% Riemann-Lebesgue

% xxx "see all at once" from trig -- not just cot(theta) but
% cos(theta)/sin(theta) and 1/tan(theta) -- see them ALL.

% Flavors of continuity ... Faris' nested 4.
% Wik:  Lips => abs cts => unif cts => cts

% Assane's Lebesgue-integrability criterion.

% Difference between *measurable* function and *Lebesgue integrable* function.
% f(x)=x is the former but not the latter.
% * Mble:  preimage of all [a,b] is mble.  <==> preimage of all bo sets is
%   in X's sigma-alg.
% * Int'ble w/ std machine as usual.

% xxx borel set vs. borel function
% borel measurable fnc.  B => M ...

% note the harmonic series diverges, but the alternating one goes to ln(2).
% Perhaps use this to construct an example for which int|f| diverges, but int f
% converges.  Use this in turn as a mnemonic for int |f| cvg -> int f cvg (by
% ctrap).

% fourier nits:  L1 / L2 in out l1 l2 etc. etc.

% not-to-be-confused-with's (borel, etc.).

% point masses and counting measures somewhere.

%% ----------------------------------------------------------------

% Q7

% E1 #5

% weak* convergence ...

% base & subbase ... notes ch 25 p. 258

% Lp duality ... ch 19

% image integrals ...

%% ----------------------------------------------------------------
\section{Principles}
\label{sec:principles}

Royden (\cite{bib:royden}, section 3.6) attributes to Littlewood the following
three principles of analysis:
\begin{enumerate}
\item Every measurable set is nearly a finite union of intervals;
\item every measurable function is nearly continuous;
\item every convergent sequence of measurable functions is nearly
	uniformly convergent.
\end{enumerate}
The point here is that when we don't know what to do, we can
\emph{approximate}.

Here are some themes I see in the qual packets and qual review sessions:

\begin{itemize}
\item Know the main theorems with \bfidx{precise statements} of their
	\bfidx{hypotheses} and \bfidx{conclusions}.  On the qual we will not only
	need to state the name of the theorem we're using (e.g. when interchanging
	the order of iterated integrals) but also we will need to write little
	lemmas to \emph{prove} that the theorem applies.
\item \bfidx{Name-dropping}:  If I've written up a response which does not
	make use of a theorem named after someone, I've probably missed something:
	either I've done something incorrect, or I've done a step without
	sufficient justification.
\item \bfidx{Work avoidance}:  Before performing a computation, see if it's
	necessary.  For example (this is a geometry example), don't compute the
	projection of a vector onto a plane if a quick dot-product computation
	reveals that the vector is already parallel or perpendicular to the plane.
	Don't worry about computing finite change-of-variable scale factors in an
	integral if all you need to show is that the integral is infinite.  This
	save valuable time, and will also demonstrate competence.
\item \bfidx{Memorization}:  Know some standard integrals, power-series
	expansions, etc. so that they may be recognized on sight.  [xxx insert
	xrefs].
\item \bfidx{Association}:  A lower-level exam might tell you the name of a
	theorem, merely asking you to state it.  A qual requires you to
	free-associate among many possibilities.  Furthermore, some experimentation
	is often required in a qual response.  Throughout my qual-prep notes I look
	for opportunites to populate my free-association list.  For example,
	whenever I see a double integral, I should think Tonelli and Fubini
	(theorems \ref{thm:tonelli} and \ref{thm:fubini}); whenever I see an
	integral inequality I should think of H\"older (proposition
	\ref{prop:hoelder}); etc.
\end{itemize}

%% ----------------------------------------------------------------
\newpage
\section{Topology}

\begin{defn}
Let $X$ be a set.  Let $\mcT$ be a subset of $P(X)$.  Then $\mcT$ is a
\bfidx{topology} if it is closed under:
\begin{itemize}
\item arbitrary (even uncountable) unions and
\item finite intersections
\end{itemize}
and if
\begin{itemize}
\item $X \in \mcT$ and
\item $\emptyset \in \mcT$.
\end{itemize}
If $X$ has a topology $\mcT$, then $(X,\mcT)$ is called a \bfidx{topological
space}.
\end{defn}

\begin{defn}
Let $\Gamma$ be a collection of closed subsets of a topological space $X$.
Then $\Gamma$ has the \bfidx{finite intersection property} if every finite
subcollection of $\Gamma$ has non-empty intersection.
\end{defn}

