\classheader{2012-10-23}

There will be no class on Thursday; there will instead be office hours in case anyone has questions before the exam, which is next week (October 30).

Let $\eucal{M}$ denote the collection of all measurable functions on a set $X$, and by $\eucal{M}^+$, the collection of all non-negative measurable functions $f:X\to[0,\infty]$.

Consider an operator $M:\eucal{M}\to\eucal{M}^+$, such as the maximal operator $M(f)=\sup_{B\ni x}\frac{1}{|B|}\int_B|f|$. Suppose that $\|Mf\|_\infty\leq\|f\|_\infty$ for all $f\in\eucal{M}$ and $M(f+g)\leq Mf+Mg$, and also suppose that, for some measure $\mu$ on $X$, there is some $c>0$ such that
\[\mu(\{x: Mf>t\})<\frac{c\|f\|_1}{t}.\]
(This last condition is called the ``weak 1-1 inequality''.)

\begin{claim}Then there are constants $c_p,c'$ such that $\|Mf\|_p\leq c_p\|f\|_p$ for all $p>1$ and every $f\in\eucal{M}$, and
\[\int\limits_{\{Mf>1\}}Mf(x)\,d\mu\leq c'\int_X|f|(1+\log^+|f|)\,d\mu.\]
\end{claim}

\begin{lemma}
We have that
\[\mu(\{x\mid Mf(x)>t\})\leq\frac{2c}{t}\int\limits_{|f|>\frac{t}{2}}|f|\,d\mu.\]
\end{lemma}
\begin{proof}
``Cut'' $f$ at $\pm\frac{t}{2}$, and call this function $f_1$; in other words take
\[f_1=\max\{\min\{f,\tfrac{t}{2}\},-\tfrac{t}{2}\}.\]
Let $f_2=f-f_1$. Then $|f_1|\leq\frac{t}{2}$ everywhere, so that by our assumptions about $M$, we have $\|Mf_1\|_\infty\leq\frac{t}{2}$, and hence $Mf_1\leq\frac{t}{2}$ a.e. We also have
\[Mf\leq Mf_1+Mf_2.\]
Let $A=\{x\mid Mf(x)>t\}$. Then for almost all $x\in A$,
\[t<Mf(x)\leq\frac{t}{2}+Mf_2(x)\]
so for almost all $x\in A$, $\frac{t}{2}\leq Mf_2(x)$. Now note that
\[\mu(A)\leq\mu(\{x\mid Mf_2(x)>\tfrac{t}{2}\})\stackrel{\text{weak 1-1}}{\leq}\;\frac{c\|f_2\|_1}{t/2}=\frac{2c}{t}\int_X|f_2|=\frac{2c}{t}\int\limits_{|f|>\frac{t}{2}}|f_2|\leq\frac{2c}{t}\int\limits_{|f|>\frac{t}{2}}|f|.\qedhere\]
\end{proof}
\begin{lemma}
For any $g\in\eucal{M}^+$,
\[\int_Xg\,d\mu=\int_0^\infty\mu(\{x\mid g(x)>t\})\,dt.\]
\end{lemma}
\begin{proof}
This is clear from our intuition about integrals.
\end{proof}
\begin{proof}[Proof of our claim.]
We have
\[\int_X|Mf|^p\,d\mu=\int_0^\infty\mu(\{x\mid Mf(x)>t^{1/p}\})\,dt.\]
Making the change of variables $y=t^{1/p}$, this is equal to
\[\int_0^\infty py^{p-1}\mu(\{x\mid Mf(x)>y\})\,dy.\]
From our first lemma, we have an inequality
\[\int_0^\infty py^{p-1}\mu(\{x\mid Mf(x)>y\})\,dy\leq\int_0^\infty py^{p-1}\frac{2c}{y}\int\limits_{|f|>\frac{y}{2}}|f|\,d\mu\,dy\]
and then letting $c_p=2cp$,
\[\int_0^\infty py^{p-1}\mu(\{x\mid Mf(x)>y\})\,dy\leq\int_0^\infty py^{p-1}\frac{2c}{y}\int\limits_{|f|>\frac{y}{2}}|f|\,d\mu\,dy=\iint\limits_{0<y<2|f|}c_py^{p-2}|f|\,d\mu\,dy\]
\[=c_p\int\limits_{|f|>0}|f|\left(\int_0^{2|f|}y^{p-2}\,dy\right)\,d\mu\leq c_p\int|f|^p\]
where, in the final expression above, the value of $c_p$ has changed by a constant factor.
\end{proof}

Recall that a strong Lebesgue point just means a Lebesgue point with respect to the strong basis.

