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Besicovitch's covering theorem and the differentiation of Radon measures

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

For Lebesgue measure on \(\mathbb R^r\), the averages of a locally integrable function over the balls \(B(y,\delta)\) converge to its value at almost every \(y\) as \(\delta\to0\). The usual proof uses a covering argument of Vitali type, which needs a bound \(\lambda B(y,2\delta)\le C\lambda B(y,\delta)\) for the measure \(\lambda\). An arbitrary Radon measure on \(\mathbb R^r\) has no such bound. Besicovitch found a covering principle that uses only the geometry of Euclidean balls: from any family of balls centred at the points of a bounded set one can select \(5^r\) disjoint subfamilies that still cover the set (Theorem 3.1). From it follows the differentiation theorem for every Radon measure on \(\mathbb R^r\) (Theorem 5.1) and the density theorem (Corollary 5.2).

The covering principle and the differentiation theorem are due to Besicovitch. The form of the covering lemma with the constant \(5^r\), and the arrangement of the proofs, follow [Fremlin, Volume 4, Section 472], which is free to read.

We use from the core course Measure and Integration: Lebesgue measure on \(\mathbb R^r\), which is translation invariant (Fremlin, Volume 1, 134A) and multiplies by \(|\det T|\) under a linear map \(T\) (Fremlin, Volume 2, 263A); Radon measures on \(\mathbb R^r\) and their outer regularity (Fremlin, Volume 2, 256A–256B); and the absolute continuity of the integral (Fremlin, Volume 2, 225A).

1. Radon measures on \(\mathbb R^r\)

Throughout, \(r\ge1\), \(\|\cdot\|\) is the Euclidean norm, and a ball is a closed ball \(B(x,\delta)=\{y:\|y-x\|\le\delta\}\) with \(\delta>0\). Write \(\mathcal L\) for Lebesgue measure and \(\beta_r=\mathcal L B(0,1)\), which is positive and finite. Since \(x\mapsto\delta x\) has determinant \(\delta^r\), \(\mathcal LB(x,\delta)=\beta_r\delta^r\).

A Radon measure on \(\mathbb R^r\) is a complete measure \(\lambda\), with domain \(\Sigma\), that is defined on all open sets, finite on bounded sets and inner regular with respect to the compact sets: \(\lambda E=\sup\{\lambda K:K\subseteq E\text{ compact}\}\) for \(E\in\Sigma\). It is outer regular: for \(E\in\Sigma\) and \(\varepsilon>0\) there is an open \(G\supseteq E\) with \(\lambda(G\setminus E)\le\varepsilon\). The outer measure of \(A\subseteq\mathbb R^r\) is \(\lambda^*A=\inf\{\lambda E:A\subseteq E\in\Sigma\}\). A function is locally integrable if it is integrable over every bounded measurable set.

Lemma 1.1 (Outer measure). Let \(\lambda\) be a Radon measure and \(A\subseteq\mathbb R^r\) with \(\lambda^*A<\infty\).

  1. There is \(E\in\Sigma\) with \(A\subseteq E\) and \(\lambda(F\cap E)=\lambda^*(F\cap A)\) for every \(F\in\Sigma\). (An envelope of \(A\).)
  2. If \(A_0\subseteq A_1\subseteq\cdots\) and \(\lambda^*\bigl(\bigcup_nA_n\bigr)<\infty\), then \(\lambda^*\bigl(\bigcup_nA_n\bigr)=\lim_n\lambda^*A_n\).
  3. For \(F\in\Sigma\), \(\lambda^*A=\lambda^*(A\cap F)+\lambda^*(A\setminus F)\).

