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\section{The Concept of Measure}

The purpose of the Lebesgue integral is to enlarge the class of functions
for which $\int_a^bf(x)dx$ has a well-defined value.  The Riemann
integral works well for continuous functions and uniform limits: if
$f_n\in C[a,b]$ and $f_n\to f$ uniformly then $\int_a^bf_n(x)\,dx
\to \int_a^bf(x)\,dx$. It also works for bounded functions with a
finite number of discontinuities but has problems with functions like
Dirichlet's function,
  \[ f(x) = \left\{\begin{array}{ll} 1 & x\ \mbox{rational}\\ 0 & x\
  \mbox{irrational} \end{array}\right. \]
Let $\{r_1,r_2,\dots\} = \Q\cap[a,b]$ be an enumeration of the rational
numbers in $[a,b]$. Define
  \[ f_n(x) = \left\{\begin{array}{ll} 1 & x = r_1, r_2, \dots, r_n\\
  0 & \mbox{otherwise} \end{array}\right. \]
Then $f_n$ is integrable ($\int_a^bf_n(x)\,dx = 0$) and for each
$x\in[a,b]$, $f_n(x) \to f(x)$, and but $f$ is not Riemann-integrable. The
Lebesgue integral can handle functions and limits like this.

Lebesgue's own description of his method was in terms of a parable. A
merchant could add up the day's receipts by
  \begin{enumerate}
  \item adding them in the order received, say,
    \[ 5 + 10 + 1 + 1 + 25 + 5 + 10 + 50+ 25 + 10 = 142 \]
    or
  \item sorting the values first and adding them by type,
    \[ 2\times1 + 2\times5 + 3\times10 + 2\times25 + 1\times50 = 142 \]
  \end{enumerate}
The Riemann integral is like (1); it adds values of $f(x)$ in the order they
occur by partitioning the \emph{domain}, $\{x_0,x_1,\dots,x_n\}$, and
calculating the integral through approximations of the form 
  \[\sum_{j=1}^n f(x_j)(x_j-x_{j-1}) \]
The Lebesgue integral is like (2); it sorts the values of $f(x)$ by
partitioning the \emph{range}, $\{y_0,y_1,\dots,y_n\}$, and
calculating the integral through approximations of the form 
  \[\sum_{j=1}^n y_j|A_j| \]
where
  \[ A_j = \{ x\in[a,b]\ | \ y_{j-1} < f(x) \le y_j \} \]
and $|A_j|$ is the \emph{size} of $A_j$. The hard part of the theory is
arriving at a precise notion of $|A_j|$, and this leads to the concept
of \emph{Lebesgue measure}.

\subsection{Properties of Length}
Any method of measuring the size of sets of real numbers should be an
extension of the concept of the length of an interval.

\begin{definition} If $I = (a,b)$, $[a,b]$, $(a,b]$, or $[a,b)$, then the
\emph{length} of $I$ is $|I| = b - a$. If $a = -\infty$ or $b = \infty$,
then $|I| = \infty$. If $a = b$ or $I = \emptyset$, then $|I| = 0$.
\end{definition}

The following properties of length are easy to verify.

\emph{Additivity:} If $I = I_1\cup\dots\cup I_n$ (disjoint
intervals), then $|I| = \sum_{j=1}^n |I_j|$.

\emph{Subadditivity:} If $I \subset I_1\cup\dots\cup I_n$ (not
necessarily disjoint intervals), then $|I| \le \sum_{j=1}^n |I_j|$.

It is tempting to extend these properties to arbitrary unions, but this
cannot work: If $I = [0,1]$ and $I_c = [c,c]$, then
  \[ I = \bigcup_{c\in[0,1]} I_c \]
but $|I_c| = 0$ and $|I| = 1$. Countable unions, on the other hand,
are more tractable. In fact, the following holds.

\emph{$\sigma$-Additivity:} If $I = \bigcup_{j=1}^\infty I_j$ (disjoint
intervals), then $|I| = \sum_{j=1}^\infty |I_j|$.

\begin{proposition}\label{11:INTERVALS} If $I$ is any interval then
  \[ |I| = \inf \Big\{\sum_{j=1}^\infty |I_j|\ \Big|\ I \subset
  \bigcup_{j=1}^\infty I_j\ \mbox{(intervals)}\Big\} \]
\end{proposition}

