Order Turns Preimages into Intervals
A monotone function can have jump discontinuities and still be Borel-measurable. The reason is order: if a nondecreasing function exceeds a fixed threshold at one point, it exceeds that threshold at every larger point in its domain. Thus each strict threshold preimage has a simple interval form. We will use that structure to prove measurability of preimages of all open sets, not just threshold sets.
Let \(I\subseteq\mathbb{R}\) be an interval, possibly empty, and let \(\mathcal{B}_I=\{I\cap B:B\in\mathcal{B}_{\mathbb{R}}\}\) be its relative Borel sigma-algebra. A function \(f:I\to\mathbb{R}\) is Borel-measurable if \(f^{-1}(B)\in\mathcal{B}_I\) for every \(B\in\mathcal{B}_{\mathbb{R}}\). It is enough to show that preimages of open sets are in \(\mathcal{B}_I\), because the collection of sets \(B\subseteq\mathbb{R}\) whose preimages belong to \(\mathcal{B}_I\) is a sigma-algebra, and open sets generate \(\mathcal{B}_{\mathbb{R}}\).
We call \(f\) nondecreasing if \(x\le y\) in \(I\) implies \(f(x)\le f(y)\), and nonincreasing if \(x\le y\) implies \(f(x)\ge f(y)\). Here “monotone” allows either direction; it does not mean strictly monotone.
Threshold Preimages
Proof. The empty interval \(I\) and the cases \(A=\varnothing\) or \(A=I\) have the stated form and are Borel in \(I\). Now suppose \(I\) is nonempty, \(A\ne\varnothing\), and \(A\ne I\). If \(A\) is unbounded below, then for each \(y\in I\) there is \(x\in A\) with \(x\le y\); upward closure gives \(y\in A\), contradicting \(A\ne I\). Thus \(A\) is bounded below. Put \(c=\inf A\), which is a finite real number. If \(y\in I\) and \(y<c\), then \(y\notin A\), because \(c\) is a lower bound for \(A\). If \(y\in I\) and \(y>c\), the definition of infimum gives an \(x\in A\) with \(x<y\); upward closure then gives \(y\in A\). Only membership of \(c\) remains undecided, and it matters only if \(c\in I\). Consequently \(A=I\cap[c,\infty)\) when \(c\in A\), and \(A=I\cap(c,\infty)\) otherwise. Both sets are in \(\mathcal{B}_I\). \(\square\)
A set is downward closed if \(x\in A\), \(y\in I\), and \(y\le x\) imply \(y\in A\). Reversing the order in the lemma gives the corresponding description: such a set is empty, all of \(I\), or \(I\cap(-\infty,c]\) or \(I\cap(-\infty,c)\). These sets are also relative Borel sets.
Worked Example: A Step Function with Two Jumps
Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=-1\) for \(x<0\), \(f(x)=2\) for \(0\le x<3\), and \(f(x)=5\) for \(x\ge3\). The values increase from \(-1\) to \(2\) to \(5\), so \(f\) is nondecreasing.
For the threshold \(t=2\), the strict preimage is \(\{x:f(x)>2\}=[3,\infty)\): the value on \([0,3)\) equals \(2\), so those points are not included. For \(t=1\), \(\{x:f(x)>1\}=[0,\infty)\). For \(t=5\), \(\{x:f(x)>5\}=\varnothing\). Each is Borel, including the empty-set case.
For the open interval \((1,4)\), only the value \(2\) lies in the interval, so \(f^{-1}((1,4))=[0,3)\), a Borel set. At \(0\), the left-hand limit is \(-1\), while \(f(0)=2\) and the right-hand limit is \(2\); the jump size is \(2-(-1)=3\). At \(3\), the left-hand limit is \(2\), while \(f(3)=5\) and the right-hand limit is \(5\); the jump size is \(5-2=3\). Away from \(0\) and \(3\), the function is locally constant and hence continuous there. The jumps do not prevent Borel measurability.
Every Monotone Real Function Is Borel-Measurable
Proof. First suppose \(f\) is nondecreasing. For every \(t\in\mathbb{R}\), the set \(\{x\in I:f(x)>t\}\) is upward closed: if \(f(x)>t\) and \(y\in I\) satisfies \(y\ge x\), then \(f(y)\ge f(x)>t\). By the lemma, this set is relative Borel. Likewise, \(\{x\in I:f(x)<t\}\) is downward closed, so it is relative Borel.
Every open set \(U\subseteq\mathbb{R}\) is a union of the rational open intervals contained in it: \(U=\bigcup\{(p,q):p,q\in\mathbb{Q},\ p<q,\ (p,q)\subseteq U\}\). This union is countable because \(\mathbb{Q}\times\mathbb{Q}\) is countable. For each such interval, the preimage is exactly
Both sets on the right are relative Borel, so their intersection is relative Borel. Taking the countable union over rational intervals shows that \(f^{-1}(U)\in\mathcal{B}_I\). Since the preimages of open sets are relative Borel, the sigma-algebra argument above gives the preimage property for every \(B\in\mathcal{B}_{\mathbb{R}}\).
