2.8 Densities and differentiation of integrals9563E8B823D2

Proof. We first construct the density on a half-open cube as an $L^2$ limit of averages of $\mu $ over dyadic subcubes. Let $Q=\prod _{i=1}^ d[a_ i,a_ i+s)$ be a half-open cube whose closure lies in $\Omega $, so that $\nu (Q)<\infty $, and suppose that $\mu (E)\leq C\nu (E)$ for Borel sets $E\subset Q$; in particular, $\mu (Q)\leq C\nu (Q)<\infty $. For $k\geq 0$, let $\mathcal Q_ k$ be the partition of $Q$ into $2^{kd}$ half-open dyadic subcubes of side $2^{-k}s$. On each $Q'\in \mathcal Q_ k$, let $g_ k=\mu (Q')/\nu (Q')$ if $\nu (Q')>0$ and $g_ k=0$ otherwise; in the latter case also $\mu (Q')=0$, by domination. Thus $0\leq g_ k\leq C$, and $\int _{Q'}g_ l\, \mathrm d\nu =\mu (Q')$ whenever $Q'\in \mathcal Q_ k$ and $l\geq k$, because $Q'$ is a disjoint union of cubes of $\mathcal Q_ l$. As $g_ k$ is constant on the cubes of $\mathcal Q_ k$, it follows that $\int _ Qg_ lg_ k\, \mathrm d\nu =\int _ Qg_ k^2\, \mathrm d\nu $, and hence, in $L^2(Q,\nu )$,

The squared norms increase and are bounded by $C^2\nu (Q)$, so $(g_ k)$ is a Cauchy sequence in $L^2(Q,\nu )$. Choose $k_1<k_2<\cdots $ with $\| g_{k_{j+1}}-g_{k_ j}\| _2\leq 2^{-j}$. By the Cauchy–Schwarz inequality, $\int _ Q|g_{k_{j+1}}-g_{k_ j}|\, \mathrm d\nu \leq \nu (Q)^{1/2}2^{-j}$, so by monotone convergence $\sum _ j|g_{k_{j+1}}-g_{k_ j}|$ has finite integral; hence it is finite, and $(g_{k_ j})$ converges, $\nu $-almost everywhere. Let $g$ be the limit of $(g_{k_ j})$ on the Borel set where it converges, and $0$ elsewhere. Then $g$ is a Borel function with values in $[0,C]$, $g_{k_ j}\to g$ in $L^2(Q,\nu )$ by dominated convergence, and, the sequence being Cauchy, $g_ k\to g$ in $L^2(Q,\nu )$ and hence in $L^1(Q,\nu )$. Consequently $\int _{Q'}g\, \mathrm d\nu =\mu (Q')$ for every dyadic subcube $Q'$ of $Q$.

The dyadic subcubes of $Q$, together with $\varnothing $, form a family closed under finite intersections, since two of them are nested or disjoint. If $U$ is open, every point of $U\cap Q$ lies in arbitrarily small dyadic subcubes, hence in one contained in $U$, so $U\cap Q$ is a countable union of dyadic subcubes. Thus these cubes generate the Borel $\sigma $-algebra of $Q$, and lemma 2.2.6(2) shows that the finite measures $\mu $ and $g\nu $, which agree on all dyadic subcubes including $Q$, agree on every Borel subset of $Q$.

For uniqueness, let $g,h\geq 0$ be Borel functions with $g\nu =h\nu =\mu $ on the Borel subsets of such a cube $Q$. The set $A=Q\cap \{ g>h\} $ satisfies $\int _ A(g-h)\, \mathrm d\nu =\mu (A)-\mu (A)=0$, because $\mu (A)<\infty $, so $\nu (A)=0$; by symmetry, $g=h$ $\nu $-almost everywhere on $Q$. Since $\Omega $ is a countable union of such cubes, densities on $\Omega $ are unique up to $\nu $-null sets.