There are multiple characterization of \bfidx{compactness}.  A topological
space $X$ is compact if:
\begin{itemize}
\item Every open cover has a finite subcover.
\item Every sequence has a convergent subsequence.
\item Every collection of closed sets with the finite intersection property has
a common point.
\item For every collection of sets with the finite intersection property,
there exists a point near (in the closure of) each set.
\item $X$ is the continuous image of another compact topological space.
\item it is a subset of a metric space, and is \bfidx{complete} and
	\bfidx{totally bounded}.  (The latter means is can be covered by finitely
	many sets of finite diameter.)
\end{itemize}

% cpct: notes 25 p. 260
% cts image of cpct is cpct

\begin{thm}[Tychonoff product theorem]
An\index{Tychonoff product theorem} arbitrary product of compact topological
spaces is compact.
\end{thm}

\begin{thm}
\label{thm:cts_img_cpct}
cts image of cpct is cpct
\end{thm}

\begin{thm}
\label{thm:cts_img_connd}
cts image of connd is connd
\end{thm}

\begin{defn}
A topological space is \bfidx{separable} if it contains a countable
dense subset.
\end{defn}

\begin{ex}
The rationals, which are countable, are dense in the reals (with
the usual topology).  Likewise $\R^n$ (finite-dimensional Euclidean space).
(Triv topo on any set since any singleton is dense.)  (Cite Wikipedia here \ldots)
\end{ex}

xxx need a non-example as well.  Use the reals with the discrete topology?
Infinite-dimensional Hilbert space --- separable iff it has a countable
orthonormal basis?

%% ----------------------------------------------------------------
\newpage
\section{Metric spaces}

\begin{defn}
\bfidx{metric space} \ldots.
\end{defn}

\begin{defn}
A \bfidx{Borel function} is a pointwise limit of a sequence of continuous
real-valued functions on a metrizable space.
\end{defn}

%% ----------------------------------------------------------------
\newpage
\section{Calculus}

\begin{thm}[Fundamental theorem of calculus]
\label{thm:first_fund}
If\index{fundamental theorem of calculus} $F$ is an antiderivative of $f$, then
$$ \int_a^b f(x) dx = F(b) - F(a).$$
\end{thm}

\begin{cor}
\label{cor:f_and_f_prime}
Writing $F'=f$ and applying the FTC to $f'$, we have
\begin{eqnarray*}
	\int_a^x f'(t)dt &=& F'(x)-F'(a) \\
	&=& f(x)-f(a).
\end{eqnarray*}
In particular, if $f(0)=0$, then
$$
	\int_0^x f'(t)dt = f(x).
$$
\end{cor}

\begin{thm}[Second fundamental theorem of calculus]
\label{thm:second_fund}
If\index{second fundamental theorem of calculus}
$$ F(x) = \int_a^x f(t) dt,$$
then
$$ F'(x) = f(x).$$
\end{thm}

\begin{cor}
$$
	\frac{d}{dx} \int_a^x f(t) dt = f(x).
$$
\end{cor}

The following is called the Extreme Value Theorem in \cite{bib:anton}; at the
graduate level, it becomes topological.

\begin{thm}[Extreme Value Theorem]
A\index{Extreme Value Theorem} continuous real-valued function from a closed
interval in $\R$ attains its maximum and minimum values.
\begin{proof}
This is an elementary statement of the fact (theorem \ref{thm:cts_img_cpct})
that the continuous image of a compact set is compact.
\end{proof}
\end{thm}

\begin{thm}[Intermediate Value Theorem]
Let\index{Intermediate Value Theorem} $f$ be a continuous function from the
closed interval $[a,c]$ to $\R$.  Let $x=f(a)$ and $z=f(c)$.  If $x<y<z$ or
$x>y>z$ then there is $a<b<c$ such that $y=f(b)$.
\begin{proof}
This is an elementary statement of the fact (theorem \ref{thm:cts_img_connd})
that the continuous image of a connected set is connected.
\end{proof}
\end{thm}

Rolle

Cor: MVT

xxx 1st, series $\sum a_n$ and tests.  Then, power series.

\begin{defn}
A \bfidx{power series} is a formal expression of the form
$$
	\sum_{n=0}^\infty a_n x^n.
$$
Coefficients are taken to be in $\R$ or $\C$.  If we think of a power series as
a \emph{function} to $\R$ or $\C$, respectively, we immediately ask for what
$x$'s the series converges to a real number.  It is shown in \cite{bib:cb}, for
the complexes, that there is a \bfidx{radius of convergence} (perhaps 0,
perhaps infinite) within which the series converges for all $x$ and outside of
which it converges for no $x$.  Behavior on the border circle is less
simple and is not addressed here.
\end{defn}

