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For $f\in L^1({\mathbb{R}}^d)$ and a positive threshold $\lambda$, take maximal dyadic cubes with mean of $|f|$ exceeding $\lambda$. They are disjoint, have total volume at most $\Vert f{\Vert }_1/\lambda$, and their means are at most $2^d\lambda$. Replace $f$ inside each selected cube by its mean to obtain $g$; write $b_Q=(f-f_Q)1_Q$. Then $\Vert g{\Vert }_{\infty }\le 2^d\lambda$, $\Vert g{\Vert }_1\le \Vert f{\Vert }_1$, $\int b_Q=0$, and ${\sum }_Q\Vert b_Q{\Vert }_1\le 2\Vert f{\Vert }_1$. The outside bound follows from the Lebesgue differentiation theorem. This separates a bounded part from localized cancellation errors.

For locally integrable real $u$ with nonnegative distributional derivative $\Delta u$, its ball means increase with radius. To see this, mollify: for a smooth subharmonic function the derivative of its spherical mean is the integral of $\Delta u$ over the ball divided by the sphere area. The ball means inherit monotonicity; local $L^1$ convergence passes this to $u$. Their decreasing-radius limit exists, may be minus infinity at exceptional points, and equals $u$ almost everywhere by the Lebesgue differentiation theorem. It is upper semicontinuous as an infimum of continuous ball means and satisfies the mean-value inequality. Arbitrary changes of a representative on a null set can destroy its pointwise maximum principle.

For $p>n$, the Hölder continuous function representative $u^{\ast }$ of $u\in W^{1,p}({\mathbb{R}}^n)$ has a classical Fréchet derivative at almost every point, equal to its weak derivative. At a point $x$ where the $p$-mean oscillation of $Du$ tends to zero, apply the Morrey inequality on a cube to $u^{\ast }(y)-u^{\ast }(x)-Du(x)\cdot (y-x)$. On a cube of side comparable to $|y-x|$, its oscillation is at mostThe required points have full measure by the Lebesgue differentiation theorem.

The centered normal densitiesform an approximate identity: for every $f\in L^1(\mathbb{R})$, $f\ast g_t\to f$ at almost every point where the Lebesgue differentiation theorem applies. Their Fourier transforms are $e^{-t{\xi }^2/2}$ in the angular-frequency convention.

A point $x\in {\mathbb{R}}^n$ is a Lebesgue point of a locally Lebesgue integrable function $f$ whenThe Lebesgue differentiation theorem says that almost every point is a Lebesgue point.

An increasing right-continuous function $F:\mathbb{R}\to \mathbb{R}$ determines a measure byThe Radon-Nikodym theorem and Lebesgue differentiation theorem identify the density of the absolutely continuous part of this measure with $F^{\prime }$ almost everywhere.

Let $a_n=2^{-2^n}$ and, in angular coordinates near zero, take a bounded indicator alternating between one and zero on the rings $a_{n+1}<|t|\le a_n$. Since $a_{n+1}/a_n\to 0$, the mean over $[-a_n,a_n]$ approaches the value of the outermost ring. Its even and odd subsequences approach one and zero. Thus an integrable function can fail to have a limiting centered average at a specified point, although Lebesgue differentiation theorem ensures convergence almost everywhere.

A nontangential limit at $\xi \in \partial \mathbb{D}$ restricts approach to a fixed Stolz region ${\Gamma }_A(\xi )=\{z\in \mathbb{D}:|z-\xi |<A(1-|z|)\}$, where $A>1$. Radial approach is a special case. The Fatou theorem for bounded holomorphic functions says that every $f$ in H-infinity space has a finite boundary value $f^{\ast }(\xi )$ for almost every $\xi$, with convergence through every fixed Stolz region and $|f^{\ast }|\le \Vert f{\Vert }_{\infty }$ almost everywhere.