\proof If $I_1 = I$ and $I_j = \emptyset$ for $j>1$, then $I =
\bigcup_{j=1}^\infty I_j$ and $|I| = \sum_{j=1}^\infty |I_j|$, so the
infimum is \emph{at most}~$|I|$.  It remains to show
  \[ I \subset \bigcup_{j=1}^\infty I_j \Rightarrow |I| \le
  \sum_{j=1}^\infty|I_j| \]
If the $I_j$ are not disjoint, we can shrink them to $I_j'$ so they are
and still maintain the same union. For example, if $I_j = [a,b]$, $I_k =
[c,d]$ with $a<c<b<d$, then $I_j' = [a,c]$ and $I_k' = [c,d]$. Therefore,
  \[ I \subset \bigcup_{j=1}^\infty I_j' = \bigcup_{j=1}^\infty I_j \]
and $|I_j'| \le |I_j|$. Now let $I_j'' = I\cap I_j'$. Then
  \[ I = \bigcup_{j=1}^\infty I_j'' \subset \bigcup_{j=1}^\infty I_j \]
and $|I_j''| \le |I_j'| \le |I_j|$. By $\sigma$-additivity,
  \[ |I| = \sum_{j=1}^\infty |I_j''| \le \sum_{j=1}^\infty |I_j| \]
\qed

\subsection{Measurable Sets}
It turns out to be impossible to extend the definition of length
$|I|$ from intervals $I$ to arbitrary subsets $A\subset \R$ and still
retain $\sigma$-additivity. It is possible to do this, however, for a
special, but still very large, class of sets called \emph{Borel sets}.

\begin{definition} A collection of sets $\FF$ is called a \emph{field of
sets} if it contains the empty set and is closed under the set operations
of union, intersection, and complement.
\end{definition}

Thus if $A,B$ are sets in a field $\FF$, then $A\cup B$, $A\cap B$, and
$A^c = \{x\in X\ |\ x \notin A\}$ are in $\FF$. Here $X$ is some fixed
universe that contains all the sets in~$\FF$ (and must be in $\FF$ itself).

\emph{Example:} Let $X = [a,b]$ and let $\FF$ be the collection of finite
unions of intervals contained in~$X$. Then $\FF$ is a field.

\begin{definition} A field of sets $\FF$ is a \emph{$\sigma$-field}
if countable unions of sets in $\FF$ are again in $\FF$,
  \[ A_j\in\FF\Rightarrow \bigcup_{j=1}^\infty A_j\in\FF \]
\end{definition}

The previous example is a field but clearly not a $\sigma$-field. Given
any field $\FF$, we can construct a $\sigma$-field containing $\FF$ by
adding all possible countable unions, their complements, intersections,
etc., to~$\FF$. The process is somewhat technical to carry out and we
shall skip the details.

\begin{definition} If $\FF$ is a field of sets, then the
\emph{$\sigma$-field generated by $\FF$}, denoted $\FF_\sigma$, is the
intersection of all $\sigma$-fields containing~$\FF$.  \end{definition}

It is straightforward to check that $\FF_\sigma$ is indeed
a $\sigma$-field. It is clearly the smallest $\sigma$-field
containing~$\FF$.

\begin{definition} Let $\FF$ be the field of finite unions of intervals
in $\R$. Then the sets in $\FF_\sigma$ are called the \emph{Borel sets}.
\end{definition}

There is no really satisfactory description of the Borel sets, but the
class is large enough to include any set describable in conventional
mathematical terms.  Our goal is to extend the length function from
intervals to Borel sets while preserving $\sigma$-additivity. The extended
length function is called \emph{Lebesgue measure}.

\subsection{Basic Properties of Measure}
We first introduce the general concept of a measure and derive some of
its properties.

\begin{definition} A \emph{measure} is a function,
$\mu:\FF\to\R\cup\{\infty\}$, defined on a $\sigma$-field of sets $\FF$,
called the \emph{measurable sets}, that satisfies
  \begin{enumerate}

    \item \emph{Non-negativity:} If $A$ is measurable then $0 \le \mu(A)
    \le \infty$, and $\mu(\emptyset) = 0$.

    \item \emph{$\sigma$-Additivity:} If $A_j$ is a sequence of disjoint
    measurable sets, then
      \[ \mu\Big(\bigcup_{j=1}^\infty A_j\Big) = \sum_{j=1}^\infty \mu(A_j) \]

  \end{enumerate}
\end{definition}

\begin{theorem}\label{11:MEASURE} For any measure $\mu$ the following
properties hold:
  \begin{enumerate}

    \item \emph{Monotonicity:} If $A$ and $B$ are measurable sets and
    $A\subset B$, then $\mu(A) \le \mu(B)$.