Criteria \cite{bib:rosenlicht} for finding the radius of convergence of
$\sum_{n=0}^\infty a_n x^n$
\begin{itemize}
\item When the limit exists: $$
	\lim \left| \frac{a_n}{a_{n+1}}\right|.
$$
\item With the conventions that $1/\infty = 0$ and $1/0 = \infty$: $$
	\frac{1}{\limsup \sqrt[n]{|a_n|}}.
$$
\end{itemize}

You can \bfidx{integrate} and \bfidx{differentiate} \bfidx{term by term}
anywhere inside the radius of convergence.

Three handy facts:
\begin{itemize}
\item[(1)] The \bfidx{harmonic series} diverges:
	$$
		\sum_{k=1}^\infty \frac{1}{k} = \infty
	$$
\item[(2)] The \bfidx{alternating harmonic series} converges.  Specifically:
	$$
		\sum_{k=1}^\infty \frac{(-1)^k}{k} = \ln(2).
	$$
\item[(3)] The \bfidx{square harmonic series} (for lack of a better name)
	converges.  Specifically:
	$$
		\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}.
	$$
\end{itemize}

\begin{rem}
Here are some power series (all Maclaurin series) which we should recognize on
sight:
\begin{itemize}
\item Exponential function:
$$
	1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots
	= \sum_{n=0}^\infty \frac{x^n}{n!}
	= e^x.
$$
Its radius of convergence is infinite.
\item Cosine:
$$
	1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots
	= \sum_{n=0}^\infty (-1)^{n}\frac{x^{2n}}{(2n)!}
	= \cos(x).
$$
Its radius of convergence is infinite.
\item Sine:
$$
	x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots
	= \sum_{n=0}^\infty (-1)^{n}\frac{x^{2n+1}}{(2n+1)!}
	= \sin(x).
$$
\item The \bfidx{geometric series}: $$
	1 + x + x^2 + x^3 + \ldots
	= \sum_{n=0}^\infty x^n
	= \frac{1}{1-x}.
$$
Its radius of convergence is 1.
\item Log:
$$
	x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots
	= \sum_{n=1}^\infty (-1)^{n-1} \frac{x^n}{n}
	= \ln(1+x).
$$
Its radius of convergence is 1.
\item Log:
$$
	x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots
	= \sum_{n=1}^\infty \frac{x^n}{n}
	= -\ln(1-x).
$$
Its radius of convergence is 1.
\end{itemize}
\end{rem}

\begin{thm}[Implicit function theorem]
\label{thm:implicit_function}
Let\index{implicit function theorem} $E$ be an open subset of $\R^{m+n}$, and
let $\vecf:E \to \R^n$ be a vector-valued function which is continuously
differentiable, i.e. of type $C^2$.  Let
$$(\veca, \vecb) = (a_1, \ldots, a_m, b_1, \ldots, b_n) \in E$$
be such that
$$ \vecf(\veca, \vecb) = 0.$$
Let $D$ be the $n \times n$ submatrix of the Jacobian of $f$ given by
$\Df_i/\Dx_j$, for $i=1, \ldots, n$ and $j=m+1, \ldots, n$, evaluated at
$\vecb$.  If $\det(D) \ne 0$, then there exists a neighborhood $U$ of $\veca$
and a unique $C^2$ function $\vecg:U \to \R^n$ such that
$$
\vecg(\veca)=\vecb, \qie \vecf(\veca,\vecg(\veca))=0,
$$
and for all $\veca' \in U$,
$$
\vecf(\veca',\vecg(\veca')) = 0.
$$
That is, we can \emph{\bfidx{solve for}} the $\vecb$ variables at and near
$\veca$.
\end{thm}

\begin{rem}
The point is that, given a system of equations (for this course, usually a
single equation), we have an easy criterion for when we can solve for some
variables in terms of the others.  Note however that the implicit function
theorem ensures existence and uniqueness; actually \emph{finding} the function
$\vecg$ is another matter.
\end{rem}

\begin{rem}
The theorem, as stated, has the to-be-solved-for variable(s) written last.  In
practice, this may not be the case.  E.g. given a function $f(v,w,x,y,z):\R^5
\to \R^2$, we might want to solve for, say, $v$ and $x$.  In that case, we
would need to check the submatrix formed by the first and third columns of the
Jacobian of $f$.  Furthermore, we might not know ahead of time which variables
to solve for, until we apply the implicit function theorem to various
submatrices of the Jacobian.
\end{rem}

