From an Image Measure to Its Cumulative Function
A pushforward measure records how much mass a measurable map sends into each measurable target set. In explicit problems, the calculation often has two stages: first determine the mass assigned to a target set by taking its preimage, then organize those values through a cumulative function. The cumulative function also lets us recover interval masses and point masses, provided we keep track of the endpoints.
For a measurable space \((X,\mathcal M)\), a measure \(\mu\) on \(\mathcal M\), and a measurable map \(T:X\to\mathbb R\), the image measure, or pushforward, is the measure \(\nu=T_\#\mu\) on \((\mathbb R,\mathcal B_{\mathbb R})\) defined by \(\nu(B)=\mu(T^{-1}(B))\) for every Borel set \(B\). In this tutorial, the measures in the worked examples are finite. The cumulative function of a finite Borel measure \(\nu\) on \(\mathbb R\) is \(F(y)=\nu((-\infty,y])\). Its defining interval is closed at \(y\): that convention determines which increments and jumps recover masses.
For an atomic measure, the calculation is a grouping problem. If \(X\) is countable, \(\mu=\sum_j w_j\delta_{x_j}\) with nonnegative weights and finite total mass, then for every Borel set \(B\), \(\nu(B)=\sum_{j:T(x_j)\in B}w_j\). In particular, distinct source points that have the same image contribute to a single target atom. For restricted Lebesgue measure, the preimage definition is just as direct: \(F(y)\) is the Lebesgue measure of the part of the original interval on which \(T(x)\le y\).
Two Recovery Rules
Proof. The disjoint union \((-\infty,b]=(-\infty,a]\cup(a,b]\) gives \(F(b)=F(a)+\nu((a,b])\) by additivity, so subtracting \(F(a)\) proves the interval formula. For the atom formula, the sets \((-\infty,c-1/n]\) increase to \((-\infty,c)\) as \(n\to\infty\). Continuity from below for measures therefore gives \(\lim_{n\to\infty}F(c-1/n)=\nu((-\infty,c))\). Monotonicity of \(F\) shows this is the left limit \(F(c-)\): for any \(y<c\), all sufficiently large \(n\) have \(c-1/n>y\), so \(F(y)\le F(c-1/n)\le F(c-)\), and taking the supremum over \(y<c\) identifies the limit. The disjoint union \((-\infty,c]=(-\infty,c)\cup\{c\}\) now gives \(F(c)=F(c-)+\nu(\{c\})\). \(\square\)
These formulas apply to any finite Borel measure, not only one with finite support. In particular, finite total mass does not imply that \(F\) reaches that total at any finite argument. Continuity from below gives \(F(y)\to\nu(\mathbb R)\) as \(y\to\infty\), while \(F(y)\to0\) as \(y\to-\infty\). If the measure has no lower-bounded support, there need not be any finite \(y\) below all its mass; likewise, without an upper bound on the support, the total mass need not be attained at a finite \(y\).
Worked Problems
Worked Example: Grouping Atoms Under a Map
Let \(X=\{a,b,c,d\}\) with its power-set sigma-algebra, and let \(\mu=3\delta_a+2\delta_b+4\delta_c+\delta_d\). Define \(T(a)=2\), \(T(b)=0\), \(T(c)=2\), and \(T(d)=-1\). Find \(\nu=T_\#\mu\), its cumulative function, and the mass of \((0,2]\).
For any Borel set \(B\subseteq\mathbb R\), include the weight of each source point whose image lies in \(B\). The two source points \(a,c\) both map to \(2\), so their weights combine to \(3+4=7\). The other images each occur once. Thus \(\nu=\delta_{-1}+2\delta_0+7\delta_2\), with total mass \(1+2+7=10\).