    \item \emph{Continuity from below:}
    If $A_1\subset A_2\subset A_3\subset\dots$ with $A_j$ measurable, then
      \[ \mu\Big(\bigcup_{j=1}^\infty A_j\Big) = \lim_{j\to\infty} \mu(A_j) \]

    \item \emph{Conditional continuity from above:} If $B_1\supset
    B_2\supset B_3\supset \dots$ with $B_j$ measurable and $\mu(B_j)$
    finite, then
      \[ \mu\Big(\bigcap_{j=1}^\infty B_j\Big) = \lim_{j\to\infty} \mu(B_j) \]

    \item \emph{$\sigma$-Subadditivity:} If $A_j$ is a sequence of
    measurable sets, then
      \[ \mu\Big(\bigcup_{j=1}^\infty A_j \Big) \le \sum_{j=1}^\infty
      \mu(A_j) \]

  \end{enumerate}
\end{theorem}

\proof
1. $B\setminus A$ is measurable and $B = A \cup (B\setminus A)$
is a disjoint union, so by $\sigma$-additivity, $\mu(B) = \mu(A) +
\mu(B\setminus A)$.  Non-negativity then implies $\mu(B) \ge \mu(A)$.

2. The sets $B_j = A_j\setminus A_{j-1}$ are measurable and
  \[ \bigcup_{j=1}^\infty A_j = A_1\cup B_2\cup B_3\dots \]
is a disjoint union, so by $\sigma$-additivity,
  \[ \mu(\Big(\bigcup_{j=1}^\infty A_j\Big) = \mu(A_1) + \mu(B_2) +
  \mu(B_3) + \dots \]
Also, for any $n$,
  \[ A_n = A_1\cup B_2\cup B_3\dots \cup B_n \]
so
  \[ \mu(A_n) = \mu(A_1) + \mu(B_2) + \mu(B_3) + \dots + \mu(B_n) \]
Therefore
  \begin{eqnarray*}
     \mu\Big(\bigcup_{j=1}^\infty A_j\Big) &=& \mu(A_1) + \mu(B_2) +
       \mu(B_3) + \dots\\
     &=& \lim_{n\to\infty} \mu(A_1) + \mu(B_2) + \mu(B_3) + \dots + \mu(B_n)\\
     &=& \lim_{n\to\infty} \mu(A_n)
  \end{eqnarray*}

3. The sets $B = \bigcap_{j=1}^\infty B_j$ and $A_j = B_j\setminus B_{j+1}$
are measurable and
  \[ B_1 = B \cup A_1 \cup A_2 \cup \dots \]
is a disjoint union. By $\sigma$-additivity,
  \[ \mu(B_1) = \mu(B) + \mu(A_1) + \mu(A_2) + \dots \]
Since $\mu(B)$ and $\mu(B_1)$ are finite we may write
  \[ \mu(B_1) - \mu(B) =  \mu(A_1) + \mu(A_2) + \dots \]
Also, for any $n$,
  \[ B_1 = B_n \cup A_1 \cup A_2 \cup \dots \cup A_{n-1} \]
so
  \[ \mu(B_1) - \mu(B_n) =  \mu(A_1) + \mu(A_2) + \dots + \mu(A_n) \]
Therefore,
  \begin{eqnarray*}
  \mu(B_1) - \mu(B) &=&  \mu(A_1) + \mu(A_2) + \dots\\
  &=& \lim_{n\to\infty} \mu(A_1) + \mu(A_2) + \dots + \mu(A_n)\\
  &=& \lim_{n\to\infty} \mu(B_1) - \mu(B_n)\\
  &=& \mu(B_1) - \lim_{n\to\infty}\mu(B_n)
  \end{eqnarray*}
from which we conclude that $\mu(B) = \lim_{n\to\infty}\mu(B_n)$.