Proof. (1) Choose \(E_n\in\Sigma\) containing \(A\) with \(\lambda E_n\to\lambda^*A\), and put \(E=\bigcap_nE_n\); then \(\lambda E=\lambda^*A\). For \(F\in\Sigma\), \(\lambda(F\cap E)\ge\lambda^*(F\cap A)\) and \(\lambda(E\setminus F)\ge\lambda^*(A\setminus F)\), while the sum of the right sides is at least \(\lambda^*A=\lambda E\), the sum of the left sides. All terms are finite, so both inequalities are equalities.

(2) Let \(E_n\) be envelopes of \(A_n\) and \(E_n'=\bigcap_{m\ge n}E_m\). Then \(A_n\subseteq E_n'\subseteq E_n\), so \(\lambda E_n'=\lambda^*A_n\), and \(E_n'\) increases. Hence \(\lambda^*\bigl(\bigcup_nA_n\bigr)\le\lambda\bigl(\bigcup_nE_n'\bigr)=\lim_n\lambda^*A_n\le\lambda^*\bigl(\bigcup_nA_n\bigr)\).

(3) With an envelope \(E\) of \(A\), apply (1) to \(F\) and to its complement: \(\lambda^*(A\cap F)+\lambda^*(A\setminus F)=\lambda(E\cap F)+\lambda(E\setminus F)=\lambda E=\lambda^*A\). \(\square\)

Lemma 1.2 (Support). Let \(\lambda\) be a Radon measure and \(Z\) the set of \(y\) with \(\lambda B(y,\delta)>0\) for every \(\delta>0\). Then \(Z\) is closed and \(\lambda(\mathbb R^r\setminus Z)=0\).

Proof. If \(\lambda B(y,\delta)=0\), then \(B(y',\delta/2)\subseteq B(y,\delta)\) for \(\|y'-y\|<\delta/2\), so the complement of \(Z\) is open. For such \(y\), choose \(q\in\mathbb Q^r\) with \(\|q-y\|<\delta/4\) and a rational \(s\in(\delta/4,\delta/2)\). Then \(y\in B(q,s)\subseteq B(y,\delta)\), so \(\lambda B(q,s)=0\). The complement of \(Z\) is therefore covered by countably many null balls. \(\square\)

Lemma 1.3 (Measures from the Riesz theorem). Let \(\mu\) be a measure on the Borel sets of \(\mathbb R^r\) that is finite on compact sets and outer regular: \(\mu E=\inf\{\mu G:G\supseteq E\text{ open}\}\) for every Borel \(E\). Then the completion of \(\mu\) is a Radon measure.

Proof. The completion is complete, defined on open sets and finite on bounded sets. Let \(E\) be Borel with \(E\subseteq K=B(0,m)\), and \(\varepsilon>0\). Outer regularity gives an open \(U\supseteq K\setminus E\) with \(\mu U\le\mu(K\setminus E)+\varepsilon\). Then \(C=K\setminus U\) is compact, \(C\subseteq E\), and \(E\setminus C=E\cap U\subseteq U\setminus(K\setminus E)\) has measure at most \(\varepsilon\). For a general Borel set \(E\), \(\mu E=\lim_m\mu(E\cap B(0,m))\), so \(E\) is inner regular too. A set of the completion differs from a Borel set \(E_0\subseteq E\) by a null set, and has the same inner approximations. \(\square\)

2. The covering lemma

Fix \(\varepsilon>0\) with

\[ (5^r+1)(1-\varepsilon-\varepsilon^2)^r>(5+\varepsilon)^r ; \tag{2.1} \]

it exists because the left side tends to \(5^r+1\) and the right side to \(5^r\) as \(\varepsilon\to0\).

Lemma 2.1. Let \(x_0,\dots,x_n\in\mathbb R^r\) and \(\delta_0,\dots,\delta_n>0\) satisfy

\[ \|x_i-x_j\|>\delta_i\quad\text{and}\quad\delta_j\le(1+\varepsilon)\delta_i\qquad\text{whenever }i<j\le n . \]

Then at most \(5^r\) indices \(i\le n\) satisfy \(\|x_i-x_n\|\le\delta_i+\delta_n\).

