\classheader{2012-10-11}

Today we'll be working in $\R^n$, and all our measures will be Borel measures that are $\sigma$-finite.

Let $\mu_1,\mu_2$ be measures. 

\begin{definition}
We say that $\mu_1$ is absolutely continuous with respect to $\mu_2$, and we write $\mu_1\ll\mu_2$, when $\mu_2(A)=0$ implies $\mu_1(A)=0$.
\end{definition}

\begin{definition}
We say that $\mu_1$ is singular with respect to $\mu_2$, and we write $\mu_1\perp \mu_2$, if there exist disjoint $A_1,A_2\subset\R^n$ such that $\mu_1(\R^n\setminus A_1)=0$ and $\mu(\R^n\setminus A_2)=0$.
\end{definition}

\begin{theorem}[Radon-Nykodim]
If $\mu_1\ll\mu_2$, then there is an $f$ such that $\mu_1(A)=\int_A f\,d\mu_2$ for all $A$. We say that $f$ is the Radon-Nykodim derivative of $\mu_1$ with respect to $\mu_2$, and we write that $f=\frac{d\mu_1}{d\mu_2}$.
\end{theorem}

\begin{theorem}
For any $\mu_1,\mu_2$, we can decompose $\mu_1=\alpha+\beta$ such that $\alpha\ll\mu_2$ and $\beta\perp\mu_2$.
\end{theorem}

\begin{homework}
Show that $\frac{d\mu_1}{d\mu_3}=\frac{d\mu_1}{d\mu_2}\cdot\frac{d\mu_2}{d\mu_3}$ for any $\mu_1\ll\mu_2\ll\mu_3$.
\end{homework}

Last class, we defined the maximal operator $M_\mu$ of a measure $\mu$, which is a function on $\R^n$, and noted that $M_{\mu_1+\mu_2}\leq M_{\mu_1}+M_{\mu_2}$.

\begin{theorem}
For any finite measure $\mu$, we have
\[\lambda\big(\{x\in\R^n\mid M_\mu(x)>t\}\big)\leq 3^n\cdot\frac{\mu(\R^n)}{t}\]
for any $t\geq 0$.
\end{theorem}

\begin{lemma}
If $B_1,\ldots,B_n$ are finitely many balls, there exists a subset $B_{i_1},\ldots,B_{i_k}$ of pairwise disjoint balls such that
\[\bigcup_{j}\,(3B_{i_j})\supseteq\bigcup_{i}B_i\]
where $3B$ means the ball with the same center as $B$ and three times the radius.
\end{lemma}
\begin{proof}[Proof of lemma]
We use a greedy algorithm. At each step, choose the largest ball that is disjoint from all earlier chosen ones. This will obviously terminate because there are only finitely many balls.

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Any ball $B_a$ not chosen will intersect some ball $B_b$ that was chosen, and $B_a$ must have radius less than or equal to $B_b$ (otherwise $B_b$ would not have been chosen), so that expanding $B_b$ by a factor of 3 will cover all of $B_a$.
\end{proof}
Note that this statement is false if we allow infinitely many balls; for example we could have nested balls around the same center whose radii go to infinity.

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\begin{homework}
Prove that the statement of the lemma is true even with infinitely many balls, as long as the radii are bounded, allowing the chosen subcollection of balls to also be infinite, and replacing 3 with some arbitrary constant.
\end{homework}

Is an arbitrary union of closed unit balls necessarily Borel? No. An easy construction is to choose a non-Borel subset of $\R$, and place balls which touch the line at exactly those points.

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The union of the balls isn't Borel; otherwise, its intersection with the Borel set $\R$ would be Borel.

\begin{homework}
Is an arbitrary union of closed unit balls necessarily Lebesgue measurable?
\end{homework}

\begin{proof}[Proof of theorem.]
We want to show, for any finite measure $\mu$ and $t>0$, that
\[\lambda\big(\{x\in\R^n\mid M_\mu(x)>t\}\big)\leq 3^n\cdot\frac{\mu(\R^n)}{t}.\]
It is enough to show that \[\lambda(K)\leq 3^n\cdot\frac{\mu(\R^n)}{t}\] for all compact $K\subseteq \{x\mid M_\mu(x)>t\}$, because Lebesgue measure is inner regular, i.e. for any Lebesgue measurable set $S$,
\[\lambda(S)=\sup_{\substack{\text{compact}\\K\subseteq S}}\lambda(K).\]
Now note that, by definition, $M_\mu(x)>t$ if and only if there is a ball $B_x$ around $x$ such that $\mu(B_x)>t\cdot\lambda(B_x)$. 