\begin{ex}
\label{ex:impl_fcn_thm_S2}
Let
$$
	f:\R^3 \to \R: \colvecthree{x}{y}{z} \mapsto x^2+y^2+z^2-1.
$$
Here we have $m=2$ and $n=1$.
Then the kernel of $f$ is the sphere $S^2$.
Consider the north pole, $(0,0,1)$, written as
$$
	(\veca,\vecb) = (a_1, a_2, b_1) = (x, y, z) = (0,0,1).
$$
The Jacobian of $f$ is
$$
	\begin{pmatrix}
		2x & 2y & 2z
	\end{pmatrix}
$$
which evaluated at the north pole is
$$
	\begin{pmatrix}
		0 & 0 & 1.
	\end{pmatrix}
$$
Now, there is only one $1 \times 1$ submatrix of this which is non-zero,
namely, the last.  So, there is a neighborhood $U$ of $(0,0)$ and a unique
function $g:U \to \R$ such that $z = g(x,y)$.  Here, it's clear what this is:
take
$$
	z = g(x,y) = \sqrt{1-x^2-y^2}.
$$
\end{ex}

\begin{thm}[Inverse function theorem]
\label{thm:inverse_function}
Let\index{inverse function theorem} $E$ be an open subset of $\R^{m}$, and
let $\vecf:E \to \R^m$ be a vector-valued function which is continuously
differentiable, i.e. of type $C^2$.  If the Jacobian of $\vecf$ is nonsingular
at a point $\vecq$ of $E$, then there exist open neighborhoods $U$ of
$\vecq$ and $V$ of $\vecf(\vecq)$ such that $\vecf$ is a diffeomorphsim
from $U$ into $V$.
\end{thm}

\begin{rem}
The point is that, even if such a function $\vecf$ is wildly non-linear, to
check for invertibility at a point it suffices to make the much simpler check
of the invertibility of the linearization of $\vecf$.
\end{rem}


[xxx where to file] interchange limit and riemann integral

%% ----------------------------------------------------------------
\newpage
\section{Measure spaces and measures of sets}

\begin{defn}
Let $X$ be a set.  Let $\mcA$ be a subset of $P(X)$.  Then $\mcA$ is a
\bfidx{$\sigma$-algebra} if it is closed under:
\begin{itemize}
\item countable unions,
\item countable intersections, and
\item complements,
\end{itemize}
and if
\begin{itemize}
\item $X \in \mcM$ and
\item $\emptyset \in \mcM$.
\end{itemize}
\end{defn}
Note that a $\sigma$-algebra needs to tolerate \emph{more} intersections than a
topology on the same space (countably infinitely many, not just finitely many),
but needs to tolerate \emph{fewer} unions (countably infinitely many, not
uncountably many).

\begin{defn}
A \bfidx{$\sigma$-ring} is similar to a $\sigma$-algebra but need only be
closed under countable unions.  Also it need not contain all of $X$.
\end{defn}

\begin{defn} Let $(X,\mcT)$ be a topological space.  Let $\mcB$ be the smallest
$\sigma$-algebra containing $\mcT$.  Then $\mcB$ is the \bfidx{Borel
$\sigma$-algebra} for $(X,\mcT)$.  (We say $\mcB$ is the \bfidx{Borel
$\sigma$-algebra} \emph{generated by} the topology $\mcT$.)
\end{defn}

\begin{defn}
A \bfidx{Borel set} is an element of a Borel $\sigma$-algebra.  That is, a
Borel set is obtained from open sets by countable intersections, countable
unions, and complements (in any order).
\end{defn}

\begin{defn}
Let $X$ be a set with a $\sigma$-algebra $\mcM$.  A (set) \bfidx{measure} on
$(X,\mcM)$ is a function $\mu: \mcM \to [0,\infty]$ such that
\begin{itemize}
\item $\mu(\emptyset) = 0$, and
\item If $\{E_j\}_{j=1}^\infty$ is a sequence of \emph{disjoint} sets
in $\mcM$ then
$$
	\mu\left(\cup_{j=1}^\infty E_j\right) = \sum_{j=1}^\infty \mu(E_j).
$$
This is the \bfidx{countable additivity} property.
\end{itemize}
\end{defn}