Here is a proof outline. Write $f=u+iv$ with bounded real harmonic functions $u,v$. Take a sequence $R_j\uparrow 1$. The bounded functions $u(R_j\xi )$ have a subsequence converging in the weak-star topology of $L^{\infty }(\partial \mathbb{D})$ by the Banach-Alaoglu theorem and weak-star metrizability of the dual ball; its $L^1$ predual is separable. Denote the limit by $U$. For $|z|<R_j$, the Poisson integral on the unit disk givesThe Poisson kernels converge uniformly for this fixed interior point and the boundary functions stay uniformly bounded, so the weak-star topology permits passage to the limit:Repeat for $v$, obtaining $V$. Thus $f$ is the Poisson integral of $F=U+iV\in L^{\infty }$.

The Lebesgue differentiation theorem makes almost every boundary point a Lebesgue point of $U$ and $V$. The lemma that a Poisson integral converges nontangentially at Lebesgue points then gives $f(z)\to F({\xi }_0)$ in every fixed Stolz region. To see why, put $\delta =1-|z|$ and let $d(\xi ,{\xi }_0)$ be the shorter angular distance. In ${\Gamma }_A({\xi }_0)$ the kernel satisfiesSplit the circle into a central arc of radius comparable to $\delta$ and annuli with radii $2^k\delta$. For a Lebesgue point, averages of $|F-F({\xi }_0)|$ on sufficiently small arcs are arbitrarily small. The annular contributions are bounded by these averages times a summable geometric sequence $C_A{2}^{-k}$. On the remaining fixed distant arc the kernel is $O(\delta )$, so its contribution vanishes. This proves the claimed nontangential limit. The same Lebesgue points work for every integer aperture $A\ge 2$, hence for all finite apertures.

These boundary values recover the interior by the Poisson integral, so they determine $f$ uniquely. They satisfy $\Vert f^{\ast }{\Vert }_{L^{\infty }}=\Vert f{\Vert }_{\infty }$: one inequality follows by taking limits of the bounded interior values, and the other from positivity and unit integral of the Poisson kernel. The dominated convergence theorem also gives $f(r\xi )\to f^{\ast }(\xi )$ in every finite $L^p$ norm. Its negative Fourier coefficients vanish, because the Taylor series of $f$ has only nonnegative powers. This explains the boundary realization of analytic Hardy spaces.

The exceptional set cannot generally be removed. For example, $f(z)=\exp (i\mathrm{Log}(1-z))$ is bounded and holomorphic on the unit disc, using the holomorphic logarithm on the right half-plane: $|f(z)|=e^{-\arg (1-z)}$ lies between $e^{-\pi /2}$ and $e^{\pi /2}$. At the boundary point one its radial values $e^{i\log (1-r)}$ oscillate without a limit. Nor does the theorem assert convergence along arbitrary tangential paths or a continuous boundary extension. Its strength is almost-everywhere control on every fixed approach cone.

Write $dm=dt/(2\pi )$ in angular coordinates on the unit circle. The centered Hardy-Littlewood maximal function iswhere $I_r(\omega )$ is the centered circular arc of angular radius $r$, with $r=\pi$ giving the whole circle. The Hardy-Littlewood maximal inequality in this one-dimensional setting is the weak-type estimateThe precise constant is not important for differentiation, but we prove this one explicitly.

Let $E=\{Mf>\alpha \}$ and choose a compact subset $F\subset E$. For each point of $F$, choose a centered arc whose average of $|f|$ exceeds $\alpha$. A finite subfamily covers $F$. Order these arcs by decreasing length and retain an arc precisely when it is disjoint from the previously retained arcs. Every rejected arc meets a retained arc of at least its length and lies in the latter's triple, interpreted on the circle and truncated to the whole circle if necessary. If the retained disjoint arcs are $I_1,\dots ,I_N$, thenThis is the interval version of the Wiener covering lemma. The superlevel set is measurable: the average at each fixed radius is continuous in the center, by translation continuity in $L^1$, so $E$ is open. Inner regularity of Lebesgue measure now passes the compact-set bound to $E$, proving the Hardy-Littlewood maximal inequality. The same finite-interval argument gives the line version $|\{Mf>\alpha \}|\le 3\Vert f{\Vert }_1/\alpha$ for integrable functions on the real line.

