Cumulative Functions Under Monotone Transformations

From a Transformation to a Cumulative Function

A cumulative function records the mass lying at or to the left of each threshold. After a measurable map changes the locations of that mass, the new cumulative function is determined by asking which original points land at or below the new threshold. Monotonicity makes that preimage an interval, and the inverse transformation identifies its endpoint.

Throughout, \(\mu\) is a finite Borel measure on \(\mathbb{R}\), including the possibility that \(\mu(\mathbb{R})=0\). Write \(M=\mu(\mathbb{R})\) and define its cumulative function by \(F_\mu(x)=\mu((-\infty,x])\). Let \(T:\mathbb{R}\to\mathbb{R}\) be a strictly monotone bijection. Such a map is Borel measurable, so its push-forward \(\nu=T_\#\mu\), defined by \(\nu(B)=\mu(T^{-1}(B))\) for Borel sets \(B\), is a finite Borel measure. Its cumulative function is \(F_\nu(y)=\nu((-\infty,y])\). These are equalities at every real threshold, not merely almost everywhere.

Key identity: For every \(y\in\mathbb{R}\), \(F_{T_\#\mu}(y)=\mu(\{x\in\mathbb{R}:T(x)\le y\})\). The direction of monotonicity determines whether this preimage is a left ray or a right ray.

The key identity is just the definition of push-forward applied to the Borel set \((-\infty,y]\). To turn it into a formula involving \(F_\mu\), one must preserve the endpoint convention: the cumulative function uses \((-\infty,x]\), while a right ray of the form \([x,\infty)\) is the complement of \((-\infty,x)\), not of \((-\infty,x]\).

The Two Inverse-Threshold Formulas

Theorem: Let \(\mu\) be a finite Borel measure on \(\mathbb{R}\), let \(T:\mathbb{R}\to\mathbb{R}\) be a strictly monotone bijection, and set \(\nu=T_\#\mu\). If \(T\) is strictly increasing, then \(F_\nu(y)=F_\mu(T^{-1}(y))\) for every \(y\in\mathbb{R}\). If \(T\) is strictly decreasing, then \(F_\nu(y)=M-F_\mu((T^{-1}(y))-)\) for every \(y\in\mathbb{R}\), where \(F_\mu(x-)=\mu((-\infty,x))\).

Proof. First suppose that \(T\) is strictly increasing. For each \(y\), strict increase and bijectivity imply \(T(x)\le y\) exactly when \(x\le T^{-1}(y)\). Therefore \(T^{-1}((-\infty,y])=(-\infty,T^{-1}(y)]\). By the definition of push-forward and then of \(F_\mu\), \(F_\nu(y)=\mu((-\infty,T^{-1}(y)])=F_\mu(T^{-1}(y))\).

Now suppose that \(T\) is strictly decreasing. Then \(T(x)\le y\) exactly when \(x\ge T^{-1}(y)\), so \(T^{-1}((-\infty,y])=[T^{-1}(y),\infty)\). This set is the complement of \((-\infty,T^{-1}(y))\). Since \(\mu\) is finite, the measure of that complement is \(M-\mu((-\infty,T^{-1}(y)))\), giving \(F_\nu(y)=M-F_\mu((T^{-1}(y))-)\).

For completeness, the left limit in this formula agrees with the measure of the open left ray: if \(x\in\mathbb{R}\), the sets \((-\infty,x-1/n]\) increase to \((-\infty,x)\). Continuity from below for measures therefore gives \(\lim_{n\to\infty}F_\mu(x-1/n)=\mu((-\infty,x))\). This defines \(F_\mu(x-)\) and proves the stated formula at every threshold, including thresholds that correspond to atoms. \(\square\)

For increasing transformations, the mass below a transformed threshold is the original mass below its inverse threshold. For decreasing transformations, the mass below the transformed threshold is instead the original mass at or above the inverse threshold. The left limit accounts for the fact that the point \(T^{-1}(y)\) is included in this right ray.