Choose a cover of $K$ by finitely many such balls $B_{x_1},\ldots,B_{x_k}$ (we can do this because $K$ is compact), and by the lemma, we can choose $x_{i_1},\ldots,x_{i_s}$ such that $B_{x_{i_1}},\ldots,B_{x_{i_s}}$ are disjoint and
\[\bigcup_{j=1}^s \,(3B_{x_{i_j}}\!)\supseteq \bigcup_{i=1}^k B_{x_i}.\]
Then
\[\lambda(K)\leq\lambda\left(\bigcup_{i=1}^kB_{x_i}\right)\leq\lambda\left(\bigcup_{j=1}^s \,(3B_{x_{i_j}}\!)\right)= 3^n\sum_{j=1}^s\lambda(B_{x_{i_j}})\]
\[\leq 3^n\sum_{j=1}^s\frac{\mu(B_{x_{i_j}})}{t}= 3^n\cdot\frac{\mu\left(\bigcup_{j=1}^sB_{x_{i_j}}\right)}{t}\leq 3^n\cdot\frac{\mu(\R^n)}{t}.\qedhere\]
\end{proof}

\begin{corollary}
For almost every $x\in\R^n$, the upper derivative of $\mu$ at $x$ is finite, i.e.  $\overline{D}\mu(x)<\infty$.
\end{corollary}
\begin{proof}
Our theorem implies that in symmetric basis, we have $ M_\mu(x)<\infty$ Lebesgue-a.e. Now note that, by definition, $\overline{D}\mu\leq M_\mu$.
\end{proof}

Note that, for a regular basis $\cal{D}$,
\[\frac{\mu(A)}{\lambda(A)}\leq\frac{\mu(B)}{\lambda(A)}=\underbrace{\frac{\mu(B)}{\lambda(B)}}_{<\;\infty}\cdot\underbrace{\frac{\lambda(B)}{\lambda(A)}}_{<\;\frac{1}{\delta}}\]
where $(x,A)\in\cal{D}$, with $A\subseteq B(x,r(A))=B$, and $\frac{1}{\delta}$ is from the regularity of $\cal{D}$.

\begin{theorem}
If $\mu$ is singular, then $D\mu(x)=0$ a.e.
\end{theorem}
\begin{proof}
By hypothesis, there exists a Lebesgue null set $N$ such that $\mu(\R^n\setminus N)=0$. We need to show that $D\mu(x)=0$ for a.e. $x\in\R^n\setminus N$.

If $N$ is closed, there is nothing to prove, because any $x\in\R^n\setminus N$ can be separated from $N$ by a sufficiently small open ball.

Now choose a compact $K\subseteq N$ such that $\mu(N\setminus K)<\epsilon^2$.

Let $\mu=\mu_1+\mu_2$, where $\mu_1(A)=\mu(A\cap K)$ and $\mu_2(A)=\mu(A\cap K^c)$. We have that $\overline{D}\mu\leq\overline{D}\mu_1+\overline{D}\mu_2$, and that $\overline{D}\mu_1$ is 0 a.e.

Note that
\[\lambda\big(\{x\in\R^n\mid \overline{D}\mu_2(x)>t\big)\leq\lambda\big(\{x\in\R^n\mid M_{\mu_2}(x)>t\}\big)\leq\ 3^n\cdot\frac{\mu_2(\R^n)}{t}\leq 3^n\cdot\frac{\epsilon^2}{t}.\]
Letting $t=\epsilon$, and using that $\overline{D}\mu_1$ is 0 a.e.,
\[\lambda\big(\{x\in\R^n\mid \overline{D}\mu(x)>t\big)\leq\lambda\big(\{x\in\R^n\mid M_{\mu_2}(x)>t\}\big)\leq\ 3^nt.\]
Now let $t\to 0$. This shows that $\lambda(\{x\in\R^n\mid \overline{D}\mu(x)>0\})=0$.
\end{proof}

What happens when $\mu\ll \lambda$? By Radon-Nykodim, we have that $\mu=\int f\,d\lambda$ for some $f$, and it is a theorem (which we are about to prove) that
\[f=\frac{d\mu}{d\lambda}=D\mu\;\;\text{a.e.}\]

Now let $f$ be a function such that $\int |f|\,d\lambda<\infty$, i.e. $f\in L^1$.

