\classheader{2012-10-09}

Last time, we talked about measure spaces $(X,\cal{A},\mu)$.

\begin{definition}
We say that a measure space is complete if, for any $A\in\cal{A}$ such that $\mu(A)=0$, we have $B\in\cal{A}$ for every subset $B\subseteq A$. By monotonicity of measures, they will also be null.
\end{definition}

\begin{claim}
The $\sigma$-algebra generated by $\cal{A}$ and all subsets of null sets is the $\sigma$-algebra of all sets $S$ such that there are $A_1,A_2\in\cal{A}$ with $A_1\subseteq S\subseteq A_2$ and $\mu(A_2\setminus A_1)=0$. We can then define $\mu(S)=\mu(A_1)=\mu(A_2)$.
\end{claim}

\begin{definition}
An outer measure is a function $\mu:\cal{A}\to [0,\infty]$, where $\cal{A}$ is an arbitrary collection of sets, such that
\begin{itemize}
\item $\mu(\varnothing)=0$,
\item $\sigma$-\textbf{sub}-additivity: $\mu(\bigcup A_n)\leq\sum \mu(A_n)$ for any $A_1,A_2,\ldots\in\cal{A}$.
\end{itemize}
\end{definition}

We can create an outer measure as follows: for an arbitrary collection of sets $\cal{A}$, and an arbitrary function $\alpha:\cal{A}\to[0,\infty]$, we can define for any $A\in P(X)$
\[\phi_\alpha(A)\stackrel{\text{def}}{=} \inf\{\textstyle\sum \mu(A_n)\mid A\subseteq\bigcup A_n,\,A_n\in\cal{A}\}.\]
If there is no cover of $A$ by some $A_n\in\cal{A}$, then $\phi_\alpha(A)=\infty$.

We now want to create a measure from this outer measure:
\[\underset{\text{(arbitrary)}}{\alpha,\,\cal{A}}\quad\longrightarrow \quad\underset{\text{(outer measure)}}{\phi_\alpha,\, P(X)}\quad\longrightarrow\quad \underset{\text{(measure)}}{\mu,\,\cal{M}_\phi}\]
\begin{definition}
Given an outer measure $\phi$, we say that $A$ is $\phi$-measurable if, for any set $S$,
\[\phi(S)=\phi(S\cap A)+\phi(S\setminus A).\]
\end{definition}

We will see that the collection of $\phi$-measurable sets will form a complete measure space. First, note that we always have
\[\phi(S)\leq \phi(S\cap B)+\phi(S\setminus B).\]
If $\phi(A)=0$, and $B\subseteq A$, then $S\cap B\subseteq B\subseteq A$ implies that $\phi(S\cap B)=0$, and $S\setminus B\subseteq S$ implies that $\phi(S\setminus B)\leq\phi(S)$, so that
\[\phi(S)\leq 0+\phi(S\setminus B)\]
hence $\phi(S)=\phi(S\setminus B)$, hence $\phi(B)=0$.

\begin{center}
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Now we want to check that
\[\phi(S)=\phi(S\cap(A\cup B))+\phi(S\setminus (A\cup B)).\]
We have that
\[\phi(S)=\phi(S\cap A)+\phi(S\setminus A).\]
Therefore
\[\phi(S\setminus A)=\phi(\underbrace{(S\setminus A)\cap B}_{S\cap B})+\phi(\underbrace{(S\setminus A)\setminus B}_{S\setminus(A\cup B)})\]
and
\[\phi(S\cap A)+\phi(S\cap B)=\phi(S\cap (A\cup B))\]
\[\phi(S\cap (A\cup B))=\phi(\underbrace{S\cap (A\cup B)\cap A}_{S\cap A})+\phi(\underbrace{(S\cap (A\cup B))\setminus A}_{S\cap B}).\]
The rest of the argument you should study on your own.

If our initial arbitrary collection of sets $\cal{A}$ is such that $\varnothing\in \cal{A}$, and for any $A,B\in\cal{A}$, we have $A\cup B,A\cap B,A\setminus B\in\cal{A}$, and we further assume that $\alpha$ is an outer measure on $\cal{A}$ such that for any $A,B\in\cal{A}$ we have
\[\alpha(A)=\alpha(A\cap B)+\alpha(A\setminus B)\]
then the resulting $\sigma$-algebra $\cal{M}$ will contain $\cal{A}$, i.e. $\cal{A}\subseteq\cal{M}$.

