,f_3\) by the table. The four singleton sets are a common finite measurable partition for all three functions.
| Point | \(f_1\) | \(f_2\) | \(f_3\) | \(-2f_1\) | \(\max(f_1,f_2,f_3)\) | \(\min(f_1,f_2,f_3)\) |
|---|---|---|---|---|---|---|
| \(a\) | 1 | 3 | -2 | -2 | 3 | -2 |
| \(b\) | 1 | -1 | 2 | -2 | 2 | -1 |
| \(c\) | -2 | 4 | 1 | 4 | 4 | -2 |
| \(d\) | 0 | 0 | 0 | 0 | 0 | 0 |
For the scalar column, multiplying the \(f_1\) entries by \(-2\) gives \((-2)(1)=-2\), \((-2)(1)=-2\), \((-2)(-2)=4\), and \((-2)(0)=0\). For the maximum and minimum, compare all three entries in each row: at \(a\), the largest is \(3\) and the smallest is \(-2\); at \(b\), they are \(2\) and \(-1\); at \(c\), they are \(4\) and \(-2\); at \(d\), both are \(0\). The outputs take finitely many values. The zero row also checks that the maximum and minimum can coincide when all input values coincide.
A Common Mistake: Ignoring the Overlap
It is incorrect to treat overlapping supports as though they were disjoint. For instance, with the sets \(A\) and \(B\) in the preceding example, one might claim \(\chi_A+\chi_B=\chi_{A\cup B}\). This would give value \(1\) at every point of \(A\cup B\). But at \(x=\tfrac32\), which belongs to both \(A\) and \(B\), the left side is \(\chi_A(\tfrac32)+\chi_B(\tfrac32)=1+1=2\), whereas the right side is \(1\). The proposed identity is false.
The correction is to use the common partition. On \(A\cap B\), the sum of the indicators is \(2\); on \(A\cap B^c\) and \(A^c\cap B\), it is \(1\); and on \(A^c\cap B^c\), it is \(0\). Thus the sum is simple, but its values must be computed on the intersections, not inferred by treating the sets as disjoint. The same discipline applies to products, scalar multiples, maxima, and minima.
This closure result is useful because finite algebraic combinations of simple functions remain within the same class of functions. The proof needs only measurable level sets and finite intersections and unions. It does not invoke integration, and it does not assert that an infinite sum or an infinite supremum of simple functions is simple: those operations can produce infinitely many values and require separate analysis.
Check Your Understanding
Use common level-set refinements to answer these questions.
- If \(f\) takes \(r\) distinct values and \(g\) takes \(s\) distinct values, how many cells can their common refinement have at most?
- For \(f=2\chi_A\) and \(g=3\chi_B\), where \(A\) and \(B\) are measurable and may overlap, list the four possible common partition cells and the value of \(f+g\) on each.
- Why does the proof that \(cf\) is simple need a separate sentence for \(c=0\)?
- Which assumptions on the measure \(\mu\) are needed to prove that the maximum of two simple real-valued functions is simple?
- Can two distinct cells in a common refinement yield the same value of the pointwise product? Explain why this does not affect simplicity.