Worked Example: A Positive Affine Transformation

Let \(\mu\) be Lebesgue measure restricted to \([0,1]\), so \(\mu(B)=m(B\cap[0,1])\), with total mass \(M=1\). Its cumulative function is \(F_\mu(x)=0\) for \(x<0\), \(F_\mu(x)=x\) for \(0\le x\le1\), and \(F_\mu(x)=1\) for \(x>1\). Find the cumulative function of \(T(x)=3x+2\).

The map is strictly increasing and has inverse \(T^{-1}(y)=(y-2)/3\). The theorem gives \(F_{T_\#\mu}(y)=F_\mu((y-2)/3)\). If \(y<2\), then \((y-2)/3<0\), so the value is \(0\). If \(2\le y\le5\), then \(0\le(y-2)/3\le1\), so the value is \((y-2)/3\). If \(y>5\), then \((y-2)/3>1\), so the value is \(1\). Thus

F_{T_\#\mu}(y)= \begin{cases} 0,&y<2,\\ \dfrac{y-2}{3},&2\le y\le5,\\ 1,&y>5. \end{cases}

Check: at \(y=4\), the formula gives \((4-2)/3=2/3\). Directly, \(3x+2\le4\) is equivalent to \(x\le2/3\), and \(\mu([0,2/3])=2/3\). At \(y=5\), the formula gives \((5-2)/3=1\), matching the full source mass on \([0,1]\).

Worked Example: Reflection of a Uniform Measure

Again let \(\mu(B)=m(B\cap[0,1])\), and now take the strictly decreasing bijection \(T(x)=1-x\). Its inverse is \(T^{-1}(y)=1-y\), and \(M=1\). For \(0\le y\le1\), the theorem gives \(F_{T_\#\mu}(y)=1-F_\mu((1-y)-)\). The source cumulative function is continuous at every point of \([0,1]\), so this is \(1-(1-y)=y\). If \(y<0\), the inverse threshold \(1-y\) is greater than \(1\), giving \(F_\mu((1-y)-)=1\) and transformed cumulative value \(0\). If \(y>1\), the inverse threshold is below \(0\), giving left limit \(0\) and transformed cumulative value \(1\). Consequently,

F_{T_\#\mu}(y)= \begin{cases} 0,&y<0,\\ y,&0\le y\le1,\\ 1,&y>1. \end{cases}

Check: at \(y=1/4\), the formula gives \(1/4\). The preimage condition \(1-x\le1/4\) is equivalent to \(x\ge3/4\); the interval \([3/4,1]\) has Lebesgue measure \(1-3/4=1/4\). This also verifies the endpoint inclusion in the preimage.

Atoms, Left Limits, and a Nonlinear Example

Worked Example: A Decreasing Map with Atoms

Let \(\mu=2\delta_{-1}+3\delta_2\), a finite Borel measure of total mass \(M=2+3=5\), and let \(T(x)=-x\). The map is a strictly decreasing bijection with inverse \(T^{-1}(y)=-y\). Under \(T\), the mass \(2\) at \(-1\) moves to \(1\), and the mass \(3\) at \(2\) moves to \(-2\). Thus the transformed measure has mass \(3\) at \(-2\) and mass \(2\) at \(1\), giving

F_{T_\#\mu}(y)= \begin{cases} 0,&y<-2,\\ 3,&-2\le y<1,\\ 5,&y\ge1. \end{cases}

The theorem gives the same result while showing why the left limit is necessary. At \(y=-2\), the inverse threshold is \(2\). Since \(F_\mu(2-)=2\), the formula gives \(5-2=3\), including the transformed atom at \(-2\). At \(y=1\), the inverse threshold is \(-1\), and \(F_\mu((-1)-)=0\), so the formula gives \(5\), including both atoms. Check: directly, \(-x\le-2\) means \(x\ge2\), which captures mass \(3\); \(-x\le1\) means \(x\ge-1\), which captures masses \(2+3=5\).