Here is a special case. Consider ``bricks'', i.e. sets of the form
\[\prod_{i=1}^n[\hspace{0.02cm}a_i,b_i)\subseteq \R^n\]
and let $\cal{A}$ be the set of finite unions of such sets. Let $\alpha$ be the volume function. Then $\phi_\alpha$ is the outer Lebesgue measure, $\cal{M}$ consists of the Lebesgue measurable sets, and $\mu$ is the Lebesgue measure. We can do the same construction on the bricks for any additive function $\alpha$; the resulting measure will always contain at least the Borel sets.

\begin{definition}
We say that $\mu$ is a Borel measure on a $\sigma$-algebra $\cal{M}$ if
\begin{itemize}
\item The $\sigma$-algebra $\cal{M}\supseteq\{\text{Borel sets}\}$.
\item Some people require that $\mu$ is $\sigma$-finite, i.e. there exist $A_1,A_2,\ldots$ with $\mu(A_n)<\infty$ such that $\mu(X\setminus \bigcup A_n)=0$.
\item Some people require that $\mu(K)<\infty$ for any compact $K$ (in $\R^n$, this just says that bounded sets have finite measure).
\end{itemize}
\end{definition}

\begin{definition}
We say that a measure $\mu$ is regular when for any $A\in\cal{A}$,
\[\mu(A)=\inf\{\textstyle\sum \mu(G_n)\mid G_n\text{ are open, }A\subseteq \bigcup G_n\}=\inf\{\mu(G)\mid G\text{ open}, A\subseteq G\}\]
\end{definition}

For any $\epsilon>0$, we can choose $G$ such that $\mu(A)\leq \mu(G)\leq\mu(A)+\epsilon$. Letting $\epsilon\to 0$, we can see that for any $A$ there is a $G_\delta$ set containing $A$ of the same measure. By the same argument for the complement, we have an $F_\sigma\subseteq A$ of the same measure.

Now we'll start on a new topic.

Let $\lambda$ be the Lebesgue measure on $\R^n$, and $\cal{B}(\R^n)$ the Borel sets in $\R^n$.

\begin{definition}
For any $X\subseteq\R^n$, a differential basis for $X$ is a collection $\cal{D}\subseteq X\times \cal{B}(\R^n)$
such that
\begin{itemize}
\item For any $(x,A)\in\cal{D}$, we have $\lambda(A)>0$.
\item For any $x\in X$ and $r>0$, there is some $(x,A)\in\cal{D}$ such that $A\subseteq B(x,r)$, the ball of radius $r$ around $x$.
\end{itemize}
\end{definition}

Given an $x\in\R^n$ and $A\subseteq\R^n$, let $r(A)$ be the smallest radius such that $A\subseteq B(x,r(A))$.

We say that a differential basis $\cal{D}$ is regular when it satisfies the following property: for any $x\in X$, there exist a $\delta>0$ and $r_0>0$, depending on $x$, such that for any $(x,A)\in\cal{D}$ with $r(A)<r_0$, \[\lambda(A)\geq \delta\cdot \lambda(B(x,r(A))).\]
In other words, $\cal{D}$ is regular when, for any $x\in X$, there is a positive ratio $\delta$ for which all sufficiently small $(x,A)\in\cal{D}$ take up at least $\delta$ of their bounding ball.

\begin{examples}
$\text{}$
%The following are examples of regular differential bases.

\begin{itemize}
\item The symmetrical basis: all balls with center $x$. This is regular.
\item The standard basis: all cubes $Q$ that contain $x$.
\begin{center}
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We know that $\lambda(Q)=\ell^n$, and that $r(Q)\leq\sqrt{n}\ell$, so $\lambda(B)\leq c_n\ell^n$ for some constant $c_n$ depending only on $n$, which we can take as our $\delta$. This is regular.
\item The interval basis (also called the strong basis): all bricks that contain $x$. This is \textbf{not} regular.
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\end{itemize}
\end{examples}

\begin{definition}
Given a differential basis $\cal{D}$, we define
\[\overline{D}\mu(x)=\limsup_{r\to 0}\left\{\frac{\mu(A)}{\lambda(A)}\;\middle\vert\;(x,A)\in\cal{D},A\subseteq B(x,r)\right\}\]
to be the upper derivative of $\mu$ at $x$. The lower derivative $\underline{D}\mu(x)$ is the same except with $\liminf$, and the derivative is defined to be their common value if they are equal.
\end{definition}

\begin{example}
Let $f:\R\to\R$ be a differentiable function that is monotone increasing, and define $\mu([a,b))=f(b)-f(a)$. This defines a Borel measure $\mu$ on $\R$. We'll choose $\cal{D}=$ symmetrical basis, so
\[\lim_{r\to 0}\frac{\mu(B(x,r))}{\lambda(B(x,r))}=\lim_{r\to 0}\frac{f(x+r)-f(x-r)}{2r}.\]
This is the symmetric derivative. Note that $f$ did not have to be differentiable at $x$ for this to exist. In contrast, if we chose the standard basis or interval basis (which are the same in one dimension), then we'd get the limit
\[\lim_{\substack{y,z\to x\\ x\in[y,z]}}\frac{f(y)-f(x)}{y-z}\]
which exists if and only if $f$ is differentiable.
\end{example}

\begin{definition}
We define the maximal operator of a measure $\mu$ to be
\[M(x)=M_\mu(x)=\sup_{(x,A)\in\cal{D}}\frac{\mu(A)}{\lambda(A)}.\]
\end{definition}
Note that if $\mu=\mu_1+\mu_2$, then $M_\mu\leq M_{\mu_1}+M_{\mu_2}$. We only get an inequality because the set which maximizes $\mu_1+\mu_2$ need not maximize $\mu_1$ and $\mu_2$ separately.