Point Masses Are Recorded by Jumps
For a finite measure on the real line, the cumulative distribution function records the mass accumulated up to each threshold. Its value at \(x\) includes any mass located exactly at \(x\). The amount by which the function rises between just before \(x\) and \(x\) itself is therefore precisely the mass of the singleton \(\{x\}\). This connection lets us identify atoms directly from a graph or formula for a cumulative function.
Throughout, \((\mathbb{R},\mathcal{B}_{\mathbb{R}},\mu)\) is a finite Borel measure space: \(\mu(\mathbb{R})<\infty\). Its cumulative distribution function is the real-valued function \(F:\mathbb{R}\to[0,\mu(\mathbb{R})]\) defined by \(F(x)=\mu((-\infty,x])\). A point \(x\) is an atom of \(\mu\) if \(\mu(\{x\})>0\). We use the left limit notation \(F(x-)=\lim_{t\uparrow x}F(t)\). The previous material on cumulative functions gives their monotonicity and right continuity; the formula below identifies what the left-to-right jump measures.
Proof. Fix \(x\in\mathbb{R}\). For each positive integer \(n\), put \(E_n=(-\infty,x-1/n]\). The sets \(E_n\) increase with \(n\), and their union is \((-\infty,x)\): every \(y<x\) satisfies \(y\le x-1/n\) for some sufficiently large \(n\), while no \(E_n\) contains a point \(y\ge x\). By continuity from below of a measure, \(\mu(E_n)\to\mu((-\infty,x))\). This continuity follows from countable additivity by writing an increasing union as the disjoint union of its first set and the successive set differences.
Since \(F(x-1/n)=\mu(E_n)\), these values converge to \(\mu((-\infty,x))\). Monotonicity of \(F\) shows that this is the left limit: if \(t<x\), then for all sufficiently large \(n\), \(t\le x-1/n\), so \(F(t)\le F(x-1/n)\le\mu((-\infty,x))\); and as \(t\uparrow x\), the sets \((-\infty,t]\) increase to \((-\infty,x)\), so their measures converge to that same value. Thus \(F(x-)=\mu((-\infty,x))\). The disjoint union \((-\infty,x]=(-\infty,x)\cup\{x\}\) gives \(F(x)=F(x-)+\mu(\{x\})\), proving the jump formula. Since a measure is nonnegative, the jump is positive exactly when the singleton has positive measure. \(\square\)
Proof. If \(\mu(\{x\})>0\), the theorem gives \(F(x)>F(x-)\), so \(F\) is not continuous at \(x\). If \(\mu(\{x\})=0\), then \(F(x)=F(x-)\), so \(F\) is continuous from the left at \(x\). It is also continuous from the right: for \(t\downarrow x\), the sets \((-\infty,t]\) decrease to \((-\infty,x]\), and continuity from above applies because \(\mu(\mathbb{R})<\infty\). Hence \(F(t)\to F(x)\) from the right as well. The two one-sided limits equal \(F(x)\), which proves continuity at \(x\). \(\square\)
Reading Jumps in Discrete and Mixed Measures
Worked Example: A Measure Supported on Three Points
Let \(\mu=2\delta_{-1}+3\delta_2+\delta_5\) on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), where \(\delta_a(E)=1\) if \(a\in E\) and \(0\) otherwise. This is a finite measure with total mass \(2+3+1=6\). Compute its cumulative function and each atomic mass from the jumps.
The threshold set \((-\infty,x]\) contains a support point exactly when that point is at most \(x\). Consequently, $$ F(x)= \begin{cases} 0,&x<-1,\\ 2,&-1\le x<2,\\ 5,&2\le x<5,\\ 6,&x\ge5. \end{cases} $$ At \(x=-1\), \(F(-1)=2\) and \(F((-1)-)=0\), so the jump is \(2-0=2=\mu(\{-1\})\). At \(x=2\), \(F(2)=5\) and \(F(2-)=2\), so the jump is \(5-2=3=\mu(\{2\})\). At \(x=5\), \(F(5)=6\) and \(F(5-)=5\), so the jump is \(6-5=1=\mu(\{5\})\). At every other point the function is locally constant, so there is no jump and no atom. Check: adding the three jump sizes gives \(2+3+1=6\), the total mass of \(\mu\).
Worked Example: Separating Atoms from a Continuous Part
Define a finite Borel measure by $$ \mu(E)=m(E\cap[0,1])+2\delta_0(E)+\tfrac12\delta_1(E), $$ where \(m\) is Lebesgue measure on \(\mathbb{R}\) and \(E\in\mathcal{B}_{\mathbb{R}}\). The three contributions have total masses \(1\), \(2\), and \(\tfrac12\), respectively, so \(\mu(\mathbb{R})=\tfrac72\). Find \(F\) and identify all atoms.
