Reading a Measure Formula as a Density
A formula of the form \(\nu(A)=\int_A g\,d\mu\) already displays a candidate Radon–Nikodym derivative: it is \(g\), relative to the measure \(\mu\). The essential care is to keep the reference measure visible. A function that describes masses at individual points is a density with respect to counting measure; a function inside a Lebesgue integral is a density with respect to Lebesgue measure. The same numerical function need not be a density relative to both.
Here \((X,\mathcal{M},\mu)\) is a measure space, with \(\mu\) not necessarily finite. The notation \(L^1(\mu)\) means the measurable functions \(g\) for which \(\int_X |g|\,d\mu<\infty\). Functions and integrals may be real-valued or complex-valued as specified. Equalities of set functions below hold for every \(A\in\mathcal{M}\); a Radon–Nikodym derivative itself is determined only \(\mu\)-almost everywhere.
For a finite signed measure, the integral formula requires \(g\in L^1(\mu)\). The same condition gives a finite complex measure when \(g\) is complex-valued. The next result explains why these formulas genuinely define measures, as well as how to read off their derivatives. We use the standard uniqueness of Radon–Nikodym derivatives: two integrable densities giving the same measure agree \(\mu\)-almost everywhere.
Proof. We first check finiteness. For every \(A\in\mathcal{M}\), \(|\nu(A)|\le\int_A|g|\,d\mu\le\int_X|g|\,d\mu<\infty\). To check countable additivity, let \(A_1,A_2,\ldots\) be pairwise disjoint measurable sets, put \(A=\bigcup_{k=1}^{\infty}A_k\), and set \(B_n=\bigcup_{k=1}^{n}A_k\). The functions \(g\chi_{B_n}\) converge pointwise to \(g\chi_A\), and \(|g\chi_{B_n}|\le |g|\), with \(|g|\in L^1(\mu)\). The Dominated Convergence Theorem therefore gives \(\nu(B_n)\to\nu(A)\). Finite additivity of the integral gives \(\nu(B_n)=\sum_{k=1}^{n}\nu(A_k)\); taking the limit proves countable additivity. If \(\mu(A)=0\), then \(\int_Ag\,d\mu=0\), so \(\nu\ll\mu\). Finally, the formula defining \(\nu\) says that \(g\) is a Radon–Nikodym derivative, and almost-everywhere uniqueness identifies it as the derivative up to a \(\mu\)-null set. \(\square\)
The Dominated Convergence Theorem applies here because the sequence converges pointwise and the single dominating function \(|g|\) is integrable. For complex \(g\), it may equivalently be applied to the real and imaginary parts. If \(\mu\) is counting measure on a countable set, then \(\int_Ag\,d\mu=\sum_{x\in A}g(x)\), provided the sum is interpreted as the integral; for \(g\in L^1(\mu)\), the sum of absolute values is finite.
Worked Computations
Worked Example: A Signed Density on a Finite Interval
Let \(X=[0,1]\), let \(\mathcal{M}\) be the Lebesgue sigma-algebra on \(X\), and let \(\mu\) be Lebesgue measure restricted to \(X\). This is a finite measure space, with \(\mu(X)=1\). Define the signed set function \(\nu(A)=\int_A(2x-1)\,d\mu\) for \(A\in\mathcal{M}\).
The function \(g(x)=2x-1\) is measurable and integrable, since \(|2x-1|\le1\) on \(X\). The proposition shows that \(\nu\) is a finite signed measure, is absolutely continuous with respect to \(\mu\), and has derivative \(d\nu/d\mu=2x-1\) \(\mu\)-almost everywhere. For \(A=[0,\tfrac14]\), the density is nonpositive throughout \(A\), and direct integration gives
As a second calculation, \(\nu(X)=\int_0^1(2x-1)\,dx=[x^2-x]_0^1=(1-1)-0=0\). Check: the antiderivative \(x^2-x\) has derivative \(2x-1\), and its endpoint difference on \([0,\tfrac14]\) is \(-3/16\), as computed. The zero total mass is consistent with positive and negative contributions cancelling; it does not mean that \(\nu\) is the zero measure, since its value on \([0,\tfrac14]\) is nonzero.
Worked Example: A Density with Respect to Counting Measure
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), let \(\mathcal{M}\) be its power set, and let \(\mu\) be counting measure, so \(\mu(A)\) is the number of points in \(A\), possibly infinite. This measure is sigma-finite, since \(X=\bigcup_{n\ge1}\{n\}\) and each singleton has measure \(1\). Define \(g(n)=(-1)^{n+1}2^{-n}\) and \(\nu(A)=\sum_{n\in A}g(n)\).
Here \(g\in L^1(\mu)\), because \(\sum_{n=1}^{\infty}|g(n)|=\sum_{n=1}^{\infty}2^{-n}=1\). Thus \(\nu\) is a finite signed measure, \(\nu\ll\mu\), and \(d\nu/d\mu=g\) at every point of \(\mathbb{N}\). In particular, the derivative is the mass assigned to a singleton: \(\nu(\{n\})=g(n)\), because integrating against counting measure sums the values at the points.