4. Define $B_1 = A_1$ and for $j \ge 2$,
  \[ B_j = A_j\setminus (A_1\cup\dots\cup A_{j-1}) \]
Then the sets $B_j$ are measurable and disjoint, and $\mu(B_j) \le \mu(A_j)$ by
monotonicity. Therefore, since
  \[ \bigcup_{j=1}^\infty A_j = \bigcup_{j=1}^\infty B_j \]
we obtain by $\sigma$-additivity
  \[ \mu\Big(\bigcup_{j=1}^\infty A_j\Big) = \sum_{j=1}^\infty \mu(B_j)
  \le \sum_{j=1}^\infty\mu(A_j) \]
\qed

\subsection{A Formula for Lebesgue Measure}

Suppose $B \subset \bigcup_{j=1}^\infty I_j$ (intervals). If $B$ is to be
measurable with respect to a measure that extends the length function on
intervals, then $\sigma$-subadditivity  implies $|B| \le \sum_{j=1}^\infty
|I_j|$. In particular, $|B|$ is less than or equal to the infimum of
all such sums.  In Proposition~\ref{11:INTERVALS} we have seen that
the length of an \emph{interval} equals this infimum.  Lebesgue measure
declares $|B|$ to be equal to this infimum for any Borel set.

\begin{definition}
The \emph{Lebesgue measure} of a Borel set $B$ is defined to be
\[ |B| = \inf\Big\{\sum_{j=1}^\infty |I_j|\ \Big|\ B\subset
\bigcup_{j=1}^\infty I_j \ \mbox{(intervals)}\Big\} \]
\end{definition}

Lebesgue measure is clearly non-negative. The proof that it satisfies
$\sigma$-additivity is rather involved and we shall be content to simply
state this fact without proof.

\begin{theorem}
Lebesgue measure is a measure on the Borel sets.
\end{theorem}

The key observation of Lebesgue theory is that using countable covers
by intervals gives much more precise information about the size of sets
than using finite covers.

\emph{Examples:}

1) Let $B = [0,1]\cap\Q$ be the rational numbers in $[0,1]$. Since $B =
\{r_1,r_2,r_3,\dots\}$ is countable, it is a Borel set. Suppose $B$ is
contained in a finite union of intervals, $B \subset I_1\cup \dots\cup
I_n$.  Because $B$ is dense in $[0,1]$, it follows that the union must
contain the entire interval, $[0,1] \subset I_1\cup \dots\cup I_n$,
so by subadditivity, $1 \le |I_1| + \dots + |I_n|$.  Thus, the best
estimate of $|B|$ using \emph{finite} covers by intervals is $|B| \le 1$.

However, $B$ actually has Lebesgue measure zero and we can show this by
estimating $|B|$ using countable covers by intervals. Given $\epsilon>0$,
Let $I_j = (r_j-2^{-j}\epsilon, r_j+2^{-j}\epsilon)$. Then $B \subset
\bigcup_{j=1}^\infty I_j$ and by $\sigma$-subadditivity
  \[ |B| \le \sum_{j=1}^\infty |I_j| = \sum_{j=1}^\infty 2^{-j+1}\epsilon
  = 2\epsilon \]
Since $\epsilon$ is arbitrary, we must have that $|B| = 0$. A similar
argument can be used to prove:
  \begin{quote} \emph{Any countable set of real numbers has Lebesgue
  measure zero.} \end{quote}

2) Let $A = [0,1]\setminus\Q$ be the irrational numbers and $B =
[0,1]\cap\Q$ the rational numbers in $[0,1]$. Then $[0,1] = A\cup B$ is
the disjoint union of two Borel sets, so $1 = |A| + |B|$.  We know that
$|B| = 0$ by example 1), so $|A| = 1$. Thus, even though the rational
numbers are dense in $[0,1]$, they are negligible in size since they
have Lebesgue measure zero. All the ``weight'' of the interval $[0,1]$
is carried by the irrational numbers because they have the same Lebesgue
measure as the entire interval.

3) Any open set of real numbers, $A\subset\R$, is the countable
disjoint union of open intervals, $A = \bigcup_{j=1}^\infty I_j$,
see Theorem~\ref{01:OPEN}. Therefore $A$ is a Borel set and $|A| =
\sum_{j=1}^\infty |I_j|$. By taking complements, it follows that any
closed set of real numbers is a Borel set. The definition of Lebesgue
measure also implies that we can ``approximate'' any Borel set $B$
with an open set in the sense that given any $\epsilon>0$, there is an
countable open cover $A = \bigcup_{j=1}^\infty I_j\supset B$ such that
  \[ |A \setminus B| \le \epsilon \]
By making the same argument with the complement, $B^c \subset A_0 =
\bigcup_{j=1}^\infty I_j$, we see that there is also an approximation of
$B$ by a closed set $C = A_0^c\subset B$,
  \[ |B\setminus C| = |A_0\cap B| = |A_0\setminus B^c| < \epsilon \]