Proof. Replacing every \(x_i\) by \(x_i/\delta_n\) and every \(\delta_i\) by \(\delta_i/\delta_n\) changes neither the hypotheses nor the conclusion, so let \(\delta_n=1\). Then \(\delta_i\ge1/(1+\varepsilon)\) for all \(i\). Let \(I\) be the set of the indices in question; for \(i\in I\), \(\|x_i-x_n\|\le\delta_i+1\), and \(\|x_i-x_n\|>\delta_i\) if \(i<n\). For \(i\in I\) put \(x_i'=x_i\) if \(\|x_i-x_n\|\le2+\varepsilon\), and otherwise let \(x_i'\) be the point of the segment from \(x_n\) to \(x_i\) at distance \(2+\varepsilon\) from \(x_n\). In the second case \(\|x_i-x_i'\|=\|x_i-x_n\|-(2+\varepsilon)\).

Claim: \(\|x_i'-x_j'\|\ge1-\varepsilon-\varepsilon^2\) for distinct \(i<j\) in \(I\). Write \(d_k=\|x_k-x_n\|\).

Counting. Put \(\rho=\frac12(1-\varepsilon-\varepsilon^2)\). By the claim the open balls \(U_i=\{z:\|z-x_i'\|<\rho\}\), \(i\in I\), are disjoint. Each contains \(B(x_i',\rho')\) for every \(\rho'<\rho\), so \(\mathcal LU_i\ge\beta_r\rho^r\). Since \(\|x_i'-x_n\|\le2+\varepsilon\) and \(\rho\le\frac12(1-\varepsilon)\), they lie in \(B\bigl(x_n,2+\varepsilon+\frac12(1-\varepsilon)\bigr)=B\bigl(x_n,\frac12(5+\varepsilon)\bigr)\). Comparing measures, \(\#I\,(1-\varepsilon-\varepsilon^2)^r\le(5+\varepsilon)^r\), and (2.1) gives \(\#I<5^r+1\). \(\square\)

3. Besicovitch's covering theorem

Theorem 3.1. Let \(A\subseteq\mathbb R^r\) be bounded and \(\mathcal I\) a family of balls such that every point of \(A\) is the centre of a member of \(\mathcal I\). Then there are \(5^r\) countable disjoint subfamilies \(\mathcal I_0,\dots,\mathcal I_{5^r-1}\) of \(\mathcal I\) whose union covers \(A\).

Proof. For \(x\in A\) choose \(\delta_x\) with \(B(x,\delta_x)\in\mathcal I\). If \(A\) is empty there is nothing to prove. If \(\sup_{x\in A}\delta_x=\infty\), one ball \(B(x,\delta_x)\) with \(\delta_x\ge\operatorname{diam}A\) covers \(A\). So let the radii be bounded; then \(C=\bigcup_{x\in A}B(x,\delta_x)\) is bounded.

Selection. Choose balls \(B_0,B_1,\dots\), some possibly empty, as follows. If \(A\subseteq\bigcup_{i<n}B_i\), put \(B_n=\emptyset\). Otherwise let \(\alpha_n=\sup\{\delta_x:x\in A\setminus\bigcup_{i<n}B_i\}\), choose \(x_n\in A\setminus\bigcup_{i<n}B_i\) with \((1+\varepsilon)\delta_{x_n}\ge\alpha_n\), and put \(B_n=B(x_n,\delta_{x_n})\). If \(B_n\neq\emptyset\), then for \(i<j\le n\) we have \(x_j\notin B_i\), that is \(\|x_i-x_j\|>\delta_{x_i}\), and \(\delta_{x_j}\le\alpha_i\le(1+\varepsilon)\delta_{x_i}\). Two balls \(B_i\), \(B_n\) meet exactly when \(\|x_i-x_n\|\le\delta_{x_i}+\delta_{x_n}\). By Lemma 2.1, the set \(I_n\) of the \(i<n\) with \(B_i\cap B_n\neq\emptyset\) has fewer than \(5^r\) elements.