\begin{definition}
We say that $x\in\R^n$ is a Lebesgue point of $f$ if
\[\lim_{r\to 0}\frac{1}{\lambda(B(x,r))}\int_{B(x,r)}|f(t)-f(x)|\,dt =0.\]
\end{definition}
\begin{theorem}
For any $f\in L^1$, almost every $x\in\R^n$ is a Lebesgue point of $f$.
\end{theorem}
\begin{proof}
Define the measure
\[\mu_x(A)=\int_A|f(t)-f(x)|\,dt.\]
We need to show that $D{\mu_x}(x)=0$ for almost all $x$. We will assume the following lemma for the moment:
\begin{lemma}
For $f\in L^1$ and any $\epsilon>0$, there is a continuous, compactly supported $g$ such that $\int_{\R^n}|f-g|<\epsilon$.
\end{lemma}
Now fix an $\epsilon>0$, and choose $g$ from the lemma such that $\int|f-g|<\epsilon ^2$. Let $h=f-g$, so that $f=g+h$. We have
\[|f(t)-f(x)|\leq|g(t)-g(x)|+|h(t)-h(x)|\]
It is easy to see that
\[\lim_{r\to 0}\frac{1}{\lambda(B(x,r))}\int_{B(x,r)}|g(t)-g(x)|\,dt =0\]
for all $x$ (this essentially follows from $g$ being continuous), so now we just need to look at $h$. Define
\[\nu_x(A)=\int_A|h(t)-h(x)|\,dt.\]
Our work so far shows that that $\overline{D}\mu_x\leq\overline{D}\nu_x$.
Using the triangle inequality to produce a bound, and dividing by $\lambda(A)$,
\[\frac{\nu_x(A)}{\lambda(A)}\leq\frac{\int_A|h(t)|\,d\lambda}{\lambda(A)}+|h(x)|\cdot\frac{\lambda(A)}{\lambda(A)}.\]
Thus,
\begin{align*}\{x\in\R^n\mid \overline{D}{\mu_x} >2\epsilon\}&]\subseteq\{x\in\R^n\mid \overline{D}{\nu_x} >2\epsilon\}\\
&\subseteq\left\{x\in\R^n\;\middle\vert\; \substack{\text{\small there exist arbitrarily }\\\text{\small small $B\ni x$ such that } }\frac{\int_B|h(t)|\,dt}{\lambda(B)}>\epsilon\right\}\cup\{x\in\R^n\mid |h(x)|>\epsilon\}.
\end{align*}
Call the first set $S_1$ and the second set $S_2$. Note that the first set is precisely where the maximal operator of $h$ is greater than $\epsilon$. Thus, from the lemma, 
\[\lambda(S_1)\leq 3^n\cdot\frac{\epsilon^2}{\epsilon}=3^n\epsilon\]
and because $\int |f-g|=\int |h|<\epsilon^2$, we have
\[\lambda(S_2)\leq\frac{\int |h|}{\epsilon}\leq\epsilon.\]
Because $\epsilon>0$ is arbitrary, we can conclude that
\[\lambda(\{x\in\R^n\mid \overline{D}{\mu_x} >0\})= 0.\]
Thus, the set of non-Lebesgue points of $f$ is null. Now we prove our other claim. Given a measure $\mu\ll\lambda$, we can let $f=\frac{d\mu}{d\lambda}$, and then we see that as we take smaller and smaller balls $B\ni x$,
\[\left|\frac{\mu(B)}{\lambda(B)}-f(x)\right|=\left|\left(\frac{1}{\lambda(B)}\int_Bf(t)\,d\lambda\right) - f(x)\right|\]
\[=\left|\frac{1}{\lambda(B)}\int_B(f(t)-f(x))\right|\leq\frac{1}{\lambda(B)}\int_B|f(t)-f(x)|\to 0.\]
This demonstrates that we have $\frac{d\mu}{d\lambda}=\overline{D}\mu$ a.e.
\end{proof}

\newpage

\begin{proof}[Proof of lemma.]
$\text{}$

\begin{enumerate}
\item If $r$ is large enough then $\int_{B(0,r)^c}|f|<\epsilon$.
\item If $M$ is large enough then $\int_{\{x\,:\, |f(x)|>M\}}|f|<\epsilon$.
\end{enumerate}
Let $\bigcup A_m=A$, so that $A_m=A\cap\{x\in\R^n\mid m\epsilon\leq f< (m+1)\epsilon\}$. Then
\[A=B(0,r)\cap\{x\in\R^n\mid |f|\leq M\}\]
Choose compact $K_m\subseteq A_m$ such that
\[\int_{A_m\setminus K_m}|f|\leq \epsilon'\]
and also such that $\lambda(A_m\setminus K_m)\leq\epsilon'$. We can do this because $\lambda$ is inner regular, so that for each $A_m$ there is a sequence $K_{m,i}\subseteq A_m$ of compact sets such that $\lambda(A_m\setminus K_{m,i})<\frac{1}{i}$, and WLOG we can assume $K_{m,i}\subseteq K_{m,i+1}$, so letting $d\nu = f\,d\lambda$,
\[\lim_{n\to\infty}\nu(K_n)=\nu(A_n).\]



Define $g=m\epsilon$ on $K_n$. Extend it to a continuous function $g:B(0,r)\to[-M,M]$, with $g=0$ outside $B(0,r)$.
\end{proof}