\begin{theorem}
If $f\in L^p(\R^2)$ for $p>1$, then almost every point is a strong Lebesgue point of $f$.
\end{theorem}
\begin{lemma}
For a measurable function $f:[0,1]^2\to\R^+$ such that $f\in L^p$ for $p>1$, let $\varphi$ be the measure defined by $\varphi(A)=\int_A f\,d\mu$. Then
\[\int_{[0,1]^2}\overline{D}_S\varphi\,d\lambda\leq c_p\|f\|_p.\]
\end{lemma}
\begin{proof}[Proof of Lemma]
Fix some $y$. Take the maximal function in the $x$ coordinate:
\[m(x,y)=\sup_{a\,\leq\, x\,\leq\, b}\;\frac{1}{b-a}\int_a^b f(u,y)\,du\]
where $\sup_{a\,\leq\, x\,\leq\, b}$ means the supremum over all intervals $[a,b]\subseteq[0,1]$ containing $x$. We will see that $m\in L^p$. Define
\[E=\left\{(x,y)\;\middle\vert\; \lim_{\substack{c\,\leq\, y\,\leq\, d\\ (d-c)\to 0}}\;\frac{1}{d-c}\int_c^d m(x,v)\,dv=m(x,y)\right\}.\]
If $x\in[a,b]$, then
\[\frac{1}{(b-a)(d-c)}\int\limits_{[a,b]\times[c,d]}\!\!f=\frac{1}{d-c}\int_c^d\left(\frac{1}{b-a}\int_a^bf(u,v)\,du\right)dv\leq\frac{1}{d-c}\int_c^dm(x,v)\,dv.\]
If $(x,y)\in E$, then
\[\overline{D}_S\varphi(x,y)\leq m(x,y).\]
Therefore, if we show that almost every point in $[0,1]^2$ is in $E$, and additionally show that $\int_{[0,1]^2}m(x,y)$ is bounded by $c_p\|f\|_p$, we will have proved the lemma.

We know that for one-dimensional functions in $L^1$, almost every point is a Lebesgue point; thus, if we show that $m$ is $L^1([0,1]^2)$, we will have that almost every one-dimensional ``slice'' of $m$ is in $L^1([0,1])$, thereby implying that for almost every slice $\{x\}\times[0,1]$, almost every point of the slice is in $E$; this then implies that almost every point of $[0,1]^2$ is in $E$ (apply Fubini's theorem to the characteristic function of $E$).

Now we see that will be enough to show that $\int_{[0,1]^2}m\leq c_p\|f_p\|$. This follows from the claim we proved earlier today:
\[\int\limits_{[0,1]^2}m\leq\left(\int\limits_{[0,1]^2}m^p\right)^{1/p}=\left(\int_0^1\int_0^1m^p(x,y)\,dx\,dy\right)^{1/p}\leq\left(\int_0^1c_p\int_0^1f(x,y)^p\,dx\,dy\right)^{1/p}=c_p\|f\|_p.\qedhere\]
\end{proof}
\begin{proof}[Proof of Theorem]
WLOG, we can assume we are working in the unit square, because being a Lebesgue point is a local property; we are taking smaller and smaller balls around a point, so we can forget about what the function is doing far away.

Let $f\in L^p$. Choose a continuous $g$ such that $\|f-g\|_p<\epsilon^2$. Let $h=f-g$, and define $\varphi(S)=\int_S|h|\,d\lambda$. Define two sets
\[A=\{x:|h(x)|>\epsilon\},\quad B=\{x:\overline{D}_S\varphi(x)>\epsilon\}.\]
We will show that these sets are small.

Note that
\[\int\limits_{[0,1]^2}h\leq\|h\|_p<\epsilon^2\]
so that $\lambda(A)\leq\epsilon$. The lemma then implies that $\lambda(B)\leq c_p\epsilon$.

If $x\notin A\cup B$, then (letting $R$ be a sufficiently small rectangle)
\[\int_R|f(t)-f(x)|\,dt\leq\underbrace{\int_R|f(t)-g(t)|\,dt}_{\leq\,\epsilon|R|\text{ since }x\notin B}+\underbrace{\int_R|g(t)-g(x)|\,dt}_{\substack{\leq\,\epsilon|R|\text{ since }g\text{ is continuous}\\\text{and $R$ was chosen small enough}}}+\underbrace{\int_R|g(x)-f(x)|\,dx}_{\substack{=\,|g(x)-f(x)|\cdot|R|\\ \leq\,\epsilon|R|\text{ since }x\,\notin A}}.\]
Thus, for any given $\epsilon>0$, the measure of the set of points $x$ where
\[\limsup_{R\to x}\frac{1}{|R|}\int_R|f(t)-f(x)|\,dt>c\epsilon\]
is less than $c\epsilon$.
\end{proof}

We've shown that in any regular basis, we can differentiate any $L^1$ function (this fact, applied to characteristic functions, implies that almost every point of a set is a density point).

We've shown that in the strong basis, we can differentiate any $L^p$ function for $p>1$, but not necessarily $L^1$ functions (this fact, applied to characteristic functions, implies that almost every point of a set is a strong density point).

If, instead of axis-parallel rectangles, we take the basis consisting of all rectangles including rotated ones, then {\color{red}{\textbf{NOTHING IS TRUE}}}.

\begin{homework}[$\ast$]
There exists a compact $K\subseteq\R^2$ of positive measure such that, for each $x\in K$, there exists a line segment that meets $K$ at no other point. (Such a set is called a ``hedgehog''.)
\end{homework}


A hedgehog set demonstrates that the rotated rectangle basis is bad; for any $x\in K$, we can choose some other $y\in K$ arbitrarily close to it such that, taking the line from $x$ to $y$, we can find a line segment and then a very thin rectangle around that line segment where most of the rectangle is disjoint from the set $K$, making the density go to 0.