Colouring. Define \(f(n)\) inductively as the least \(k<5^r\) that differs from \(f(i)\) for all \(i\in I_n\), and let \(\mathcal I_k\) consist of the non-empty \(B_n\) with \(f(n)=k\). If \(i<n\) and \(f(i)=f(n)\), then \(i\notin I_n\), so \(B_i\cap B_n=\emptyset\): each \(\mathcal I_k\) is disjoint.

Covering. The balls of one \(\mathcal I_k\) are disjoint and lie in the bounded set \(C\), so \(\sum_n\mathcal LB_n\le5^r\mathcal L^*C<\infty\). If some \(x\in A\) lay in no \(B_n\), every \(\alpha_n\) would be defined and at least \(\delta_x\), every \(B_n\) would have radius at least \(\delta_x/(1+\varepsilon)\), and \(\sum_n\mathcal LB_n\) would diverge. \(\square\)

4. Disjoint covers up to a null set

Theorem 4.1. Let \(\lambda\) be a Radon measure on \(\mathbb R^r\), \(A\subseteq\mathbb R^r\), and \(\mathcal I\) a family of balls such that every point of \(A\) is the centre of arbitrarily small members of \(\mathcal I\). Then there is a countable disjoint \(\mathcal I_0\subseteq\mathcal I\) with \(\lambda^*\bigl(A\setminus\bigcup\mathcal I_0\bigr)=0\).

Proof. Step 1. If \(A'\subseteq A\) is bounded, there is a finite disjoint \(\mathcal J\subseteq\mathcal I\) with \(\lambda^*\bigl(A'\cap\bigcup\mathcal J\bigr)\ge6^{-r}\lambda^*A'\). Indeed, if \(\lambda^*A'=0\) take \(\mathcal J=\emptyset\). Otherwise Theorem 3.1 gives disjoint countable \(\mathcal J_k\), \(k<5^r\), covering \(A'\); since \(\lambda^*A'\le\sum_k\lambda^*(A'\cap\bigcup\mathcal J_k)\), some \(k\) has \(\lambda^*(A'\cap\bigcup\mathcal J_k)\ge5^{-r}\lambda^*A'\). Enumerate \(\mathcal J_k\) as \(B_0,B_1,\dots\). By Lemma 1.1(2), \(\lambda^*\bigl(A'\cap\bigcup_{i\le n}B_i\bigr)\) tends to \(\lambda^*(A'\cap\bigcup\mathcal J_k)\), which is finite, so it exceeds \(6^{-r}\lambda^*A'\) for some \(n\); take \(\mathcal J=\{B_0,\dots,B_n\}\).

Step 2. Fix a sequence \((m_n)\) of natural numbers in which every natural number occurs infinitely often. Put \(\mathcal K_0=\emptyset\). Given a finite disjoint \(\mathcal K_n\subseteq\mathcal I\), let \(\mathcal I'\) consist of the members of \(\mathcal I\) disjoint from the closed set \(\bigcup\mathcal K_n\), and \(A_n=A\cap B(0,m_n)\setminus\bigcup\mathcal K_n\). Every point of \(A_n\) is the centre of arbitrarily small members of \(\mathcal I'\). Step 1 gives a finite disjoint \(\mathcal J_n\subseteq\mathcal I'\) with \(\lambda^*(A_n\cap\bigcup\mathcal J_n)\ge6^{-r}\lambda^*A_n\); put \(\mathcal K_{n+1}=\mathcal K_n\cup\mathcal J_n\), again finite and disjoint. By Lemma 1.1(3), applied to the closed set \(\bigcup\mathcal J_n\),

\[ \lambda^*\Bigl(A\cap B(0,m_n)\setminus\bigcup\mathcal K_{n+1}\Bigr)=\lambda^*\Bigl(A_n\setminus\bigcup\mathcal J_n\Bigr)\le(1-6^{-r})\,\lambda^*A_n . \]

