The Maximal Function and Lebesgue Differentiation — Examples

Let $x>1$, and let $I_r=[x-r,x+r]$ with $r>0$. If $r\le x-1$, then [L1, given, algebra] $I_r\cap [0,1]=\mathrm{\varnothing }$ and the average is $0$. If $x-1<r<x$, then $I_r\cap [0,1]=[x-r,1]$, so the average is $\frac{r-x+1}{2r}=\frac{1}{2}-\frac{x-1}{2r},$ which increases with $r$. If $r\ge x$, then $I_r$ contains $[0,1]$, so the average is $1\mathrm{/}(2r)$, which decreases with $r$. The maximum is therefore attained at $r=x$, with value $1\mathrm{/}(2x)$.

If $0<x<1$, choose $r<\min \{x,1-x\}$; then $[x-r,x+r]\subseteq [0,1]$, so [L1, given, choose, algebra] the average equals $1$. Since no average of an indicator can exceed $1$, one has $Mf(x)=1$ on $(0,1)$.

The same calculation with the reflected interval shows that for $x<0$ the [step 1.1, algebra] maximum is attained at $r=1-x$ and equals $1\mathrm{/}(2(1-x))$. At the endpoints this gives $Mf(0)=Mf(1)=1\mathrm{/}2$.

Steps 1.1, 1.2, and 2.1 give the displayed formula.

Let $x\ge 1$ and take the centered interval $[0,2x]$, which has centre $x$ [L1, given, algebra] and contains $[0,1]$. Then [L1] gives $Mf(x)\ge \frac{1}{2x}{\int }_0^{1}1dy=\frac{1}{2x}\mathrm{.}$

Therefore [step 1.1, algebra] ${\int }_1^{R}Mf(x)dx\ge \frac{1}{2}{\int }_1^R\frac{dx}{x}=\frac{1}{2}\log R(R>1)\mathrm{.}$ Letting $R\to \mathrm{\infty }$ shows ${\int }_1^{\mathrm{\infty }}Mf(x)dx=+\mathrm{\infty }$. Hence $Mf\mathrm{\notin }{L}^1(\mathbb{R})$.

So the displayed claim is false.

One has [given, algebra] $\mathrm{\parallel }{f}_{\epsilon }{\mathrm{\parallel }}_1={\int }_0^{\epsilon }{\epsilon }^{-1}dx=1.$

If $x\ge \epsilon$, then the centered interval $[0,2x]$ contains the [L1, given, algebra] support of $f_{\epsilon }$, so $M{f}_{\epsilon }(x)\ge \frac{1}{2x}{\int }_0^{\epsilon }{\epsilon }^{-1}dy=\frac{1}{2x}\mathrm{.}$

If $0<t\le (2\epsilon {)}^{-1}$ and $\epsilon \le x<1\mathrm{/}(2t)$, then [step 1.2, algebra] step 1.2 gives $M{f}_{\epsilon }(x)>t$. Therefore $[\epsilon ,1\mathrm{/}(2t))\subseteq \{M{f}_{\epsilon }>t\},$ so $\lambda (\{M{f}_{\epsilon }>t\})\ge \frac{1}{2t}-\epsilon \mathrm{.}$

This explicit family matches the weak-type $t^{-1}$ scale from [L1] on a [L1, step 1.1, step 2.1] unit-$L^1$ example.

The set $\mathbb{Q}$ is countable, so [L2] gives [L2, given] ${\mathbf{1}}_{\mathbb{Q}}=0$ almost everywhere. Thus the $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ class of $f$ is the same as the class of the zero function.

For the zero representative and every $x\in \mathbb{R}$, [L1, step 1.1, algebra] $\frac{1}{\lambda (B(x,r))}{\int }_{B(x,r)}\mathrm{\mid }0-0\mathrm{\mid }d\lambda =0(r>0)\mathrm{.}$ Therefore every point is a Lebesgue point of the zero representative, so by [L1] the Lebesgue set of the class is all of $\mathbb{R}$.

For every $0<r\le 1$, [L2, given, algebra] $E\cap (-r,r)=[0,r),$ so [L2] gives $\frac{\lambda (E\cap (-r,r))}{\lambda ((-r,r))}=\frac{r}{2r}=\frac{1}{2}\mathrm{.}$

The ratio in step 1.1 is constant for all sufficiently small $r$, so [L1] [L1, step 1.1] gives $\mathrm{\Theta }(E,0)=1\mathrm{/}2$. In particular $0$ is not a density-one point of $E$.