If \(x<0\), none of the three contributions puts mass in \((-\infty,x]\), so \(F(x)=0\). At \(x=0\), the Lebesgue contribution is \(m(\{0\})=0\), while the Dirac contribution at \(0\) is \(2\), giving \(F(0)=2\). For \(0\le x<1\), the accumulated Lebesgue mass is \(m([0,x])=x\), and the mass at \(0\) is \(2\), so \(F(x)=2+x\). At \(x=1\), the Lebesgue mass is \(1\) and both point masses are included; hence \(F(1)=1+2+\tfrac12=\tfrac72\). For \(x>1\), all mass has already been included. Therefore $$ F(x)= \begin{cases} 0,&x<0,\\ 2+x,&0\le x<1,\\ \tfrac72,&x\ge1. \end{cases} $$
At \(0\), the left limit is \(0\), so the jump is \(F(0)-F(0-)=2-0=2\). At \(1\), the left limit is \(2+1=3\), so the jump is \(F(1)-F(1-)=\tfrac72-3=\tfrac12\). On \(0<x<1\), \(F\) is linear and continuous, so there are no other jumps. Thus the atoms are \(0\) and \(1\), with masses \(2\) and \(\tfrac12\). Check: the continuous part has total mass \(1\), and \(1+2+\tfrac12=\tfrac72\), agreeing with the limiting value of \(F\) as \(x\to\infty\).
Atoms of a Push-Forward Measure
The jump formula also identifies atoms when a measure is obtained by mapping another measure onto the real line. If \((X,\mathcal{M},\lambda)\) is a measure space, \(T:X\to\mathbb{R}\) is measurable, and \(\nu\) is its push-forward measure, then \(\nu(E)=\lambda(T^{-1}(E))\) for Borel \(E\). In particular, the mass at \(y\) is \(\nu(\{y\})=\lambda(T^{-1}(\{y\}))\). For finite \(\lambda\), the theorem says this is also the jump of \(\nu\)'s cumulative distribution function at \(y\). A point in the target is an atom precisely when its preimage has positive measure.
Worked Example: A Push-Forward with One Atom
Let \(\lambda\) be Lebesgue measure restricted to \([0,1]\), and define the measurable map $$ T(x)= \begin{cases} 0,&0\le x\le\tfrac13,\\ x,&\tfrac13<x\le1. \end{cases} $$ Let \(\nu(E)=\lambda(T^{-1}(E))\) for Borel \(E\subseteq\mathbb{R}\). Compute the cumulative function of \(\nu\) and its atoms.
The source interval \([0,\tfrac13]\), of length \(\tfrac13\), is mapped to \(0\), so \(\nu(\{0\})=\lambda([0,\tfrac13])=\tfrac13\). If \(x<0\), no point maps into \((-\infty,x]\), and \(F_\nu(x)=0\). If \(0\le x\le\tfrac13\), only the constant part maps into \((-\infty,x]\), giving \(F_\nu(x)=\tfrac13\). If \(\tfrac13<x\le1\), the constant part contributes \(\tfrac13\), and the identity part contributes the length of \((\tfrac13,x]\), namely \(x-\tfrac13\). Thus \(F_\nu(x)=\tfrac13+(x-\tfrac13)=x\). For \(x>1\), the entire source interval is included and \(F_\nu(x)=1\). In full, $$ F_\nu(x)= \begin{cases} 0,&x<0,\\ \tfrac13,&0\le x\le\tfrac13,\\ x,&\tfrac13<x\le1,\\ 1,&x>1. \end{cases} $$
The left limit at \(0\) is \(0\), so the jump there is \(\tfrac13\). At \(x=\tfrac13\), both the value and the left limit are \(\tfrac13\); at \(x=1\), both are \(1\). Between these points the function has no jumps. Equivalently, each \(y\in(\tfrac13,1]\) has a singleton preimage under \(T\), which has Lebesgue measure zero, and points outside \(\{0\}\cup(\tfrac13,1]\) have empty preimages. So \(0\) is the only atom. Check: \(F_\nu(x)\to1\) as \(x\to\infty\), matching the total source mass \(\lambda([0,1])=1\).
A Common Sign and Endpoint Error
The convention \(F(x)=\mu((-\infty,x])\) includes mass at the endpoint \(x\). The jump is therefore \(F(x)-F(x-)\), not the change immediately to the right of \(x\). For instance, if \(\mu=\delta_2\), then \(F(x)=0\) for \(x<2\) and \(F(x)=1\) for \(x\ge2\). Hence \(F(2)-F(2-)=1-0=1\), the mass of \(\{2\}\). But the right-hand limit is \(F(2+)=1\), so \(F(2+)-F(2)=1-1=0\). Using that latter difference would incorrectly report no atom. Right continuity is exactly why the mass is detected by the left-to-value jump.
Finiteness is also important to the stated framework: it ensures \(F\) is finite-valued and that continuity from above gives right continuity without an infinite-measure qualification. The jump identity itself compares the mass accumulated strictly before \(x\) with the mass accumulated through \(x\); it does not say that every increase over an interval is an atom. In the mixed-measure example, the linear increase on \((0,1)\) comes from the Lebesgue part, while the two jumps come only from point masses.
Check Your Understanding
Use the convention \(F(x)=\mu((-\infty,x])\) and the jump formula in each question.
- For \(\mu=4\delta_{-2}+\delta_3\), compute \(F(-2-)\), \(F(-2)\), \(F(3-)\), and \(F(3)\), then give both jump sizes.
- Suppose a finite Borel measure has \(F(a-)=\tfrac25\) and \(F(a)=\tfrac35\). What is \(\mu(\{a\})\), and is \(a\) an atom?
- Explain why an atom at \(a\) forces a discontinuity of \(F\), even though \(F\) is right-continuous.
- For the push-forward measure in the last example, calculate the measure of the preimage of \(\{0\}\) and compare it with the jump at \(0\).
- For \(\delta_2\), compute \(F(2+)-F(2)\). Explain why this is not the atomic mass at \(2\).