For fixed \(m\), the numbers \(\lambda^*(A\cap B(0,m)\setminus\bigcup\mathcal K_n)\) decrease in \(n\), are finite, and are multiplied by at most \(1-6^{-r}\) at every \(n\) with \(m_n=m\); so they tend to \(0\). With \(\mathcal I_0=\bigcup_n\mathcal K_n\), which is countable and disjoint, \(\lambda^*(A\cap B(0,m)\setminus\bigcup\mathcal I_0)=0\) for every \(m\), and hence \(\lambda^*(A\setminus\bigcup\mathcal I_0)=0\). \(\square\)

5. The differentiation theorem

Theorem 5.1 (Besicovitch). Let \(\lambda\) be a Radon measure on \(\mathbb R^r\) and \(f\) a locally \(\lambda\)-integrable complex function on \(\mathbb R^r\). For \(\lambda\)-almost every \(y\), \(\lambda B(y,\delta)>0\) for all \(\delta>0\) and

\[ \text{(a)}\ \lim_{\delta\downarrow0}\frac1{\lambda B(y,\delta)}\int_{B(y,\delta)}f\,d\lambda=f(y),\qquad \text{(b)}\ \lim_{\delta\downarrow0}\frac1{\lambda B(y,\delta)}\int_{B(y,\delta)}|f(x)-f(y)|\,\lambda(dx)=0 . \]

Proof. By Lemma 1.2 the support \(Z\) is conegligible, and at its points the averages are defined. It suffices to treat real \(f\).

(a) For \(n\in\mathbb N\) and rationals \(q<q'\) let \(A\) be the set of \(y\in Z\) with \(\|y\|<n\), \(f(y)\le q\) and \(\limsup_{\delta\downarrow0}\frac1{\lambda B(y,\delta)}\int_{B(y,\delta)}f\,d\lambda>q'\). We show \(\lambda^*A=0\). Let \(\varepsilon>0\). Since \(f\) is integrable on \(B(0,n)\), there is \(\eta\in(0,\varepsilon]\) with \(\int_F|f|\,d\lambda\le\varepsilon\) for measurable \(F\subseteq B(0,n)\) with \(\lambda F\le\eta\) (Fremlin, Volume 2, 225A). Let \(E\) be an envelope of \(A\) (Lemma 1.1(1)); replacing \(E\) by its intersection with the measurable set \(S=\{y\in Z:\|y\|<n,\ f(y)\le q\}\), which contains \(A\), we keep an envelope and get \(f\le q\) on \(E\). Choose an open \(G\) with \(E\subseteq G\subseteq\{\|y\|<n\}\) and \(\lambda(G\setminus E)\le\eta\). Let \(\mathcal I\) be the family of balls \(B\subseteq G\) with \(\int_Bf\,d\lambda\ge q'\lambda B\). Every point of \(A\) is the centre of arbitrarily small members of \(\mathcal I\), so Theorem 4.1 gives a countable disjoint \(\mathcal I_0\subseteq\mathcal I\) with \(\lambda^*(A\setminus\bigcup\mathcal I_0)=0\). Let \(V=\bigcup\mathcal I_0\subseteq G\). By the envelope property \(\lambda(E\setminus V)=\lambda^*(A\setminus V)=0\), and \(\lambda(V\setminus E)\le\eta\). Hence \(|\lambda V-\lambda E|\le\varepsilon\), and

\[ q'\lambda E\le q'\lambda V+|q'|\varepsilon=\sum_{B\in\mathcal I_0}q'\lambda B+|q'|\varepsilon\le\int_Vf\,d\lambda+|q'|\varepsilon\le\int_Ef\,d\lambda+\varepsilon+|q'|\varepsilon\le q\lambda E+(1+|q'|)\varepsilon . \]

So \((q'-q)\lambda^*A=(q'-q)\lambda E\le(1+|q'|)\varepsilon\) for every \(\varepsilon\), and \(\lambda^*A=0\). Taking the union over \(n\), \(q\), \(q'\): for almost every \(y\), the \(\limsup\) of the averages is at most \(f(y)\). Applying this to \(-f\), the \(\liminf\) is at least \(f(y)\) almost everywhere.