\begin{rem}
If $\{E_j\}_{j=1}^\infty$ is a sequence of sets in $\mcM$
which is \emph{not} necessarily disjoint, then
$$
	\mu\left(\cup_{j=1}^\infty E_j\right) \le \sum_{j=1}^\infty \mu(E_j).
$$
This is the \bfidx{countable subadditivity} property.
\end{rem}

\begin{mnem}
Measure closed intervals $[a,b]$ in $\R$ by $\mu([a,b]) = b-a$.
Let $E_1 = E_2 = [0,1]$.  Then 
$$
	\mu(E_1 \cup E_2) = 1 \le 2 = 1+1 = \mu(E_1) + \mu(E_2).
$$
The subadditivity of measure of non-disjoint sets comes from the
\emph{overcounting} that happens when sets \emph{overlap}.
\end{mnem}

\begin{defn} If
\begin{itemize}
\item $X$ is a set, and

\item $\mathcal{M}$ is a $\sigma$-algebra on $X$, and

\item $\mu$ is a non-negative countably additive set function,
\end{itemize}

then $(X, \mcM, \mu)$ is called a \bfidx{measure space}.
\end{defn}

\begin{rem} Ideally we'd want to be able to ``measure'' the size of \emph{any}
subset of \emph{any} set $X$.  That is, instead of $\sigma$-algebras of $X$,
we'd simply use all of $P(X)$.  However, it can be shown (\cite{bib:folland},
chapter 1) that this is too much to hope for.  Thus we restrict ourselves to
measuring elements of something (much) smaller than all of $P(X)$.
\end{rem}

\begin{defn}
Let $X=\R$ with the Borel $\sigma$-algebra.  Then \bfidx{Lebesgue measure} of a
closed interval $[a,b]$ is $b-a$, written
$$
	\lambda([a,b]) = b-a.
$$
Likewise, on $\R^n$, Lebesgue measure of a cube is given by its volume.
\end{defn}

\begin{rem}
Obviously this only tells us how to measure intervals and cubes.  Other sets
are measured by approximating them from within and without.  (Making this
simple, intuitive statement precise occupies entire chapters of
\cite{bib:jones}, and also takes up significant space in the other cited
references.  The most concise treatment of the four authors is Rudin's.) More
precisely (\cite{bib:jones}, ch. 2), we measure open sets by approximating them
from within by combinations of cubes.  Then a set $E$ of $X$ is measurable iff
for all $\veps > 0$ there are closed and open sets $F$, $G$ respectively such
that $$F \subseteq E \subseteq G$$ and such that $\lambda(G \setminus F) <
\veps$.
Always remember that at its simplest --- details and distractions
notwithstanding --- on Euclidean spaces, \textbf{Lebesgue measure is volume}.
\end{rem}

\begin{rem}
For $\R^n$, the class of Borel sets of $X$ is contained in the set of
Lebesgue-measurable sets.  See \cite{bib:jones}, ch. 5, for an example.  That
is, for Euclidean space, \emph{all Borel sets are measurable}.  \emph{Not all
measurable sets are Borel.}
\end{rem}

% - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
\begin{defn}
Let $(X, \mcM, \mu)$ be a measure space.  Then $\mu$ is said to be a
\bfidx{finite measure} if $\mu(X) < \infty$.
\end{defn}
For example, take $X$ to be the unit interval, with Lebesgue measure.  Then
$\mu(X) = 1$.
For a non-example, take $X$ to be $\R$, with Lebesgue measure.

\begin{defn}
Let $(X, \mcM, \mu)$ be a measure space.  Then $\mu$ is said to be a
\bfidx{$\sigma$-finite measure} if there is an increasing sequence of sets $E_j
\upto X$ (in particular I mean $\cup_j E_j = X$) such that $\mu(E_j) < \infty$.
\end{defn}

\begin{ex}
For an example, take $X$ to be $\R$, with Lebesgue measure, and $E_j = [-j,
j]$.  For a non-example, take $X=\R$ with counting measure, i.e. $\mu(\{q\}) =
1$ for all singletons.  It takes uncountably many points $q$ to cover $\R$.
\end{ex}