To prove differentiation, first recall that continuous functions are dense in $L^1(m)$ on the circle. Approximate an integrable function by simple functions; for each of their measurable sets, inner and outer regularity give a compact subset and an open superset with arbitrarily small measure difference. A continuous cutoff between these sets approximates the indicator in $L^1$. Combining finitely many cutoffs proves the claimed density.

For a representative of $f$ finite almost everywhere, define its local absolute oscillation byIf $h$ is continuous and $u=f-h$, the triangle inequality and continuity of $h$ giveThus the Hardy-Littlewood maximal inequality and the elementary integral bound for a superlevel set yield, for every $\epsilon >0$,Choose continuous $h$ with $\Vert f-h{\Vert }_1$ arbitrarily small. It follows that $\Delta f=0$ outside a null set, by using positive rational $\epsilon$. In particular,for almost every $\omega$. This proves the requested Lebesgue differentiation theorem, including the stronger absolute-oscillation formulation, rather than assuming pointwise convergence of the approximating continuous functions.

For failure at a specified point, use coordinates $-\pi <t\le \pi$ and put $a_n=2^{-2^n}$. Let $f$ be one on the annuli $a_{2j+1}<|t|\le a_{2j}$ for $j\ge 0$, and zero elsewhere, including at zero. This bounded measurable function is integrable. Since $a_{n+1}/a_n\to 0$, the average $A_n$ over the centered interval $[-a_n,a_n]$ satisfiesIndeed, the outermost annulus occupies all but the fraction $a_{n+1}/a_n$ of that interval, and has value one or zero according to parity. The even-radius subsequence tends to one and the odd-radius subsequence to zero, so no centered-average limit exists at $\omega =1$. The normalization of $m$ cancels from these averages. This is how oscillating annuli obstruct differentiation at a point without contradicting almost-everywhere differentiation.

For a real locally integrable function, the definitions are distributional:Thus the Laplacian is respectively a positive distribution, a negative distribution, or zero. A superharmonic function is the negative of a subharmonic function; being both is equivalent to being harmonic. For a $C^2$ function these conditions are $\Delta f\ge 0$, $\Delta f\le 0$, and $\Delta f=0$ pointwise.

An $L_{\mathrm{loc}}^1$ equivalence class does not itself specify point values. For the maximum principle use the canonical representative of a subharmonic distributionThese normalized integrals are ball averages. To justify this construction, first mollify $f$: its smooth local regularizations satisfy $\Delta f_{\epsilon }\ge 0$. For any smooth subharmonic function, the spherical mean $m_x(r)$ obeysThis follows by differentiating the spherical integral and applying the divergence theorem. Its ball mean also increases with radius: its derivative is $n/r$ times the spherical mean minus the ball mean. Local $L^1$ convergence of the regularizations passes the monotonicity of the ball means to $f$. The decreasing-radius limit therefore exists, possibly as minus infinity at an exceptional point, and equals $f$ almost everywhere by the Lebesgue differentiation theorem. It is upper semicontinuous as the local infimum of continuous ball averages and satisfiesFor a superharmonic function use the corresponding lower semicontinuous representative and reverse the inequality.

For a distributionally harmonic function, the smooth regularizations are harmonic and have constant spherical means. A radial mollifier ${\rho }_{\delta }$ of integral one therefore reproduces them by averaging these means. At fixed interior radius $\delta$, pass to the local $L^1$ limit to obtain $f=f\ast {\rho }_{\delta }$ almost everywhere on a smaller region. The right side is smooth and has zero distributional Laplacian, giving the smooth harmonic representative. This proves the relevant Weyl lemma. Changing a value at a single point can spoil a pointwise maximum principle, so the representative convention is essential.