To compute \(F(y)=\nu((-\infty,y])\), include an atom exactly when its location is less than or equal to \(y\). Consequently, \(F(y)=0\) for \(y<-1\); \(F(y)=1\) for \(-1\le y<0\); \(F(y)=3\) for \(0\le y<2\); and \(F(y)=10\) for \(y\ge2\). The boundary values belong to the intervals beginning at the atom: for example, \(F(0)=1+2=3\), while \(F(2)=1+2+7=10\).
Since \((0,2]\) excludes the atom at \(0\) and includes the atom at \(2\), its mass is \(F(2)-F(0)=10-3=7\). Directly, the only source points whose images lie in \((0,2]\) are \(a,c\), with total weight \(3+4=7\). The jump at \(2\) is \(F(2)-F(2-)=10-3=7\), which also verifies the target atom there. The total mass check \(F(y)=10\) for \(y\ge2\) agrees with \(\nu(\mathbb R)=10\); here the image has finite support.
Worked Example: Squaring on a Restricted Interval
Let \(X=[-2,1]\) with its Borel sigma-algebra, let \(\mu(A)=m(A\cap[-2,1])\), and set \(T(x)=x^2\). This is a finite measure with \(\mu(X)=3\), and \(T\) is continuous, hence measurable. Compute the cumulative function of \(\nu=T_\#\mu\).
By the preimage formula, \(F(y)=m(\{x\in[-2,1]:x^2\le y\})\). If \(y<0\), the set is empty and \(F(y)=0\). At \(y=0\), the preimage is the singleton \(\{0\}\), whose Lebesgue measure is zero. For \(0<y<1\), the preimage is \([-\sqrt y,\sqrt y]\), entirely inside \([-2,1]\), so its length is \(2\sqrt y\). At \(y=1\), it is \([-1,1]\), of length \(2\).
For \(1<y<4\), the inequality \(x^2\le y\) gives \(-\sqrt y\le x\le\sqrt y\); intersecting with \([-2,1]\) gives \([-\sqrt y,1]\), of length \(1+\sqrt y\). For \(y\ge4\), every \(x\in[-2,1]\) satisfies \(x^2\le y\), so the length is \(3\). Combining the cases, including their endpoints, gives \(F(y)=0\) for \(y<0\); \(F(y)=2\sqrt y\) for \(0\le y\le1\); \(F(y)=1+\sqrt y\) for \(1<y<4\); and \(F(y)=3\) for \(y\ge4\). At \(y=1\), both neighboring formulas yield \(2\); at \(y=4\), the left limit and the value are both \(3\).
As a check using an interval increment, \(\nu((1,4])=F(4)-F(1)=3-2=1\). Directly, \(x^2\in(1,4]\) for \(x\in[-2,-1)\) within the domain; this interval has length \(1\). The point \(x=-1\) is excluded because the target interval excludes \(1\), and the endpoint \(x=-2\) is included but has no effect on length. There are no atoms: for any \(y\ge0\), the preimage of \(\{y\}\) contains at most two points, hence has Lebesgue measure zero, and for \(y<0\) it is empty. In particular the continuity of the displayed \(F\) agrees with the absence of jumps.
Worked Example: Reading Atoms and Interval Masses from a Cumulative Function
Consider the finite Borel measure \(\nu=\tfrac12\,m|_{[0,2]}+2\delta_1+\delta_3\) on \(\mathbb R\), where \(m|_{[0,2]}(B)=m(B\cap[0,2])\). The restricted Lebesgue component has total mass \(\tfrac12\cdot2=1\), so the total mass is \(1+2+1=4\). Find \(F\), then recover \(\nu((\tfrac12,2])\) and the atoms.