%% ----------------------------------------------------------------
\newpage
\section{Integration (measures of functions)}

\begin{defn}
The \bfidx{extended real number system} is $\tilde\R = \R\cup\{-\infty,
+\infty\}$.
\end{defn}

\begin{defn}
\label{defn:lebesgue_measurable_function}
Let $(X, \mcM, \mu)$ be a measure space, and let $f:X \to \tilde\R$.  Note that while
$X$ has its $\sigma$-algebra $\mcM$, $\tilde\R$ has the standard Borel
$\sigma$-algebra $\mcB$. The function $f$ is said to be \bfidx{Lebesgue
measurable} if any of the following (which are equivalent) hold:
\begin{itemize}
\item For all $B$ in $\mcB$, $f^{-1}(B)$ is a measurable set in $X$.
	(That is, preimages of measurable subsets are measurable subsets.)
\item For all closed intervals $[a,b]$ of $\tilde\R$, $f^{-1}([a,b])$ is a
measurable set in $X$.
\item For all $a \in \R$, $\{x \in X: f(x)  >  a\}$ is a measurable set in $X$.
\item For all $a \in \R$, $\{x \in X: f(x) \ge a\}$ is a measurable set in $X$.
\item For all $a \in \R$, $\{x \in X: f(x)  <  a\}$ is a measurable set in $X$.
\item For all $a \in \R$, $\{x \in X: f(x) \le a\}$ is a measurable set in $X$.
\end{itemize}
\end{defn}

\begin{rem}
Lebesgue measurability is distinct from Lebesgue integrability (below).
\end{rem}

\begin{rem}
The set of measurable functions is closed under pointwise sums, differences,
and products, pointwise absolute value, pointwise max and min, pointwise
limits, but not composition.
\end{rem}

\begin{defn}
Let $(X,\mcM,\lambda)$ be as above.  A function $f:X \to \tilde\R$ is a
\bfidx{simple function} iff $f(X)$ is finite.  (Think of step functions, and
indicator functions of sets.)
\end{defn}

\begin{defn}
Let $(X,\mcM,\lambda)$ be as above, and let  $E:X \to \tilde\R$ be a simple
function.  Let $e_1, \ldots, e_n$ be the finite number of points in $E(X)$.
The \bfidx{Lebesgue integral} of $E$ is defined to be
$$
	\lambda(E) = \int_X E d\lambda = \sum_{j=1}^n e_j \lambda(f^{-1}(e_j))
$$
where the $\lambda$ on the left-hand side is a function measure, and the
$\lambda$ on the right-hand side is a set measure.
(Think of computing the areas of step functions.  This seems not too different
from the Riemann integral, until you consider, say, the simple functions which
is the indicator of the rationals on the unit invterval.)
\end{defn}

\begin{defn}
Let $(X,\mcM,\lambda)$ be as above and let  $f:X \to \tilde\R$.  The
\bfidx{Lebesgue integral} of $f$ is obtained by approximating $f$ by simple
functions:
$$
	\sup_{L \le f} \lambda(L) =
	\inf_{U \ge f} \lambda(U),
$$
with the sup and inf taken over \emph{all} simple functions $L$ and $U$,
then
$$
	\lambda(f) = \int_X f d\lambda
$$
is equal to this common value.  More practically,
Namely, let $L_k \upto f$ and $U_k \downto f$.
If
$$
	\lim_{k \to \infty} \lambda(L_k) =
	\lim_{k \to \infty} \lambda(U_k)
$$
then we write
$$
	\lambda(f) = 
	\lim_{k \to \infty} \lambda(L_k) =
	\lim_{k \to \infty} \lambda(U_k).
$$
I believe that this is the essential picture, modulo some fuss about details
and corner cases.  Note in particular, though, that we only attempt any
of this for functions which are Lebesgue measurable.
\end{defn}

Construction of the Lebesgue integral (Assane Lo calls this the \bfidx{standard
machine}):

\begin{itemize}
\item Characteristic functions
\item Simple functions
\item Non-negative measurable functions
\item Arbitrary measurable functions:  $\lambda(f) = \lambda(f_+) -
	\lambda(f_-)$.  This is defined if at most one of $\lambda(f_+)$ and
	$\lambda(f_-)$ are finite.  Otherwise, we have an $\infty-\infty$
	situation.
\end{itemize}

\begin{rem}
When the qual problems ask to show that something is \emph{not}
\plainidx{Lebesgue integrable}, typically one of two techniques
will suffice:
\begin{itemize}
\item Show that $\int f_+$ and $\int f_-$ are both infinite.
	(An example of this is below.)
\item For two-variable integrals, if you can show that
	$\int\int f(x,y)dx\,dy$ and
	$\int\int f(x,y)dy\,dx$ are different, then $f$ is not
	Lebesgue integrable.
\end{itemize}
\end{rem}

\begin{ex}
The function $f(x)=x$ is Lebesgue measurable but not Lebesgue integrable:
$f^{-1}([a,b])=[a,b]$ which is measurable in $\R$.  However,
$\int_{-\infty}^\infty f_+(x)dx$ and
$\int_{-\infty}^\infty f_-(x)dx$ are both infinite.
\end{ex}