The relevant version of Wiener covering lemma is the finite ball selection lemma: from a finite collection of open Euclidean balls one can select pairwise disjoint balls $B_j$ such that the original union is contained in ${\bigcup }_{j}3B_j$, where $3B_j$ has the same centre and three times the radius. To prove it, repeatedly retain a largest-radius remaining ball and discard every ball meeting it. If a discarded ball has radius $r\le R$ and meets a retained ball of radius $R$, every point of the former is at distance less than $r+(r+R)\le 3R$ from the latter's centre. This proves the containment and, by scaling Lebesgue measure,This is a covering result; no Fourier-algebra version of Wiener's lemma is involved.

We use it for a disjoint ball decomposition modulo null sets of $V$. The empty set needs only the empty family. Otherwise put $V_0=V$. For an open residual set $V_k$ of positive finite measure, inner regularity of Lebesgue measure supplies a compact $K_k\subseteq V_k$ with ${\lambda }_d(K_k)\ge {\lambda }_d(V_k)/2$. Cover $K_k$ by finitely many open balls whose closures lie in $V_k$, and use the Wiener covering lemma to select disjoint balls. Their total measure is at least ${\lambda }_d(V_k)/(2\cdot 3^d)$. Remove their closures to obtain the open residual set $V_{k+1}$. Ball boundaries are null sets, soAll retained balls, including those from different stages, are pairwise disjoint and contained in $V$. They form a countable family. The uncovered set is contained in ${\bigcap }_k{V}_k$ together with their countably many boundaries, and both have measure zero. Enumerating the retained balls gives

Next consider the uncentered maximal function of a finite measure. For every real $\alpha$, its strict superlevel set isIndeed, membership in the ball is exactly the strict condition $\Vert x-y\Vert <r$ in the supremum. This union is open, proving that $m_u$ is lower semicontinuous as an extended nonnegative function; it may take the value infinity.

For $\alpha >0$, take a compact subset $K$ of this superlevel set and choose a finite cover by balls satisfying the displayed density inequality. The Wiener covering lemma selects disjoint balls $B_j$ whose triples cover $K$. Since $\mu$ is a probability measure,Take the supremum over compact $K$ using inner regularity of Lebesgue measure. This works even if the open superlevel set was initially unbounded, and proves the uncentered maximal weak-type inequalityFor a finite positive measure of mass $M$, scaling gives $3^{d}M/\alpha$ instead.

One important use is the Lebesgue differentiation theorem. For $f\in L^1({\mathbb{R}}^d)$, approximate it in $L^1$ by a compactly supported continuous $g$, and set $h=|f-g|$. The limiting mean oscillation of $f$ at $x$ is at most $m_{h{\lambda }_d}(x)+h(x)$, since that of $g$ vanishes by continuity. The preceding weak-type estimate and the elementary integral bound on $\{h>t\}$ giveLet the approximation error tend to zero and then take countably many $t\downarrow 0$. The local averages of $f$ converge to $f(x)$ almost everywhere. Localizing extends this to locally integrable functions. Thus the maximal estimate turns norm approximation by continuous functions into almost-everywhere information about local averages.

Choose $x$ at which $Du(x)$ is finite and the $p$-mean version of the Lebesgue differentiation theorem holds:These conditions hold almost everywhere. On a cube about $x$, subtract the affine function and put $v(y)=u^{\ast }(y)-u^{\ast }(x)-Du(x)\cdot (y-x)$. Its weak derivative is $Dv=Du-Du(x)$ and $v(x)=0$. Approximate $u$ as in part (b) and subtract the same affine function from the approximants. Uniform convergence and convergence of their weak derivatives extend part (a)'s estimate to $v$, givingFor $h\ne 0$, choose $r=4|h|$ and $y=x+h$. It follows thatThis is precisely classical Fréchet derivative differentiability, not merely existence of coordinate derivatives. Therefore the differentiability almost everywhere of supercritical Sobolev functions gives

For an integrable function $f:{\mathbb{R}}^n\to \mathbb{C}$, the Lebesgue differentiation theorem states thatfor almost everywhere $x\in {\mathbb{R}}^n$. In particular,at every Lebesgue point of $f$.