Enumerate $\mathbb{Q}\cap (0,1)$ as $(q_k{)}_{k\ge 1}$ using [L1]. For each [L1, given, choose, construct] $k$, choose an open interval $I_k$ centred at $q_k$ of length below $2^{-k-2}$, and put $U:={\bigcup }_{k\ge 1}{I}_k\mathrm{.}$ The set $U$ is open and dense in $(0,1)$ because every nonempty open [L1, L2, L3, given, choose, construct, algebra] interval in $(0,1)$ contains a rational point $q_k$ and hence meets $I_k$. Also [L3] and [L2] give $\lambda (U)\le {\sum }_{k=1}^{\mathrm{\infty }}\lambda (I_k)<{\sum }_{k=1}^{\mathrm{\infty }}{2}^{-k-2}=\frac{1}{4}\mathrm{.}$

Define [L4, L5, step 1.1, algebra] $K:=[0,1]\setminus U\mathrm{.}$ Then $K$ is closed and bounded, hence compact by [L4]. By [L5] and step 1.1, $\lambda (K)\ge \lambda ([0,1])-\lambda (U)>1-\frac{1}{4}=\frac{3}{4},$ so $K$ has positive measure. Let $J$ be a nonempty open interval in $\mathbb{R}$. If $J$ is not [L4, L5, step 1.1, algebra] contained in $[0,1]$, then $J$ already contains a point outside $K\subseteq [0,1]$. If $J\subseteq (0,1)$, then step 1.1 gives $J\cap U\mathrm{\ne }\mathrm{\varnothing }$, so $J$ contains a point outside $K$. Thus every nonempty open interval contains a point outside $K$.

The compact set $K$ therefore has positive measure while missing part of [step 2.1] every interval.

By [L1], choose a density-one point $x\in E$. Then there is $r>0$ such that [L1, L3, given, choose] $\lambda (E\cap B(x,r))>\frac{3}{4}\lambda (B(x,r))\mathrm{.}$ Because [L3] makes the ball measure continuous in the radius, choose $0<\delta <r$ with $\lambda (B(x,r+\delta )\setminus B(x,r-\delta ))<\frac{1}{4}\lambda (B(x,r))\mathrm{.}$ If $\mathrm{\parallel }h{\mathrm{\parallel }}_2<\delta$, then $B(x,r-\delta )\subseteq B(x+h,r)\subseteq B(x,r+\delta ),$ so $B(x+h,r)\mathrm{△}B(x,r)\subseteq B(x,r+\delta )\setminus B(x,r-\delta )\mathrm{.}$ Therefore $\lambda (B(x+h,r)\mathrm{△}B(x,r))<\frac{1}{4}\lambda (B(x,r))\mathrm{.}$

Fix $\mathrm{\parallel }h{\mathrm{\parallel }}_2<\delta$. Translation invariance [L2] gives [L2, step 1.1, algebra] $\lambda ((E-h)\cap B(x,r))=\lambda (E\cap B(x+h,r))\ge \lambda (E\cap B(x,r))-\lambda (B(x+h,r)\mathrm{△}B(x,r))>\frac{1}{2}\lambda (B(x,r))\mathrm{.}$ Together with step 1.1, both $E\cap B(x,r)$ and $(E-h)\cap B(x,r)$ have measure greater than half of $\lambda (B(x,r))$, so they intersect. Choose $z\in E\cap (E-h)$. Then $z\in E$ and $z+h\in E$, so $h=(z+h)-z\in E-E$.

Every $h$ with $\mathrm{\parallel }h{\mathrm{\parallel }}_2<\delta$ lies in $E-E$, so $E-E$ contains the open [step 2.1] ball $B(0,\delta )$.

Applying the hypothesis to $I=[0,1]$ and [L2], one gets [L2, given, algebra] $\lambda (E\cap [0,1])=\frac{1}{2}\mathrm{.}$ So $E$ has positive measure.

By [L1], almost every point of $E$ is a density-one point of $E$. Choose [L1, step 1.1, given, choose, contradiction: density at x, discharge-contradiction] such a point $x\in E$. But the hypothesis applied to every interval $(x-r,x+r)$ gives $\frac{\lambda (E\cap (x-r,x+r))}{2r}=\frac{1}{2}(r>0),$ so the density of $E$ at $x$ is $1\mathrm{/}2$, not $1$. This contradiction refutes the claim.

The function $f$ takes only the values $0$ and $1$, so it is measurable and [given, algebra] bounded by $1$. Hence it is locally integrable on $\mathbb{R}$.

For $r_m:=2^{-2m}$, the set on which $f=1$ inside $(-r_m,r_m)$ is [L1, given, algebra] exactly the disjoint union ${\bigcup }_{k\ge m}(2^{-2k-1},2^{-2k}),$ whose total length is ${\sum }_{k\ge m}{2}^{-2k-1}=\frac{2^{-2m-1}}{1-2^{-2}}=\frac{2^{-2m+1}}{3}\mathrm{.}$ Therefore $A_{r_m}f(0)=\frac{1}{2r_m}{\int }_{-r_m}^{r_m}f(x)dx=\frac{1}{3}\mathrm{.}$

For $s_m:=2^{-2m-1}$, the set on which $f=1$ inside $(-s_m,s_m)$ is [L1, step 1.2, algebra] exactly ${\bigcup }_{k\ge m+1}(2^{-2k-1},2^{-2k}),$ whose total length is ${\sum }_{k\ge m+1}{2}^{-2k-1}=\frac{2^{-2m-1}}{3}\mathrm{.}$ Hence $A_{s_m}f(0)=\frac{1}{2s_m}{\int }_{-s_m}^{s_m}f(x)dx=\frac{1}{6}\mathrm{.}$

Steps 1.2 and 2.1 give two sequences of radii tending to $0$ along which [L2, step 1.2, step 2.1] $A_{r}f(0)$ tends to different values. So $A_{r}f(0)$ has no limit as $r\to 0^{+}$. This does not contradict [L2], because the exceptional set here is the singleton null set $\{0\}$.