To construct the density on $\Omega $ in case (1), cover $\Omega $ by countably many half-open cubes $Q_1,Q_2,\ldots $ with closures in $\Omega $, for instance all half-open dyadic cubes with this property, let $g^{(k)}$ be the density on $Q_ k$ just constructed, and put $g=g^{(k)}$ on $P_ k=Q_ k\setminus \bigcup _{l<k}Q_ l$. For every Borel set $E\subset \Omega $,

so $\mu =g\nu $ with $0\leq g\leq C$. In case (2), every cube $Q$ with closure in $\Omega $ lies in a relatively compact open set $U\subset \Omega $, for instance a set of points at distance less than $\delta $ from $\overline Q$ with $\delta $ small, and $\mu \leq C_ U\nu $ on Borel subsets of $Q$; the same construction gives $\mu =g\nu $ with $g\geq 0$ Borel. If $U\subset \Omega $ is relatively compact and open, the sets $A_ j=U\cap \{ g>C_ U+1/j\} $ satisfy $(C_ U+1/j)\nu (A_ j)\leq \mu (A_ j)\leq C_ U\nu (A_ j)$ and $\nu (A_ j)<\infty $, so $\nu (A_ j)=0$; thus $g\leq C_ U$ $\nu $-almost everywhere on $U$. The final assertion is case (1) for Lebesgue measure, which is a Radon measure by proposition 2.2.2. $\square$

Proof. (1) Denote also by $C\nu $ the positive distribution defined by integration against the Radon measure $C\nu $. For every nonnegative test function $\chi $ we have $\langle C\nu \pm T,\chi \rangle =C\int \chi \, \mathrm d\nu \pm \langle T,\chi \rangle \geq 0$, so the real distributions $C\nu \pm T$ are positive, and by theorem 2.5.6 they are given by Radon measures $\lambda _+$ and $\lambda _-$. The Radon measures $\lambda _++\lambda _-$ and $2C\nu $ define the same distribution, so they are equal by proposition 2.5.3(4). In particular, $\lambda _+(E)\leq 2C\nu (E)$ for every Borel set $E\subset \Omega $, and lemma 2.8.1(1) gives $\lambda _+=h\nu $ for a Borel function $h$ with values in $[0,2C]$. Hence $\langle T,\chi \rangle =\int \chi \, \mathrm d\lambda _+-C\int \chi \, \mathrm d\nu =\int \chi (h-C)\, \mathrm d\nu $ for every test function $\chi $, and $g=h-C$ has the required properties.

Uniqueness. Let $g$ and $g'$ be real Borel functions, locally integrable with respect to $\nu $, with $\int \chi g\, \mathrm d\nu =\int \chi g'\, \mathrm d\nu $ for every test function $\chi $, and put $w=g-g'$. The Borel measures $w^+\nu $ and $w^-\nu $, where $w^\pm =\max \{ \pm w,0\} $, are finite on compact sets, hence Radon by proposition 2.2.2, and they define the same distribution. By proposition 2.5.3(4) they are equal, and evaluating them on the set $\{ w>0\} $ gives $\int _{\{ w>0\} }w\, \mathrm d\nu =0$, so that $\nu (\{ w>0\} )=0$. Similarly $\nu (\{ w<0\} )=0$, and $g=g'$ $\nu $-almost everywhere.

(2) Let $\mathcal U$ be the family of the open sets $U$ in the hypothesis. Each set of a compact exhaustion of $\Omega $ is covered by finitely many members of $\mathcal U$, so countably many members $U_1,U_2,\ldots $ of $\mathcal U$ cover $\Omega $. Let $U\in \mathcal U$. The restriction of $\nu $ to the Borel subsets of $U$ is a Radon measure on $U$ by proposition 2.2.2, and test functions on $U$, extended by zero, are test functions on $\Omega $. Hence (1), applied on $U$, gives a Borel function $g_ U$ on $U$ with values in $[-C_ U,C_ U]$ which represents $T$ on $U$. By the uniqueness assertion, applied on the open sets $U_ k\cap U_ l$, we have $g_{U_ k}=g_{U_ l}$ $\nu $-almost everywhere on $U_ k\cap U_ l$. Define $g=g_{U_ k}$ on $P_ k=U_ k\setminus \bigcup _{l<k}U_ l$. Since $P_ l\cap U_ k=\varnothing $ for $l>k$, we have $g=g_{U_ k}$ $\nu $-almost everywhere on $U_ k$ for every $k$. Hence $g$ is locally integrable with respect to $\nu $, and $\langle T,\chi \rangle =\int \chi g\, \mathrm d\nu $ for every test function $\chi $ supported in some $U_ k$; by lemma 2.5.2, this holds for every test function. Finally, for every $U\in \mathcal U$, the uniqueness assertion on $U$ gives $g=g_ U$ $\nu $-almost everywhere on $U$, so $|g|\leq C_ U$ $\nu $-almost everywhere on $U$. $\square$