For \(y<0\), no component contributes, so \(F(y)=0\). For \(0\le y<1\), the restricted interval contributes length \(y\) with weight \(\tfrac12\), and neither atom is included; hence \(F(y)=y/2\). For \(1\le y<2\), the continuous contribution is \(y/2\) and the atom at \(1\) contributes \(2\), so \(F(y)=y/2+2\). At \(y=2\), the continuous component has contributed \(\tfrac12\cdot2=1\), and the atom at \(1\) contributes \(2\), giving \(F(2)=3\). For \(2\le y<3\), that value remains \(3\); for \(y\ge3\), the atom at \(3\) adds \(1\), giving \(F(y)=4\). Thus the endpoint-sensitive description is \(F(y)=0\) for \(y<0\), \(F(y)=y/2\) for \(0\le y<1\), \(F(y)=y/2+2\) for \(1\le y<2\), \(F(y)=3\) for \(2\le y<3\), and \(F(y)=4\) for \(y\ge3\). The formulas agree at \(2\), where there is no jump.
The interval increment gives \(\nu((\tfrac12,2])=F(2)-F(\tfrac12)=3-\tfrac14=\tfrac{11}{4}\). Direct calculation gives continuous mass \(\tfrac12\,m((\tfrac12,2]\cap[0,2])=\tfrac12\cdot\tfrac32=\tfrac34\), together with the atom of mass \(2\) at \(1\); their sum is \(\tfrac34+2=\tfrac{11}{4}\). The jump at \(1\) is \(F(1)-F(1-)=\tfrac52-\tfrac12=2\), and the jump at \(3\) is \(F(3)-F(3-)=4-3=1\). At \(2\), \(F(2)-F(2-)=3-3=0\). These are exactly the point masses in the defining measure.
Endpoint Pitfalls and Support
A common error is to use \(F(b)-F(a)\) for the closed interval \([a,b]\). For the measure in the third example, this difference with \(a=1\), \(b=2\) is \(F(2)-F(1)=3-\tfrac52=\tfrac12\). That is \(\nu((1,2])\), not \(\nu([1,2])\): the latter also includes the atom of mass \(2\) at \(1\), so \(\nu([1,2])=\tfrac12+2=\tfrac52\). The correction is to use \(F(b)-F(a-)\) for \([a,b]\), since \(F(a-)=\nu((-\infty,a))\).
Another error is to infer bounded support from finiteness. For example, on \(X=[0,\infty)\) with its Borel sigma-algebra, let \(\mu(A)=\int_A e^{-x}\,dx\) and \(T(x)=x\). This is a finite measure of total mass \(1\), but its image cumulative function is \(F(y)=0\) for \(y<0\) and \(F(y)=1-e^{-y}\) for \(y\ge0\). For each finite \(y\ge0\), \(e^{-y}>0\), so \(F(y)<1\); the total is approached as \(y\to\infty\), not attained at a finite point. By contrast, if a finite measure is explicitly known to have support bounded above by \(M\), then \(F(y)=\nu(\mathbb R)\) for \(y\ge M\). The bounded-support hypothesis, not finiteness alone, justifies that conclusion.
In explicit pushforward problems, a reliable order of work is to calculate preimages, include exactly the atoms satisfying the chosen inequality, form the cumulative function, and only then take increments or jumps. Finally, compare at least one recovered mass with a direct preimage calculation. This check catches both grouping errors in atomic images and endpoint errors in interval calculations.
Check Your Understanding
For each question, state which endpoint convention you are using and verify the result from the preimage or the defining measure.
- Let \(\mu=2\delta_a+5\delta_b\) and suppose \(T(a)=1\), \(T(b)=1\). Find \(T_\#\mu\), its cumulative function, and its mass at \(1\).
- For the image measure under \(T(x)=x^2\) from \([-2,1]\), compute \(\nu((0,1])\) using the cumulative function and directly from its preimage.
- For the measure \(\tfrac12m|_{[0,2]}+2\delta_1+\delta_3\), find \(\nu([1,2])\) and explain why \(F(2)-F(1)\) does not give that mass.
- Prove that if \(F(c)=F(c-)\) for a finite Borel measure on \(\mathbb R\), then \(\nu(\{c\})=0\).
- Give a reason that a finite total mass alone does not imply that the cumulative function is constant at the total mass for all sufficiently large finite arguments.