From a Cumulative Function to an Explicit Measure
A nondecreasing, right-continuous function can encode a measure on the real line: its increments give the masses of intervals, and its jumps give the masses of single points. For particular piecewise functions, one can often do more than invoke the general construction: an explicit pushforward or atomic formula gives the measure directly. We work through a clipped-linear function, a step function, and an exponential cumulative function.
Throughout, the target measurable space is \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), where \(\mathcal{B}_{\mathbb{R}}\) is the Borel sigma-algebra. All target measures below are finite Borel measures. For a finite Borel measure \(\nu\), its cumulative function is the real-valued function \(F_\nu(x)=\nu((-\infty,x])\). In particular, the left endpoint is excluded and the right endpoint is included in the interval convention \((a,b]\).
For any finite Borel measure \(\nu\) with cumulative function \(F\), finite additivity gives the interval-increment formula. If \(a<b\), then \((-\infty,b]=(-\infty,a]\cup(a,b]\), a disjoint union, so
There is also a direct formula for atoms. Define \(F(x-)=\lim_{y\uparrow x}F(y)\), which exists for a nondecreasing \(F\). The sets \((-\infty,x-1/n]\), for positive integers \(n\), increase to \((-\infty,x)\). Continuity from below of a measure therefore gives \(\nu((-\infty,x))=F(x-)\). Since \((-\infty,x]=(-\infty,x)\cup\{x\}\), disjointly,
Thus an interval mass can be calculated by subtracting cumulative values, while an atom is exactly the jump from the left limit to the value at the point.
A Clipped-Linear Cumulative Function
Worked Example: A Uniform Measure on a Bounded Interval
Define \(T:(0,1)\to\mathbb{R}\) by \(T(u)=2u\). Let \(m\) be Lebesgue measure on \(\mathbb{R}\), and let \(\lambda\) be its restriction to the measurable space \(((0,1),\mathcal{B}_{\mathbb{R}}|_{(0,1)})\). Construct the pushforward measure \(\nu=T_\#\lambda\), defined by \(\nu(B)=\lambda(T^{-1}(B))\) for Borel \(B\subseteq\mathbb{R}\).
The map \(T\) is continuous and hence Borel measurable. Its pushforward is a measure: if \((B_j)\) are pairwise disjoint Borel sets, then the preimages \(T^{-1}(B_j)\) are pairwise disjoint measurable sets, and countable additivity of \(\lambda\) gives \(\nu(\bigcup_jB_j)=\sum_j\nu(B_j)\). The total mass is \(\nu(\mathbb{R})=\lambda((0,1))=1\).
To find the cumulative function, calculate the preimage of \((-\infty,x]\). If \(x\le0\), no \(u\in(0,1)\) satisfies \(2u\le x\), so the preimage has measure \(0\). If \(0<x<2\), the preimage is \((0,x/2]\), of length \(x/2\). If \(x\ge2\), the preimage is all of \((0,1)\), of measure \(1\). Therefore
This is the clipped-linear function \(F(x)=\min(\max(x/2,0),1)\). It is nondecreasing and right-continuous, has limit \(0\) at \(-\infty\), and has limit \(1\) at \(+\infty\). For example, \(F(1)=1/2\) and \(F(-1)=0\), so the interval formula gives \(\nu((-1,1])=1/2-0=1/2\). Directly, \(T^{-1}((-1,1])=(0,1/2]\), which has Lebesgue measure \(1/2\). Also, for each \(x\in\mathbb{R}\), the preimage of \(\{x\}\) is either empty or a singleton, hence has Lebesgue measure zero; consequently \(\nu(\{x\})=0\) for every \(x\).
Check: at \(x=3/2\), the formula gives \(F(3/2)=(3/2)/2=3/4\). Directly, \(T^{-1}((-\infty,3/2])=(0,3/4]\), whose length is \(3/4\). At \(x=2\), the preimage is \((0,1)\), so \(F(2)=1\), agreeing with the clipped formula.
Step Functions and Their Atoms
Worked Example: A Three-Atom Measure
On \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), define the finite Borel measure \(\nu=\tfrac14\delta_{-1}+\tfrac12\delta_1+\tfrac14\delta_3\), where \(\delta_c(B)=1\) if \(c\in B\) and \(0\) otherwise. Its total mass is \(1/4+1/2+1/4=1\).
The cumulative value is the sum of the weights at points not exceeding \(x\). Counting each endpoint according to the condition \(c\le x\), we obtain
The function is nondecreasing and right-continuous. Its limit at \(-\infty\) is \(0\), and its limit at \(+\infty\) is \(1\). The jumps occur at \(-1,1,3\). At \(-1\), the left limit is \(0\), so the atom is \(F(-1)-F(-1-)=1/4-0=1/4\). At \(1\), the left limit is \(1/4\), so the atom is \(F(1)-F(1-)=3/4-1/4=1/2\). At \(3\), the left limit is \(3/4\), so the atom is \(F(3)-F(3-)=1-3/4=1/4\). At every other point the function has no jump and the atom is zero.