4) \emph{The Cantor Set.} 
Let $C_0 = [0,1]$, and let $C_1$ be the result of removing the segment
$(1/3,2/3)$ from $C_0$,
  \[ C_1 = [0,1/3] \cup [2/3,1] \]
Let $C_2$ be the result of removing the middle thirds of the intervals in
$C_1$,
  \[ C_2 = [0,1/9]\cup[2/9,3/9]\cup[6/9,7/9]\cup[8/9,1] \]
Continuing in this way we obtain a sequence of compact sets $C_n$
such that
  \begin{itemize}
    \item $C_1 \supset C_2 \supset C_3 \supset \dots$ \item $C_n$ is the
    union of $2^n$ intervals, each of length $1/3^n$.  In particular,
    $|C_n| = 2^n/3^n$.
  \end{itemize}
The intersection of these sets is called the \emph{Cantor set},
  \[ C = \bigcap_{n=1}^\infty C_n \]
It is clearly non-empty and bounded; it is also closed
(by Theorem~\ref{01:CLOSED}) and therefore compact (by
Theorem~\ref{01:COMPACTNESS}).

The Cantor set has no point in common with the intervals
  \[ \Big(\frac{3k+1}{3^m},\frac{3k+2}{3^m}\Big)\quad 0 \le k < 3^{m-1} - 1 \]
since these are the ones that are deleted. Any
$(a,b)\subset[0,1]$ contains such an interval (choose $m$ such that
$1/3^m < (b-a)/6$), so the Cantor set itself contains no intervals.

The Cantor set can also be described as the set of numbers in $[0,1]$
that have a representation in base-3 using only the numbers 0 and~2. An
argument similar to the one that shows $[0,1]$ is uncountable can be used
to prove the Cantor set is uncountable: If $C$ were countable, then $C =
\{c_1,c_2,\dots\}$. We construct a number $x\in C$ that is not on the
list as follows. The $i$-th digit of $x$ (base 3) is 0 if the $i$-th
digit of $c_i$ is 2, and the $i$-th digit of $x$ is 2 if the $i$-th
digit of $c_i$ is~0. The digits for $x$ involve only 0's and 2's, so $x\in
C$. Therefore, $x = c_i$ for some $i$. But, for any $i$, $x \ne c_i$
because their $i$-th digits are not equal.  This contradiction proves
that $C$ is uncountable.

Finally we remark that the Cantor set has Lebesgue measure zero. By
Theorem~\ref{11:MEASURE},
\[ |C| = \Big|\bigcap_{n=1}^\infty C_n\Big| = \lim_{n\to\infty}|C_n|
= \lim_{n\to\infty}\frac{2^n}{3^n} = 0 \]

\subsection{Other Examples of Measures}

\

1) \emph{Lebesgue Measure on $\R^n$.}
Lebesgue measure can be easily extended to $\R^n$ by replacing intervals
with rectangles, $R = I_1\times \cdots\times I_n$, where $I_j\subset\R$
is an interval. The measure (volume) of $R$ is
  \[ |R| = |I_1|\cdot|I_2|\cdots|I_n| \]
If $\FF$ denotes the field of finite unions of rectangles, then the sets
in the $\sigma$-field generated by $\FF$ are again called the Borel sets.
If $B$ is a Borel set, then the Lebesgue measure of $B$ is
  \[ |B| = \inf\Big\{ \sum_{j=1}^\infty |R_j| \ \Big|\ B \subset
  \bigcap_{j=1}^\infty R_j \ \mbox{(rectangles)} \Big\} \]
This extended Lebesgue measure satisfies $\sigma$-additivity and so is
a measure on the Borel sets in~$\R^n$.

2) \emph{Counting Measure.}
For any subset of natural numbers, $A\subset \N$, let $\mu(A)$ denote
the number of elements in $A$ ($\mu(A) = \infty$ if $A$ has an infinite
number of elements). The measure axioms are trivially satisfied, so
$\mu$ is a measure on the subsets of $\N$, called the \emph{counting
measure}. We shall see that the integration theory associated with this
measure is the theory of convergent series.

3) \emph{Probability Measure.} Let $X$ be a finite set $\{x_1,\dots,x_n\}$
and let $p_1,\dots,p_n$ be any non-negative real numbers or $\infty$. For
any subset $A\subset X$, we define
\[ \mu(A) = \sum_{x_j\in A} p_j \]
The measure axioms are again trivially verified. In fact, it is not hard
to show that \emph{any} measure on $X$ must have this form.
If the values $p_j$ also satisfy 
  \[ \mu(X) = \sum_{j=1}^\infty p_j = 1 \]
then we can interpret them as probabilities: $X$ is the sample space,
$p_j$ is the probability that $x_j$ occurs, and $\mu(A)$ is the probability
of an event $A\subset X$ (i.e., that an outcome $x$ is in $A$).
In fact, any space $X$ that has a measure defined on a $\sigma$-field
of subsets of $X$ satisfying $\mu(X) = 1$ can be interpreted as defining
probabilities on events ($=$ measurable sets) $A\subset X$.