Proof. We prove a maximal inequality on $\mathbb R^ d$, deduce the theorem for integrable functions by approximation with continuous ones, and then localize. The set $N$ where $f$ is infinite is a null set. Changing $f$ to $0$ on $N$ changes neither the averages in the statement nor the values $f(x)$ for $x\notin N$, so the modified function has the same Lebesgue points outside $N$ and the same mollifications. We may therefore assume that $f$ is finite everywhere.

Consider first $f\in L^1(\mathbb R^ d)$, and define its maximal function $Mf(x)$ as the supremum over $r>0$ of the averages of $|f|$ over $Q(x,r)$. For fixed $r$, the average of $|f|$ over $Q(x,r)$ depends continuously on $x$, because $\int _{Q(x,r)}|f|$ and $\int _{Q(x',r)}|f|$ differ by at most $\| f(\cdot +x-x')-f\| _1$, which tends to zero as $x'\to x$ by the continuity of translation in $L^1$ (lemma 2.5.1(2)). As a supremum of continuous functions, $Mf$ is lower semicontinuous, so $\{ Mf>t\} $ is open. We claim the maximal inequality

By inner regularity of Lebesgue measure, it suffices to bound $|L|$ for compact $L\subset \{ Mf>t\} $. Each point of $L$ is the center of an open cube over which the average of $|f|$ exceeds $t$, and finitely many of these cubes cover $L$. Select one of largest side length, discard all cubes meeting it, and repeat with the remaining ones. The selected cubes are disjoint, and each discarded cube meets a selected cube of at least the same side length, so it lies in the concentric cube with three times that side length. These enlarged cubes cover $L$, and therefore $|L|$ is at most $3^ d$ times the total volume of the selected cubes, which is less than $t^{-1}\| f\| _1$.

Now let $g\in C_ c(\mathbb R^ d)$. For every $x$ and $r$,

and the last term tends to zero as $r\downarrow 0$ because $g$ is continuous at $x$. Hence the upper limit $\Lambda f(x)$ of the left side satisfies $\Lambda f\leq M(f-g)+|f-g|$, and for $t>0$ the set $\{ \Lambda f>2t\} $ is contained in $\{ M(f-g)>t\} \cup \{ |f-g|>t\} $, whose measure is at most $(3^ d+1)t^{-1}\| f-g\| _1$ by the maximal inequality and Chebyshev's inequality. By proposition 2.3.5, $\| f-g\| _1$ can be made arbitrarily small, so $\{ \Lambda f>2t\} $ is contained in measurable sets of arbitrarily small measure and is therefore a null set. Letting $t$ run through the positive rationals shows that $\Lambda f=0$ almost everywhere.

Next, let $f$ be locally integrable on $\Omega $, and let $(K_ j)$ be a compact exhaustion of $\Omega $. The function equal to $f$ on $\operatorname{Int}K_{j+1}$ and to $0$ elsewhere belongs to $L^1(\mathbb R^ d)$, and for $x\in \operatorname{Int}K_ j$ and small $r$ the averages in the statement are the same for $f$ and for this function. Hence almost every point of each $\operatorname{Int}K_ j$, and therefore of $\Omega $, is a Lebesgue point of $f$.