For an interval calculation, \((-1,3]\) excludes the atom at \(-1\) and includes those at \(1\) and \(3\). Thus its mass is \(1/2+1/4=3/4\). The cumulative calculation agrees: \(F(3)-F(-1)=1-1/4=3/4\). This also checks why the half-open interval convention matters: changing either endpoint convention can change the answer when an endpoint is an atom.
An Exponential Cumulative Function
Worked Example: The Exponential Cumulative Measure
Define \(T:(0,1)\to\mathbb{R}\) by \(T(u)=-\tfrac12\ln(1-u)\), and let \(\nu=T_\#\lambda\), where \(\lambda\) is Lebesgue measure restricted to \((0,1)\) as above. The map is continuous on \((0,1)\), so this defines a finite Borel pushforward measure of total mass \(\lambda((0,1))=1\).
For \(x<0\), \(T(u)>0\) for every \(u\in(0,1)\), so the preimage of \((-\infty,x]\) is empty. For \(x\ge0\), solve the inequality defining the preimage:
Here \(1-e^{-2x}\in[0,1)\), so the preimage has Lebesgue measure \(1-e^{-2x}\). The cumulative function is therefore
It is nondecreasing and right-continuous, with limits \(0\) and \(1\) at \(-\infty\) and \(+\infty\), respectively. At \(x=0\), the formula gives \(F(0)=1-e^0=0\), so there is no atom at \(0\). For any \(x>0\), the preimage of the singleton \(\{x\}\) consists of the single point \(u=1-e^{-2x}\), which has Lebesgue measure zero. For \(x<0\), the preimage is empty. Hence \(\nu(\{x\})=0\) for every \(x\).
If \(0\le a<b\), the interval mass is \(\nu((a,b])=F(b)-F(a)=(1-e^{-2b})-(1-e^{-2a})=e^{-2a}-e^{-2b}\). For instance, substituting \(a=1/2\) and \(b=2\) gives \(\nu((1/2,2])=e^{-1}-e^{-4}\). Directly, the preimage is \((1-e^{-1},1-e^{-4}]\), whose length is \((1-e^{-4})-(1-e^{-1})=e^{-1}-e^{-4}\), confirming the interval calculation.
Check: for \(x=1\), the cumulative formula gives \(F(1)=1-e^{-2}\). The preimage condition gives \(0<u\le1-e^{-2}\), a set of exactly that Lebesgue measure. As \(x\to+\infty\), \(e^{-2x}\to0\), so \(F(x)\to1\), in agreement with the total pushforward mass.
Reading the Formulas Reliably
The three constructions illustrate a practical rule. First check that a proposed cumulative function has the required endpoint limits, monotonicity, and right continuity. Then use \(\nu((a,b])=F(b)-F(a)\) for intervals and \(\nu(\{x\})=F(x)-F(x-)\) for atoms. If a measurable map gives a convenient pushforward, computing the preimage provides an independent direct check of the cumulative values.
A frequent error is to use left limits at both endpoints when calculating a half-open interval. In the step-function example, this incorrect calculation for \((-1,1]\) would be \(F(1-)-F((-1)-)=1/4-0=1/4\). It omits the atom at \(1\), even though \(1\) belongs to \((-1,1]\). The correct calculation uses the actual cumulative values at the included right endpoint and the excluded left endpoint: \(F(1)-F(-1)=3/4-1/4=1/2\). Direct counting confirms that \((-1,1]\) contains the atom at \(1\) of mass \(1/2\), but not the atom at \(-1\).
Another error is to require only finite endpoint limits. The constant function \(F(x)=1\) has finite limits at both ends, but \(\lim_{x\to-\infty}F(x)=1\), not \(0\). It cannot be the cumulative function of a finite measure: for any finite measure \(\nu\), the sets \((-\infty,-n]\) decrease to the empty set, and continuity from above gives \(\nu((-\infty,-n])\to0\). The left endpoint condition rules out this contradiction.
Check Your Understanding
For each question, keep track of the half-open interval convention and verify the endpoint conditions before using a cumulative function.
- For the clipped-linear cumulative function in the first example, calculate \(\nu((1/2,3/2])\) from its cumulative values and from a preimage under \(T(u)=2u\).
- For the three-atom measure, calculate \(\nu([-1,1])\). Which endpoint atoms are included, and why is the interval-increment formula for \((a,b]\) not directly in that form?
- For the exponential cumulative function, calculate \(\nu((0,1])\) and \(\nu((1,2])\), then add them and compare with \(\nu((0,2])\).
- Explain why a jump of size \(c\) at \(x\) contributes mass \(c\) to \(\{x\}\), rather than to a point immediately to the left of \(x\).
- Does \(G(x)=2\) for every real \(x\) satisfy the endpoint conditions for a finite cumulative measure? Identify the failed condition.