\begin{rem}
Why isn't $$\int_{-\infty}^\infty x\,dx = 0?$$  Isn't this just
$$\lim_{a\to\infty} \int_{-a}^a x\,dx,$$
which is clearly zero?  The problem is in \emph{how} we go
to infinity; the integral balances on a knife edge.
If we take
$$\lim_{a\to\infty} \int_{-a}^{2a} x\,dx,$$
then we get something different.
Going to infinity symmetrically, via
$$\lim_{a\to\infty} \int_{-a}^a x\,dx,$$
is the \bfidx{Cauchy principal value} of the integral.
This becomes more clear when we consider
$$
	\sgn(x) = \begin{cases}
		1 & x > 0 \\
		0 & x = 0 \\
		-1 & x < 0.
	\end{cases}
$$
Clearly the principal value of the integral is zero,
via
$$
	\lim_{a\to\infty} \int_{-a}^a \sgn(x) dx,
$$
but it could be any $t$ via
$$
	\lim_{a\to\infty} \int_{-a}^{a+t} \sgn(x) dx.
$$
Specifically, this example shows that the Cauchy principal value is not
\bfidx{translation invariant}.
\end{rem}

\begin{ex}
The function $f(x) = \sin(x)/x$ is not Lebesgue integrable on $(0,\infty)$.  We
need $f_+$ and $f_-$ to each be integrable.  [xxx type up the details about
bounding $\sin(x)/x$ below by $\sin(x)/2n\pi$ on $[2n\pi, (2n+1)\pi]$ and
obtaining the (divergent) harmonic series.]
\end{ex}

%% ----------------------------------------------------------------
\newpage
\section{Theorems and inequalities for integration}

\textbf{Fact}:
$$
	\left| \mu(f) \right| \le \mu(|f|).
$$

\begin{mnem}
How to remember which way the signs go?  Use sine on
$[0,2\pi]$:
$$
	\left| \int_{0}^{2\pi} \sin(x) dx \right| = 0 \le
	\int_{0}^{2\pi} |\sin(x)|dx =
	2 \int_{0}^{\pi} \sin(x) dx =
	4.
$$
On the left, the up-hump and the down-hump cancel to zero.
On the right, the pair of up-humps gives a positive area.
\end{mnem}

\begin{mnem}
\label{mnem:toothpaste}
Another mnemonic:  I call this the \bfidx{toothpaste theorem}, since when you
squeeze it, more comes out:
$$
	\left|\int f(x)dx\right| \le \int \left|f(x)\right|dx.
$$
\end{mnem}

xxx need \bfidx{Chebyshev inequality}?

\bfidx{Fundamental norm identity}:
\begin{eqnarray*}
	\|u+v\|^2&=&\|u\|^2 + \langle u,v \rangle + \langle v,u \rangle + \|v\|^2;\\
	\|u-v\|^2&=&\|u\|^2 - \langle u,v \rangle - \langle v,u \rangle + \|v\|^2.
\end{eqnarray*}
Real case:
\begin{eqnarray*}
	\|u+v\|^2&=&\|u\|^2 + 2 \langle u,v \rangle \rangle + \|v\|^2;\\
	\|u-v\|^2&=&\|u\|^2 - 2 \langle u,v \rangle \rangle + \|v\|^2.
\end{eqnarray*}

\bfidxaux{Triangle inequality}{triangle inequality}:
\begin{eqnarray*}
	\|u \pm v\| \le \|u\| + \|v\|.
\end{eqnarray*}

\bfidxaux{Reverse triangle inequality}{reverse triangle inequality}:
\begin{eqnarray*}
	\|u \pm v\| \ge |\; \|u\| - \|v\| \;|.
\end{eqnarray*}
Together:
\begin{eqnarray*}
	|\; \|u\| - \|v\| \;| \le  \|u \pm v\| \le \|u\| + \|v\|.
\end{eqnarray*}

\bfidx{Parallelogram law}:
\begin{eqnarray*}
	\|u+v\|^2 + \|u-v\|^2 = 2\|u\|^2 + 2\|v\|^2.
\end{eqnarray*}

\bfidx{Cauchy-Schwartz inequality}:
\begin{eqnarray*}
	|\langle \vecu, \vecv\rangle|^2
		&\le& \langle \vecu, \vecu \rangle \langle \vecv, \vecv \rangle \\
	|\langle \vecu, \vecv\rangle|
		&\le& \|\vecu\|_2 \, \|\vecv\|_2 \\
\end{eqnarray*}
See also Cauchy-Schwarz for functions: proposition
\ref{prop:cauchy_schwarz_for_functions}.