Finally, let $x$ be a Lebesgue point of $f$ and let $\varepsilon >0$ be so small that $x\in \Omega _\varepsilon $. The function $\rho _\varepsilon (x-\cdot )$ is bounded by $\varepsilon ^{-d}\sup \rho =2^ d\sup \rho /|Q(x,\varepsilon )|$ and vanishes outside $B(x,\varepsilon )\subset Q(x,\varepsilon )$. Since $\int \rho _\varepsilon =1$,

which tends to zero as $\varepsilon \to 0$. $\square$

Proof. Suppose first that $F$ is continuously differentiable near $Q$. For $x,y\in Q$, the points $z^{(k)}=(y_1,\ldots ,y_ k,x_{k+1},\ldots ,x_ d)$, $0\leq k\leq d$, lead from $x$ to $y$ by changing one coordinate at a time. If $I_ k$ is the $k$th side interval of $Q$, the fundamental theorem of calculus gives

Integrate over $(x,y)\in Q\times Q$. The $k$th term does not depend on the $d+1$ coordinates $x_1,\ldots ,x_ k,y_ k,\ldots ,y_ d$, which contribute a factor $l^{d+1}$, and integration over the remaining coordinates and $t$ gives $\int _ Q|\partial _ kF|\, \mathrm dx$. Since $|F(x)-F_ Q|\leq |Q|^{-1}\int _ Q|F(x)-F(y)|\, \mathrm dy$ and $|Q|=l^ d$, the inequality follows. In general, the mollifications $F_\varepsilon $ are smooth on $V_\varepsilon =\{ x\in V:\operatorname {dist}(x,\mathbb R^ d\setminus V)>\varepsilon \} $, which is a neighborhood of $Q$ for small $\varepsilon $. By lemma 2.5.1(2) and proposition 2.5.3(1), $F_\varepsilon \to F$ and $\partial _ kF_\varepsilon =(\partial _ kF)_\varepsilon \to \partial _ kF$ in $L^1(Q)$. Since $|(F_\varepsilon )_ Q-F_ Q|\leq |Q|^{-1}\| F_\varepsilon -F\| _{L^1(Q)}$, the left sides of the inequality for $F_\varepsilon $ and for $F$ differ by at most $2\| F_\varepsilon -F\| _{L^1(Q)}$, and the inequality for $F_\varepsilon $ passes to the limit. $\square$

Proof. Steps 1 and 2 show by mollification that $f\in C^{1,1}_{\mathrm{loc}}(\Omega )$, Step 3 proves the Taylor expansion at the common Lebesgue points of the $A_{ab}$ by means of lemma 2.8.4, and Step 4 proves the converse.

Step 1: $f\in C^1(\Omega )$. Let $B=B(x_0,R)$ and $B'=B(x_0,R+s)$, where $R,s>0$ are such that $\overline{B(x_0,R+2s)}\subset \Omega $, and let $C$ bound the functions $|A_{ab}|$ on $B(x_0,R+2s)$. For $\varepsilon <s$, the closed ball $\overline{B'}$ lies in $\Omega _\varepsilon $, the mollification $f_\varepsilon $ is smooth near $\overline{B'}$, and $\partial _ a\partial _ bf_\varepsilon =(\partial _ a\partial _ bf)_\varepsilon =A_{ab}*\rho _\varepsilon $ there by proposition 2.5.3(1). At points of $\overline{B'}$ this average involves only values of $A_{ab}$ on $B(x_0,R+2s)$, so $|\partial _ a\partial _ bf_\varepsilon |\leq C$ on $\overline{B'}$. By lemma 2.5.1(1), $f_\varepsilon \to f$ uniformly on $\overline{B'}$, so $|f_\varepsilon |\leq M$ on $\overline{B'}$ for some $M$ and all small $\varepsilon $. For $x\in \overline B$ and a unit vector $e$, Taylor's formula

in which $|\partial _ e^2f_\varepsilon |=|\sum _{a,b}e_ ae_ b\partial _ a\partial _ bf_\varepsilon |\leq dC$, gives $|\partial _ ef_\varepsilon (x)|\leq 2M/s+dCs/2$. Thus the gradients $\nabla f_\varepsilon $ are bounded on $\overline B$ uniformly in small $\varepsilon $. They are also Lipschitz on the convex set $\overline B$ with constant $dC$. Indeed,

where $D^2f_\varepsilon $ is the Hessian matrix, and a $d\times d$ matrix $H$ with entries bounded by $C$ satisfies $|Hv|\leq \sqrt d\, C\sum _ b|v_ b|\leq dC|v|$ for $v\in \mathbb R^ d$, by the Cauchy–Schwarz inequality.