If \(f\) is nonincreasing, then \(g=-f\) is real-valued and nondecreasing. The result just proved makes \(g\) Borel-measurable. For any open \(U\subseteq\mathbb{R}\), reflection \(-U=\{-u:u\in U\}\) is open, and \(f^{-1}(U)=g^{-1}(-U)\). Therefore \(f^{-1}(U)\in\mathcal{B}_I\), and the same sigma-algebra argument proves Borel measurability. This also covers constant functions, which are both nondecreasing and nonincreasing. \(\square\)
Two More Preimage Calculations
Worked Example: The Floor Function
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=\lfloor x\rfloor\), the greatest integer less than or equal to \(x\). If \(x\le y\), every integer less than or equal to \(x\) is also less than or equal to \(y\), so \(\lfloor x\rfloor\le\lfloor y\rfloor\). Thus \(f\) is nondecreasing.
For the strict threshold \(t=2.4\), the integer \(\lfloor x\rfloor\) exceeds \(2.4\) exactly when it is at least \(3\). That is equivalent to \(x\ge3\), so \(\{x:\lfloor x\rfloor>2.4\}=[3,\infty)\). At the threshold \(t=3\), strictness changes the endpoint: \(\lfloor x\rfloor>3\) means \(\lfloor x\rfloor\ge4\), which is equivalent to \(x\ge4\). Hence \(\{x:\lfloor x\rfloor>3\}=[4,\infty)\), not \([3,\infty)\).
For \(U=(1.2,3.7)\), the integer values in \(U\) are exactly \(2\) and \(3\). The preimage of the value \(2\) is \([2,3)\), and the preimage of \(3\) is \([3,4)\). Therefore \(f^{-1}(U)=[2,3)\cup[3,4)=[2,4)\). In particular, \(4\) is excluded because \(\lfloor4\rfloor=4\notin U\), while \(2\) is included because \(\lfloor2\rfloor=2\in U\).
At an integer \(k\), the left-hand limit of \(\lfloor x\rfloor\) is \(k-1\), and the value at \(k\) and right-hand limit are both \(k\), giving a jump of \(k-(k-1)=1\). At every noninteger \(x\), there is a positive distance from \(x\) to the nearest integer, and the floor function is constant in a sufficiently small neighborhood of \(x\). These jump discontinuities are consistent with the Borel preimages computed above.
Worked Example: A Jump on a Closed Interval
On \(I=[0,1]\), define \(f(x)=x\) for \(0\le x<\tfrac12\), and \(f(x)=x+1\) for \(\tfrac12\le x\le1\). Within each piece the function increases. Across the break, values on the first piece are less than \(\tfrac12\), while values on the second are at least \(\tfrac32\); hence \(f\) is nondecreasing on all of \(I\).
For the open interval \(U=(1,2)\), the first piece contributes no points because its values are below \(\tfrac12\). On the second piece, \(1<x+1<2\) is equivalent to \(0<x<1\). Intersecting with the second piece \([\tfrac12,1]\) gives \(f^{-1}(U)=[\tfrac12,1)\). The left endpoint is included because \(f(\tfrac12)=\tfrac32\in(1,2)\); the right endpoint is excluded because \(f(1)=2\notin(1,2)\).
The strict threshold \(t=1\) gives \(\{x\in I:f(x)>1\}=[\tfrac12,1]\). At \(t=\tfrac32\), the value at \(\tfrac12\) equals the threshold and must be excluded, so \(\{x\in I:f(x)>\tfrac32\}=(\tfrac12,1]\). At the jump, the left-hand limit is \(\tfrac12\), and the function value and right-hand limit are \(\tfrac32\); the jump size is \(\tfrac32-\tfrac12=1\). Each calculated preimage is Borel relative to \(I\).
A Common Endpoint Mistake
A threshold preimage of a monotone function is not necessarily an open ray. For \(f(x)=\chi_{[0,\infty)}(x)\), which is nondecreasing, \(\{x:f(x)>0\}=[0,\infty)\). A mistaken calculation might replace this with \((0,\infty)\) on the grounds that the threshold set should be open; that loses the point \(0\), where \(f(0)=1>0\). The correct conclusion is that the preimage is a ray with either endpoint convention, as the upper-set lemma states. It need not be open to be Borel.
The argument depends on the order structure of the domain and the real-valued range: threshold preimages are upper or lower sets, and rational intervals reduce arbitrary open preimages to countable unions of their intersections. Jump discontinuities change whether a boundary point belongs to a preimage, but they do not disrupt this interval structure.
Check Your Understanding
For each question, keep track of whether the threshold inequality is strict and whether the domain is all of \(\mathbb{R}\) or a relative interval.
- For a nondecreasing \(f:I\to\mathbb{R}\), explain why \(\{x\in I:f(x)>t\}\) is upward closed for every real \(t\).
- Let \(f(x)=\lfloor x\rfloor\). Compute \(\{x:f(x)>-1\}\), including its endpoint.
- For the step function equal to \(0\) on \((-\infty,1)\) and \(4\) on \([1,\infty)\), compute \(f^{-1}((0,4))\).
- Why does proving that preimages of open sets are Borel imply the preimage property for every Borel set?
- Give an example of a monotone function whose strict threshold preimage is not open, and state why this does not contradict Borel measurability.