Almost every $x\in \mathbb{R}$ is a Lebesgue point of $g$. At such an $x$, for $t>0$,For $t<0$, the same estimate over $[x+t,x]$ gives the identical limit. Thereforeat every Lebesgue point of $g$, so $G$ is differentiable $\lambda$-almost everywhere. This is the differentiation of an indefinite Lebesgue integral.

The Lebesgue differentiation theorem states that if $f\in L_{\mathrm{loc}}^1({\mathbb{R}}^n)$, thenfor Lebesgue almost every $x$. The Radon-Nikodym theorem states that if two sigma-finite measures satisfy $\nu \ll \mu$, then there is a nonnegative measurable function $h=d\nu /d\mu$, unique $\mu$-almost everywhere, such that $\nu (E)={\int }_{E}hd\mu$.

For any $t\in [0,1]$, take $z=0$ and let $B$ be a planar sector of angle $2\pi t$. Every disc centred at the origin meets this sector in the proportionso its measure density at $0$ is $t$.

Apply the differentiation theorem to the indicator function $1_A$. Its averages over balls are exactly ${\rho }_{\lambda ,A}(x)$, so the Lebesgue density theorem givesalmost everywhere. If $\lambda (A)=0$, every numerator vanishes, so the density is zero wherever its denominator is nonzero. Conversely, if the density vanishes almost everywhere, the displayed identity gives $1_A=0$ almost everywhere, hence $\lambda (A)=0$.

Finally suppose $\nu$ and $\lambda$ are mutually absolutely continuous measures. Write $d\nu =hd\lambda$. Then $0<h(x)<\infty$ almost everywhere by the positive Radon-Nikodym derivative result. At almost every $x$, the differentiation theorem applied to both $h$ and $h{1}_A$ givesThus the $\nu$-density also exists and belongs to $\{0,1\}$ at $\lambda$-almost every point.

Let $Df(x)$ denote the upper limit of the displayed mean oscillation as $r\downarrow 0$. For a continuous integrable approximation $g$, continuity gives $Dg(x)=0$. With $h=f-g$, the triangle inequality yieldsTherefore the maximal inequality and Chebyshev's inequality giveDensity makes the right side arbitrarily small. For each positive rational $\epsilon$ the exceptional set is null, proving $Df=0$ almost everywhere, the Lebesgue differentiation theorem. Values where the chosen representative is not finite already form a null set.

For the fundamental theorem define $F(x)={\int }_a^{x}f(t)dt$, with oriented integration for $x<a$. This function is absolutely continuous: for disjoint intervals the sum of $|F(b_j)-F(a_j)|$ is at most the integral of $|f|$ over their union, which is small when its measure is small by absolute continuity of the integral. At a Lebesgue point of $f$, for $h\ne 0$,Thus $F^{\prime }=f$ almost everywhere and $F(b)-F(a)={\int }_a^{b}f$. Differentiability at every point is not claimed for an arbitrary integrable function.

Let $c$ be a bounded real measurable coefficient on an open set. Nonnegativity of $c$ clearly makes its gradient form nonnegative. Conversely, if $c<0$ on a set of positive measure, choose an interior Lebesgue point $x_0$ where $c(x_0)<0$. For a nonconstant compactly supported smooth function $\psi$, set $v_{\epsilon }(x)=\epsilon \psi ((x-x_0)/\epsilon )$. Change variables and use the Lebesgue differentiation theorem to obtain ${\epsilon }^{-n}\int c|\nabla v_{\epsilon }{|}^2\to c(x_0)\int |\nabla \psi {|}^2<0$, a contradiction. Thus the linear operator represented by $a(u,v)=\int c\nabla u\cdot \nabla v$ is monotone exactly when $c\ge 0$ almost everywhere.

For a finite positive measure on Euclidean space, cover a compact subset of the maximal superlevel set by finitely many high-density balls. Wiener covering lemma selects disjoint balls whose triples cover it. Their total measure mass is at most the mass of the original measure. Inner regularity of Lebesgue measure yields the displayed bound. Applied to approximation errors, it proves the Lebesgue differentiation theorem.