\subsection{Hausdorff Measure and Dimension}
Lebesgue measure on $\R$ satisfies a \emph{scaling property}: if a Borel
set $B$ is dilated by a factor of $t$,
  \[ t B = \{tx\ |\ x \in B \} \]
then the Lebesgue measure of $t B$ dilates by the same factor,
  \[ |t B| = t |B| \]
If $B$ is a Borel set in $\R^n$, then the scaling factor is $t^n$,
  \[ |t B| = t^n |B| \]
So, for example, if we double the side of a cube, then its volume
increases by a factor of $2^3 = 8$. It is possible to construct measures
on $\R$ and on $\R^n$ that scale by a factor of $t^\alpha$ for any $\alpha
\ge 0$ by considering sums of the form $\sum_{j=1}^\infty |I_j|^\alpha$
where $B \subset\bigcup_{j=1}^\infty I_j$.  For technical reasons, the
infimum of such sums does not lead to a measure directly. The definition
must be modified as follows. A countable collection of intervals $\{I_j\}$
is called an \emph{$\epsilon$-cover} of a set $B$ if  $B \subset \bigcup
I_j$ and $|I_j|< \epsilon$, $\forall\,j$.

\begin{definition} For any set $B\subset\R$ define
  \[ \mu_\epsilon^\alpha(B) = \inf\Big\{ \sum_{j=1}^\infty
  |I_j|^\alpha\ \Big|\  \{I_j\}\ \mbox{is an $\epsilon$-cover of
  $B$}\Big\} \]
The Hausdorff measure of $B$ is defined to be
  \[ \mu^\alpha(B) = \lim_{\epsilon\to 0} \mu_\epsilon^\alpha(B)
   = \sup_{\epsilon> 0} \mu_\epsilon^\alpha(B) \]
\end{definition}

As $\epsilon$ decreases, the collection of $\epsilon$-covers of $B$
is reduced, and hence the infimum, $\mu_\epsilon^\alpha(B)$,
increases. Therefore, the limit of $\mu_\epsilon^\alpha(B)$
always exists (possibly $\infty$) and equals the supremum of
$\mu_\epsilon^\alpha(B)$.  The Hausdorff measure $\mu^\alpha$ is,
in fact, a measure on the Borel sets for any $\alpha\ge0$, although we
shall omit the proof. It is easy to see that $\mu^0$ is the counting
measure and $\mu^1$ is the usual Lebesgue measure.  More generally,
Hausdorff measure can be extended to $\R^n$ by replacing the intervals
$I_j$ with closed sets $R_j$ and $|I_j|$ with the diameter of $R_j$,
  \[ d(R_j) = \sup\{|x - y|\ |\ x,y\in R_j\} \]
It can be shown that $\mu^n$ is a multiple of Lebesgue measure on the
Borel sets in $\R^n$. In fact, $\mu^n(B) = c_n |B|$, where $c_n$ is the
volume of the $n$-dimensional ball of diameter~1.

The desired scaling property for Hausdorff measure is clear from
the definition,
  \[ \mu^\alpha(t B) = t^\alpha\mu^\alpha(B) \]
Since we can take the exponent of $t$ as an indication of the dimension of
the set, Hausdorff measures with varying $\alpha$ lead, via the following
lemma, to the concept of ``fractional dimensions.''

\begin{lemma}\label{11:HAUSDORFF}
  \[ \mu^\alpha(B) < \infty \Rightarrow \mu^\beta(B) = 0, \forall\,\beta >
  \alpha \]
  \[ \mu^\alpha(B) > 0 \Rightarrow \mu^\beta(B) = \infty, \forall\,\beta <
  \alpha \]
\end{lemma}