Applying proposition 2.7.4 successively to the components of the maps $\nabla f_{1/k}$, $k>1/s$, we find $\varepsilon _ k\to 0$ such that $\nabla f_{\varepsilon _ k}$ converges uniformly on $\overline B$ to a continuous map $g=(g_1,\ldots ,g_ d)$. Since $\partial _ af_\varepsilon =(\partial _ af)_\varepsilon \to \partial _ af$ in the sense of distributions by proposition 2.5.3(1) and (2), the distribution $\partial _ af$ on $B$ is the continuous function $g_ a$. Letting $\varepsilon =\varepsilon _ k\to 0$ in the identity

we obtain the same identity for $f$ and $g$. Since $g$ is continuous, it follows that $f$ is differentiable on $B$ with $\nabla f=g$. Covering $\Omega $ by such balls $B$, we conclude that $f\in C^1(\Omega )$ and that its distributional first derivatives coincide with the continuous functions $\partial _ af$ on each such ball, hence on $\Omega $ by the last assertion of lemma 2.5.2. Consequently, $\partial _ af_\varepsilon =(\partial _ af)_\varepsilon \to \partial _ af$ uniformly on compact subsets of $\Omega $, by lemma 2.5.1(1).

Step 2: the Lipschitz bound. Let $V\subset \Omega $ be convex and open with $|A_{ab}|\leq C$ almost everywhere on $V$, and let $x,y\in V$. The segment $[x,y]$ is a compact subset of $V$; choose $\delta >0$ smaller than its distance to $\mathbb R^ d\setminus V$. For $\varepsilon <\delta $ and $z\in [x,y]$, the closed ball $\overline{B(z,\varepsilon )}$ lies in $V$, so $z\in \Omega _\varepsilon $ and $|\partial _ a\partial _ bf_\varepsilon (z)|=|(A_{ab}*\rho _\varepsilon )(z)|\leq C$. As in Step 1, $|\nabla f_\varepsilon (y)-\nabla f_\varepsilon (x)|\leq dC|y-x|$, and letting $\varepsilon \to 0$ gives $|\nabla f(y)-\nabla f(x)|\leq dC|y-x|$. Every point of $\Omega $ lies in a ball $V\Subset \Omega $, on which the $A_{ab}$ are bounded, so $\nabla f$ is locally Lipschitz and $f\in C^{1,1}_{\mathrm{loc}}(\Omega )$.

Step 3: the Taylor expansion. By theorem 2.8.3, almost every point of $\Omega $ is a common Lebesgue point of the functions $A_{ab}$. Let $x$ be such a point, let $Q=(-1,1)^ d$, and choose $r_0>0$ with $x+3r_0\overline Q\subset \Omega $. The functions $|A_{ab}|$ are bounded on the compact set $x+3r_0\overline Q$ by some constant $C$, so by Step 2, $\nabla f$ is Lipschitz with constant $\lambda =dC$ on the convex open set $x+3r_0Q$, which contains $x+2r_0\overline Q$. For $0<r\leq r_0$, define on $2Q$

We show that $G_ r\to 0$ uniformly on $\overline Q$ as $r\downarrow 0$; the expansion then follows by integrating $\nabla f$ along segments. The maps $G_ r$ vanish at $0$ and are Lipschitz with the common constant $\lambda +|A(x)|$, where $|A(x)|$ is the operator norm, and hence also with $\Lambda =\lambda +|A(x)|+1>0$.

