Differentiation of measures studies local ratios of measures on shrinking balls. It generalizes differentiation of functions and is a key bridge to geometry and statistics.
Measure ratios
Local densities compare a measure to a reference measure at small scales.
Definition. For Radon measures \(\nu,\mu\) on \(\mathbb R^d\), a derivative of \(\nu\) with respect to \(\mu\) at \(x\) is a limit, when it exists, \[ \lim_{r\downarrow0}\frac{\nu(B(x,r))}{\mu(B(x,r))}. \]
Remark. When \(\mu\) is Lebesgue measure and \(\nu=f\mu\), this derivative is \(f(x)\) at Lebesgue points of \(f\).
Vitali covering theorem
A fine cover offers arbitrarily small balls at every point; disjoint selection turns its local information into an almost-everywhere statement.
Theorem. Let \(E\subset\mathbb R^d\) have finite outer measure, and let \(\mathcal V\) be a family of closed balls such that for every \(x\in E\) and every \(\varepsilon>0\), some \(B\in\mathcal V\) is centered at \(x\) and has radius below \(\varepsilon\). Then \(\mathcal V\) contains a countable pairwise disjoint subfamily \(\{B_i\}\) such that \[ m\!\left(E\setminus\bigcup_iB_i\right)=0. \]
Proof outline. Restrict first to balls in a bounded neighborhood of \(E\). Choose greedily a disjoint sequence whose radii are within a fixed factor of the largest radius still available. Finite total volume forces the selected radii to tend to zero. If the uncovered remainder had positive outer measure, the fine-cover property would supply a disjoint ball from \(\mathcal V\) of a scale still represented in the greedy selection, contradicting maximality. Applying the argument successively to the uncovered remainder and summing a geometric loss gives a disjoint family covering \(E\) up to a null set. The stronger almost-everywhere coverage, unlike the \(5r\) lemma, uses both fineness and finite outer measure.
Besicovitch differentiation for Radon measures
Radon-Nikodym derivatives can be recovered from ball ratios.
Theorem (Besicovitch differentiation). If \(\mu\) is a locally finite Borel measure on \(\mathbb R^d\) and \(f\in L^1_{\mathrm{loc}}(\mu)\), then \[ \lim_{r\downarrow0} \frac{1}{\mu(B(x,r))} \int_{B(x,r)}f\,\mathrm{d}\mu=f(x) \] for \(\mu\)-almost every \(x\) for which the denominator is positive. Consequently, if \(\nu\ll\mu\) and \(f=\mathrm{d}\nu/\mathrm{d}\mu\), then \[ \frac{\nu(B(x,r))}{\mu(B(x,r))}\to f(x) \] for \(\mu\)-almost every \(x\).
Proof outline. The Besicovitch covering theorem replaces the Lebesgue \(5r\) argument: from balls centered in a bounded set it selects at most \(N_d\) disjoint layers whose union covers the centers. This yields the weak estimate \[ \mu\!\left( \sup_{r>0}\frac1{\mu(B(x,r))} \int_{B(x,r)}|g|\,\mathrm{d}\mu>\lambda \right) \le\frac{N_d}{\lambda}\|g\|_{L^1(\mu)} \] locally. Approximate \(f\) in \(L^1(\mu)\) by continuous compactly supported functions, whose local averages converge pointwise by continuity, and repeat the maximal-function argument from the Lebesgue differentiation theorem. For \(\nu=f\mu\), the mass ratio is exactly the \(\mu\)-average of \(f\).
Density of measurable sets
A measurable set is locally almost all or almost nothing at almost every point.
Theorem. For measurable \(E\subset\mathbb R^d\), almost every \(x\in E\) has density \(1\) and almost every \(x\notin E\) has density \(0\).
Proof. Apply the differentiation theorem to \(\mathbf 1_E\). The averages over balls are exactly the relative densities \(m(E\cap B(x,r))/m(B(x,r))\), and they converge almost everywhere to \(\mathbf 1_E(x)\).
Local ratios of the singular part
Differentiation recovers only the part visible to the reference measure.
Theorem. Let \(\nu=f\mu+\nu_s\) be the Lebesgue decomposition of locally finite Borel measures on \(\mathbb R^d\). Then \[ \lim_{r\downarrow0}\frac{\nu(B(x,r))}{\mu(B(x,r))}=f(x) \quad\text{for }\mu\text{-almost every }x. \] In particular, \[ \lim_{r\downarrow0}\frac{\nu_s(B(x,r))}{\mu(B(x,r))}=0 \quad\text{for }\mu\text{-almost every }x. \]
Proof. Apply differentiation of measures to \(\rho=\mu+\nu_s\). Write \[ g=\frac{\mathrm{d}\mu}{\mathrm{d}\rho}, \qquad h=\frac{\mathrm{d}\nu_s}{\mathrm{d}\rho}. \] Singularity gives disjoint supports on which respectively \(g=1,h=0\) and \(g=0,h=1\), \(\rho\)-almost everywhere. At \(\mu\)-almost every point, differentiation recovers \(g=1\) and \(h=0\) from ball averages, so the ratio of the two ball masses tends to \(0\). Adding the absolutely continuous part, whose ratio tends to \(f\), proves the first identity.