\proof Let $B \subset \bigcup_{j=1}^\infty I_j$ with $I_j\le \epsilon$. If
$\beta > \alpha$, then
  \[ \sum_{j=1}^\infty |I_j|^\beta = 
  \sum_{j=1}^\infty |I_j|^{\beta-\alpha} |I_j|^\alpha \le
  \epsilon^{\beta-\alpha}\sum_{j=1}^\infty |I_j|^\alpha \]
On the other hand, if $\beta < \alpha$, then similarly
  \[ \sum_{j=1}^\infty |I_j|^\beta \ge
  \epsilon^{\beta-\alpha}\sum_{j=1}^\infty |I_j|^\alpha \]
Therefore, if $\beta > \alpha$,
  \[ \mu_\epsilon^\beta(B) \le
  \epsilon^{\beta-\alpha}\mu_\epsilon^\alpha(B) \to 0
  \ \mbox{as}\ \epsilon\to 0\]
so $\mu^\beta(B) = 0$; while if $\beta < \alpha$,
  \[\mu_\epsilon^\beta(B) \ge
  \epsilon^{\beta-\alpha}\mu_\epsilon^\alpha(B) \to \infty
  \ \mbox{as}\ \epsilon\to 0\]
so $\mu^\beta(B) = \infty$. \qed

\begin{definition} The \emph{Hausdorff dimension} of a Borel set
$B$ is the unique value $\alpha$ such that $\mu^\beta(B) = \infty$,
$\forall\,\beta<\alpha$ and $\mu^\beta(B) = 0$, $\forall\,\beta>\alpha$.
A Borel set whose Hausdorff dimension is not an integer is called a
\emph{fractal}. \end{definition}

Note that if we can find a value of $\alpha$ for which $0 < \mu^\alpha(B) <
\infty$, then the Hausdorff dimension of $B$ is~$\alpha$.

\emph{Example:} Let $C$ be the Cantor set (see example~4 on p.~87).
We saw that the Lebesgue measure of $C$ is $|C|=0$. However, the Hausdorff
measure of $C$ is $\mu^\alpha(C) = 1$ where
  \[ \alpha = \log(2)/\log(3) \cong 0.6308 \]
This implies that Hausdorff dimension of $C$ is~$\alpha$ and shows
that $C$ is a fractal.  To prove $\mu^\alpha(C) = 1$ we first show that
$\mu^\alpha(C) \le 1$. Recall that
  \[ C = \bigcap_{n=1}^\infty C_n \subset C_n = \bigcup_{i=1}^{2^n}
  E_n^i \]
for certain disjoint intervals $E_n^i$ with $|E_n^i| = 1/3^n$.
Given $\epsilon>0$, choose $n$ such that $1/3^n < \epsilon$. Then
  \[ \mu_\epsilon^\alpha(C) \le \sum_{i=1}^{2^n} |E_n^i|^\alpha
  = \sum_{j=1}^{2^n} \big(\frac{1}{3^n}\big)^\alpha =
  2^n\big(\frac{1}{3^n}\big)^\alpha = \big(\frac{2}{3^\alpha}\big)^n = 1\]
since $3^\alpha = 2$. So $\mu^\alpha(C) = \lim_{\epsilon\to0}
\mu_\epsilon^\alpha(C) \le 1$.

It is much harder to prove the reverse inequality, $\mu^\alpha(C) \ge 1$.
We need the following inequality.

\begin{lemma}
Let $F$ be an interval with $|F| = 1/3^n$ and let $F_L$, $F_R$ be adjacent
intervals on the left and right of $F$, respectively, with $|F_L|$,
$|F_R| \le 1/3^n$.  Let $U = F_L\cup F\cup F_R$ and let $\alpha =
\log(2)/\log(3)$. Then
  \[ |F_L|^\alpha + |F_R|^\alpha \le |U|^\alpha \]
\end{lemma}

\proof Let $|F_L| = 1/3^n - x$ and $|F_R| = 1/3^n - y$ for some $0 \le
x,y\le 1/3^n$. Then $|U| = 1/3^{n-1} - x - y$ and we must prove that the
function
  \[ f(x,y) = \Big(\frac{1}{3^{n-1}} - x - y\Big)^\alpha
  - \Big(\frac{1}{3^n} - x\Big)^\alpha
  - \Big(\frac{1}{3^n} - y\Big)^\alpha \ge 0 \]
on the domain $0 \le x,y\le 1/3^n$. Note that
  \[ f(0,0) = \Big(\frac{1}{3^{n-1}}\Big)^\alpha -
  2\Big(\frac{1}{3^n}\Big)^\alpha = \frac{1}{2^{n-1}} - 2\frac{1}{2^n}
  = 0  \]
since $3^\alpha = 2$. Also,
  \[ f_x(x,y) = -\alpha\Big(\frac{1}{3^{n-1}} - x - y\Big)^{\alpha-1}
  + \alpha\Big(\frac{1}{3^n} - x\Big)^{\alpha-1} = 0\]
only for $y = 2/3^n$ which lies outside the domain.  Since
  \[ f_x(0,0) = -\alpha\Big(\frac{1}{3^{n-1}}\Big)^{\alpha-1}
  + \alpha\Big(\frac{1}{3^n}\Big)^{\alpha-1} =
\alpha\Big(\frac{3^n}{2^n}-\frac{3^{n-1}}{2^{n-1}}\Big) > 0 \]
it follows that $f_x(x,y) > 0$ on the domain. By symmetry, the same
holds for $f_y(x,y)$. Therefore, $f(0,0) = 0$ is the minimum of $f$
on the domain and hence $f(x,y) \ge 0$ as claimed. \qed