For the finite set \(A=\{1,2,4\}\),
The total mass is also exact. The geometric series has first term \(1/2\) and ratio \(-1/2\), so \(\nu(X)=\sum_{n=1}^{\infty}(-1)^{n+1}2^{-n}=(1/2)/(1-(-1/2))=1/3\). Check: summing the first four terms gives \(1/2-1/4+1/8-1/16=5/16\), and the remaining alternating tail starts with \(1/32\); the infinite geometric-series formula gives \(1/3\). For the finite-set calculation, the singleton masses are \(1/2,-1/4,-1/16\), whose sum is \(3/16\), agreeing with the integral formula.
Worked Example: A Probability Density on the Half-Line
Let \(X=[0,\infty)\), let \(\mathcal{M}\) be the Lebesgue sigma-algebra restricted to \(X\), and let \(\mu\) be Lebesgue measure. This measure space is sigma-finite but not finite. Define \(\nu(A)=\int_A xe^{-x}\,d\mu\). The function \(g(x)=xe^{-x}\) is nonnegative and integrable, so \(\nu\) is a finite positive measure and \(d\nu/d\mu=xe^{-x}\) \(\mu\)-almost everywhere.
Integration by parts, with \(u=x\), \(dv=e^{-x}dx\), \(du=dx\), and \(v=-e^{-x}\), yields the antiderivative \(-(x+1)e^{-x}\). Therefore
The total mass is \(\nu(X)=\lim_{R\to\infty}\int_0^Rxe^{-x}\,dx=\lim_{R\to\infty}\bigl(1-(R+1)e^{-R}\bigr)=1\), since \((R+1)e^{-R}\to0\). This limit computes the integral over the increasing sets \([0,R]\), using continuity from below for the measure defined by the nonnegative density. Check: the derivative of \(-(x+1)e^{-x}\) is \(xe^{-x}\), and its values at \(1\) and \(0\) give \(1-2/e\). The density is nonnegative and has total integral \(1\), as required for a probability measure.
What the Formula Does—and Does Not—Say
When the measure is given as an integral, the computation follows a short sequence of checks. Identify the reference measure in the integral; verify that the proposed density is measurable and integrable when a finite signed or complex measure is intended; then use the integral formula to identify the derivative. To calculate \(\nu(A)\), integrate the density over \(A\), retaining its sign if the measure is signed. In a counting-measure formula, this last step is a sum over the points in \(A\), not an integral with respect to Lebesgue measure.
The derivative is relative to the stated reference measure. In the counting example, \(d\nu/d\mu\) is the sequence of singleton masses. Replacing \(\mu\) with a different measure changes the question, and the same function cannot simply be carried over without checking the defining integral again. Also, \(d\nu/d\mu\) is not generally a pointwise ratio of two set values: it is a measurable function characterized by its integrals over every measurable set. Its value on a \(\mu\)-null set can be changed without changing \(\nu\).
Common mistake: replacing a signed integral by an integral of the absolute value. In the first example, \(2x-1\le0\) on \([0,\tfrac14]\). The signed measure of that set is \(-3/16\), not \(3/16\). The incorrect calculation is \(\int_0^{1/4}|2x-1|\,dx=3/16\); it computes the integral of the absolute density on that set, not \(\nu([0,\tfrac14])\). The correction is to use the signed density itself: \(\int_0^{1/4}(2x-1)\,dx=-3/16\). An absolute value is relevant to total variation, which is a different measure, not to the value of the signed measure \(\nu\).
Finally, an explicit formula may provide a density immediately, but the reference measure and the hypotheses still matter. Integrability ensures finiteness of the resulting signed or complex measure; the formula itself ensures absolute continuity. The derivative is then identified almost everywhere by uniqueness, while the setwise identity \(\nu(A)=\int_Ag\,d\mu\) holds for every measurable set \(A\).
Check Your Understanding
For each question, name the reference measure before identifying or using a density.
- On \([0,1]\) with Lebesgue measure, define \(\nu(A)=\int_A(3x^2-1)\,dm\). What is \(d\nu/dm\), and what is \(\nu([0,\tfrac12])\)?
- On \(\mathbb{N}\) with counting measure, suppose \(\nu(A)=\sum_{n\in A}5^{-n}\). What is \(d\nu/d\mu\), and what is \(\nu(\{2,4\})\)?
- Why does an integrable real-valued \(g\) in the formula \(\nu(A)=\int_Ag\,d\mu\) give \(\nu\ll\mu\)?
- For \(g(x)=2x-1\) on \([0,\tfrac14]\), compare \(\int g\,dm\) and \(\int|g|\,dm\). Which one gives the signed measure of the interval?
- In the half-line example, why is the Radon–Nikodym derivative stated only up to equality almost everywhere, even though a specific formula \(xe^{-x}\) was supplied?