To compute the derivatives of $G_ r$, let $\psi $ be a test function on $2Q$. The function $z\mapsto \psi ((z-x)/r)$ is a test function on $\Omega $, and the distributional derivative $\partial _ b$ of the continuous function $\partial _ af$ is $\partial _ b\partial _ af=\partial _ a\partial _ bf=A_{ab}$, since partial derivatives of distributions commute. Hence the change of variables $z=x+ry$ gives

Thus $\partial _{y_ b}(G_ r)_ a(y)=A_{ab}(x+ry)-A_{ab}(x)$ in the sense of distributions on $2Q$, and since $x$ is a Lebesgue point,

Lemma 2.8.4, applied on the cube $\overline Q$ of side length $2$ to each component of $G_ r$, gives

We deduce that $\sup _{\overline Q}|G_ r-(G_ r)_ Q|\to 0$. Otherwise there would be $\kappa >0$, radii $r_ k\to 0$ and points $y_ k\in \overline Q$ with $|G_{r_ k}(y_ k)-(G_{r_ k})_ Q|\geq \kappa $, and by the common Lipschitz bound, $|G_{r_ k}-(G_{r_ k})_ Q|\geq \kappa /2$ on $Q\cap B(y_ k,\kappa /(2\Lambda ))$. For $y\in \overline Q$ and $0<\sigma \leq 1$, the open cube $\prod _ lI_ l$, with $I_ l=(y_ l-\sigma ,y_ l)$ if $y_ l>0$ and $I_ l=(y_ l,y_ l+\sigma )$ if $y_ l\leq 0$, lies in $Q$ and within distance $\sqrt d\sigma $ of $y$. With $\sigma =\min \{ 1,\kappa /(2\Lambda \sqrt d)\} $, this shows that $Q\cap B(y_ k,\kappa /(2\Lambda ))$ has measure at least $\sigma ^ d$ for every $k$, so that $\int _ Q|G_{r_ k}-(G_{r_ k})_ Q|\, \mathrm dy\geq \sigma ^ d\kappa /2$, contradicting the convergence of these integrals to zero. Since $G_ r(0)=0$, also $|(G_ r)_ Q|=|G_ r(0)-(G_ r)_ Q|\to 0$, and hence $\sup _{\overline Q}|G_ r|\to 0$ as $r\downarrow 0$.

Now let $0<|h|\leq r_0$, and put $r=|h|$ and $y=h/|h|\in \overline Q$. For $0<t\leq 1$ we have $\nabla f(x+th)-\nabla f(x)=rt\, G_{rt}(y)+tA(x)h$, so, as $f$ is continuously differentiable,

The absolute value of the right side is at most $|h|^2\sup _{0<\varrho \leq |h|}\sup _{\overline Q}|G_\varrho |=o(|h|^2)$.

Step 4: the converse. Let $B\Subset \Omega $ be a ball, and let $f_ j$ be as in the statement, with $|\partial _ a\partial _ bf_ j|\leq C_ B$ on $B$ for all $j$. For every nonnegative test function $\chi $ supported in $B$, two integrations by parts give

so $|\langle \partial _ a\partial _ bf,\chi \rangle |\leq C_ B\int \chi \, \mathrm dx$. The distribution $\partial _ a\partial _ bf$ is real, because $f$ is real, and every point of $\Omega $ lies in such a ball $B$. By corollary 2.8.2(2), applied with $\nu $ equal to Lebesgue measure, $\partial _ a\partial _ bf$ is given by a Borel function $A_{ab}$ with $|A_{ab}|\leq C_ B$ almost everywhere on every such ball $B$. Each set of a compact exhaustion of $\Omega $ is covered by finitely many such balls, so countably many such balls $B_ k$ cover $\Omega $. Redefine $A_{ab}$ to be $0$ on the null set $\bigcup _ k\{ x\in B_ k:|A_{ab}(x)|>C_{B_ k}\} $. The new function still represents $\partial _ a\partial _ bf$, and it is bounded on every compact set, because a compact set is covered by finitely many of the balls $B_ k$. Finally, since $f_\varepsilon \to f$ in the sense of distributions (proposition 2.5.3(2)), the mollifications $f_ j=f_{\varepsilon _ j}$ with $\varepsilon _ j\to 0$ satisfy the hypothesis of the converse whenever their second derivatives are bounded on each ball $B\Subset \Omega $ uniformly in small $\varepsilon $. $\square$