Now suppose $C \subset \bigcup_{j=1}^\infty I_j$ is an $\epsilon$-cover
of~$C$. Since $C$ is compact, $\exists\,N$ such that $C \subset
\bigcup_{j=1}^N I_j$. Each interval $E_n^i$ of $C_n$ is adjacent to exactly
one open interval $G_n^i$ of length $1/3^n$ (a middle third that was
removed) and another interval $E_n^{i\pm1}$ on the other side of $G_n^i$
such that
  \[ E_n^i \cup G_n^i\cup E_n^{i\pm1} = E_{n-1}^k \]
for some~$k$. By the lemma, we have for each $1\le j\le N$,
  \[ |E_n^i\cap I_j|^\alpha + |E_n^{i\pm1}\cap I_j|^\alpha \le
  |E_{n-1}^k\cap I_j|^\alpha \]
Repeating this construction recursively with the intervals $E_m^i$
in $C_m$ for $0 \le m \le n$, we find that
  \[  \sum_{i=1}^{2^n} |E_n^i\cap I_j|^\alpha \le \sum_{i=1}^{2^{n-1}}
  |E_{n-1}^i\cap I_j|^\alpha \le \dots \le |E_0\cap I_j|^\alpha =
  |I_j|^\alpha \]
Now $|E_n^i\cap I_j| \ne 0$ only if $E_n^i\cap I_j\ne\emptyset$ and
in this case $|E_n^i\cap I_j|  = |E_n^i|$ (i.e., $E_n^i\subset I_j$)
with at most two exceptions---those $E_n^i$ that overlap the endpoints
of the interval~$I_j$. Therefore,
  \[ \sum_{E_n^i\cap I_j\ne\emptyset} |E_n^i|^\alpha \ - \ \frac{2}{3^n}\
  \le \ |I_j|^\alpha \]
Since the intervals $I_j$, $1\le j\le N$, cover $C$, together they must
intersect all the sets $E_n^i$. So, adding the above inequalities for
$1\le j\le N$, gives
  \[ \sum_{i=1}^{2^n} |E_n^i|^\alpha - \frac{2N}{3^n} \ = \
  \sum_{j=1}^N \Big(\sum_{E_n^i\cap I_j\ne\emptyset} |E_n^i|^\alpha -
  \frac{2}{3^n}\Big) \ \le \ \sum_{j=1}^N |I_j|^\alpha \]
Since $\sum_{i=1}^{2^n} |E_n^i|^\alpha  = 1$ (see above), we find
  \[ 1 - \frac{2N}{3^n} \le \mu^\alpha_\epsilon(C) \le \mu^\alpha(C) \]
for all~$n$. Therefore, $1 \le \mu^\alpha(C)$, as claimed.

For sets that are self-symmetric, like the Cantor set, there are
often simple heuristic arguments to determine their dimension using
the scaling property. If we scale a set by a factor of $t$ and the
resulting set gives of $k$ copies of the original set, then the scaling
property implies $t^\alpha = k$ and so its dimension should be $\alpha =
\log(k)/\log(t)$.  For example, if we multiply the Cantor set by three, we
get two copies of the original Cantor set (one on the interval $[0,1]$,
and the other on the interval $[2,3]$). Therefore, its dimension is
$\alpha = \log(2)/\log(3)$. This argument is heuristic because we do
not know \emph{a priori} that there is a way to measure the size of
the Cantor set. That is, we need a function that satisfies the scaling
property $\mu(tC) = t^\alpha\mu(C)$ with $\mu(C) \ne 0$ so that we
can argue that $\mu(3C) = 3^\alpha\mu(C) = 2\mu(C) \Rightarrow
3^\alpha = 2$.

 \end{document}