The left limit cannot in general be replaced by the cumulative value at the inverse threshold. In this example, at \(y=1\), using \(M-F_\mu(-1)\) would give \(5-2=3\), incorrectly excluding the transformed atom at \(1\). The correct expression is \(M-F_\mu((-1)-)=5-0=5\). At a point with no atom, \(F_\mu(x-)=F_\mu(x)\), so the distinction disappears; at an atom, it is essential.

Worked Example: A Cubic Transformation

Let \(\mu(B)=m(B\cap[-1,1])\), so \(M=2\) and \(F_\mu(x)=0\) for \(x<-1\), \(F_\mu(x)=x+1\) for \(-1\le x\le1\), and \(F_\mu(x)=2\) for \(x>1\). Take \(T(x)=x^3\), a strictly increasing bijection with inverse \(T^{-1}(y)=\sqrt[3]{y}\). Applying the increasing formula gives

F_{T_\#\mu}(y)= \begin{cases} 0,&y<-1,\\ \sqrt[3]{y}+1,&-1\le y\le1,\\ 2,&y>1. \end{cases}

Check: at \(y=1/8\), the inverse threshold is \(\sqrt[3]{1/8}=1/2\), so the formula gives \(1/2+1=3/2\). Directly, \(x^3\le1/8\) is equivalent to \(x\le1/2\); the portion of \([-1,1]\) satisfying this has length \(1/2-(-1)=3/2\). At \(y=1\), the formula gives \(2\), the full mass of the source interval.

How to Use the Formulas Reliably

For a finite measure, these formulas separate the geometric task from the measure calculation. First solve the inequality \(T(x)\le y\). For increasing \(T\), this yields \(x\le T^{-1}(y)\); for decreasing \(T\), it yields \(x\ge T^{-1}(y)\). Then evaluate the measure of that preimage, taking care with whether the endpoint is included. In the decreasing case, the original cumulative function uses a closed left ray, so the complement calculation requires its left limit.

Finite total mass is used in the decreasing formula to write the mass of a complement as \(M\) minus the mass of the complementary set. The formula as stated therefore applies to finite measures; it should not be used as an expression of the form \(\infty-\infty\) for an infinite measure. Strict monotonicity and bijectivity ensure that every threshold has a unique inverse threshold and that the preimage has the indicated ray form. When a transformation is only monotone on a restricted domain, the preimage must be intersected with that domain, and thresholds outside its image must be handled separately.

The formulas also explain the effect of reversing order. An increasing transformation preserves the order of locations and reads the original cumulative function at the inverse threshold. A decreasing transformation reverses the order: mass that ends up below \(y\) came from at or to the right of the inverse threshold. Atoms remain atoms under a bijection, but their jumps occur at transformed locations; the left-limit term ensures that the jump is counted on the correct side of the threshold.

Check Your Understanding

For each question, identify the preimage of the closed left ray before applying a cumulative-function formula.

If \(T(x)=2x-3\) and the source measure is uniform on \([0,1]\), what is \(F_{T_\#\mu}(y)\) on each range of \(y\)?
For a finite measure with cumulative function \(F_\mu\), explain why a strictly decreasing bijection gives \(M-F_\mu((T^{-1}(y))-)\), rather than \(M-F_\mu(T^{-1}(y))\).
Let \(\mu=\delta_0+4\delta_3\) and \(T(x)=-x\). Find the transformed atoms and write the cumulative function at all thresholds, specifying which endpoints are included.
For \(\mu(B)=m(B\cap[-2,2])\) and \(T(x)=x^3\), compute \(F_{T_\#\mu}(8)\) by the inverse-threshold formula and by measuring the preimage directly.
Which part of the decreasing-map formula requires \(\mu(\mathbb{R})<\infty\), and why can the same complement subtraction be invalid for an infinite measure?