(b) Apply (a) to the functions \(|f-q|\), \(q\in\mathbb Q\), to get a conegligible set \(D\) on which all their averages converge. For \(y\in D\) and \(\varepsilon>0\) choose \(q\) with \(|f(y)-q|\le\varepsilon\); for small \(\delta\) the average of \(|f-q|\) over \(B(y,\delta)\) is at most \(|f(y)-q|+\varepsilon\le2\varepsilon\), and the average of \(|f-f(y)|\) is at most that plus \(|q-f(y)|\), hence at most \(3\varepsilon\). \(\square\)

Corollary 5.2 (Density theorem). Let \(\lambda\) be a Radon measure on \(\mathbb R^r\) and \(K\in\Sigma\). For \(\lambda\)-almost every \(y\in K\), \(\lambda(K\cap B(y,\delta))/\lambda B(y,\delta)\to1\), and for \(\lambda\)-almost every \(y\notin K\) it tends to \(0\), as \(\delta\downarrow0\).

Proof. Theorem 5.1(a) with \(f=1_K\). \(\square\)

6. Exercises

Exercise 6.1. Show that one disjoint subfamily does not always suffice in Theorem 3.1: take \(r=1\), \(A=\{0,1,2\}\) and the balls \([-\tfrac32,\tfrac32]\), \([0.9,1.1]\) and \([\tfrac12,\tfrac72]\).

Solution. The first ball is the only one containing \(0\), and the third the only one containing \(2\). Any subfamily covering \(A\) contains both, and they meet.

Exercise 6.2. (a) Let \(\lambda\) be the Dirac measure at \(0\) on \(\mathbb R^r\), defined on all subsets. Check Theorem 5.1 directly. (b) On \(\mathbb R\) let \(\lambda E=\sum\{1/k!:k\ge1,\ 1/k\in E\}\) for all \(E\subseteq\mathbb R\). Show that \(\lambda\) is a Radon measure and that, for \(k\ge4\), \(y=1/k\) and \(\delta=1/(2k(k-1))\), one has \(\lambda B(y,2\delta)\ge k\,\lambda B(y,\delta)\). So no bound \(\lambda B(y,2\delta)\le C\lambda B(y,\delta)\) holds at the points of the support, and Theorem 5.1 is needed.

Solution. (a) The support is \(\{0\}\), and every average at \(0\) equals \(f(0)\). (b) Every subset is measurable, so \(\lambda\) is complete and defined on open sets; the total mass is finite; and \(\lambda E\) is the supremum of \(\lambda\) over the finite, hence compact, subsets of \(E\cap\{1/k:k\ge1\}\). The atoms nearest to \(1/k\) are at distances \(\frac1{k(k+1)}\) and \(\frac1{k(k-1)}\). For \(k\ge4\), \(\delta=\frac1{2k(k-1)}<\frac1{k(k+1)}\), so \(B(y,\delta)\) contains only the atom \(1/k\), of mass \(1/k!\), while \(B(y,2\delta)\) contains \(1/(k-1)\), of mass \(1/(k-1)!=k/k!\).

Where this leads

The lesson Vector-valued functions, tensor products with \(L^p\), and preduals uses Theorem 5.1 and Corollary 5.2 to construct selectors for every Radon measure on \(\mathbb R^n\) (the remark after its Proposition 3.4). The maximal inequality for Radon measures on \(\mathbb R^r\) is in [Fremlin, Volume 4, 472E–472F].

References

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