An elementary but handy inequality:
$$
	ab \le a^2 + b^2.
$$
\begin{proof}
{$^{}$}
\begin{eqnarray*}
	0 &\le & (a-b)^2\\
	0 &\le & a^2 - 2ab + b^2 \\
	2ab &\le & a^2 + b^2 \\
	ab  &\le & \frac{a^2 + b^2}{2} \le a^2 + b^2.
\end{eqnarray*}
\end{proof}

\begin{defn}
A function is \bfidx{absolutely continuous} if it is continuous and is an
indefinite integral.
\end{defn}

\begin{thm}[Monotone convergence theorem]
If\index{monotone convergence theorem} $f_n \ge 0$ and $f \ge 0$,
and if $f_n \upto f$ pointwise,
then $\mu(f_n) \upto f$.
\end{thm}

xxx note rel'nship with countable additivity (1-11 notes).

\begin{thm}[Improved monotone convergence theorem]
If\index{improved monotone convergence theorem} $-\infty < f_1$,
and if $f_n \upto f$ pointwise,
then $\mu(f_n) \upto f$.
\end{thm}

\begin{thm}[Fatou's lemma]
Let\index{Fatou's lemma} $(X, \mcM, \mu)$ be a measure space.  [xxx do we
assume $\mu$ is Lebesgue measure $\lambda$?] Let $f_n$ be a sequence of
non-negative measurable functions.  If
$$
	f(x) = \liminf_{n \to \infty} f_n(x)
$$
then
$$
	\liminf_{n \to \infty} \int_X f_n d\mu \ge \int_X f d\mu
$$
i.e.
$$
	\liminf_{n \to \infty} \int_X f_n d\mu
	\ge
	\int_X \liminf_{n \to \infty} f_n d\mu.
$$
\end{thm}

\begin{mnem}
One can only \emph{lose mass} by moving the limit inside the integral.  One may
make this mnemonic embarrassingly more memorable by matching the \emph{fat} in
\emph{Fatou} with \emph{weight loss}.
\end{mnem}

\begin{mnem}
To remember which way the signs go, let
$$
	f_n = 1_{[n,n+1]}.
$$
These are little squares with height one, starting at $x=n$.  The integral of
each is of course one, but the pointwise limit of the $f_n$'s is $f=0$.  (The
unit mass gets shifted off to infinity and so is lost in the limit.)
\end{mnem}


\begin{thm}[Dominated convergence theorem (DCT), or Lebesgue dominated
convergence theorem (LDCT)]
Let $(X,\mcM, \mu)$ be a measure space.  Let\index{dominated convergence
theorem}\index{Lebesgue dominated convergence theorem} $f_n$ be a sequence of
measurable functions on $X$.  Let $g \ge 0$ be in $L^1(X)$.  Suppose the
pointwise limit of $f_n$ exists for all $x \in X$, say to $f$.  If
$$
	|f_n(x)| \le g(x)
$$
for all $x \in X$, then $f \in L^1(X)$ and
$$
	\int f d\mu = \lim \int f_n d\mu.
$$
\end{thm}

\begin{rem}
That is, if the $f_n$'s are dominated in absolute value by a function in $L^1$,
then we can move the limit through the integral.
\end{rem}

\begin{thm}[Tonelli's theorem]
\label{thm:tonelli}
Let\index{Tonelli's theorem} $(X, \mu)$ and $(Y, \nu)$ be $\sigma$-finite
measure spaces.  If $f$ is a non-negative real-valued function on $X \times Y$,
then
$$
	\int_X \left[ \int_Y f d\nu \right] d\mu =
	\int_{X \times Y} f d(\mu\times\nu) =
	\int_Y \left[ \int_X f d\mu \right] d\nu.
$$
\end{thm}

\begin{thm}[Fubini's theorem]
\label{thm:fubini}
Let\index{Fubini's theorem} $(X, \mu)$ and $(Y, \nu)$ be complete measure
spaces.  If $f \in L^1(X \times Y)$,
i.e. if
$$
	\int_{X \times Y} |f d(\mu \times \nu)| < +\infty,
$$
then 
$$
	\int_X \left[ \int_Y f d\nu \right] d\mu =
	\int_{X \times Y} f d(\mu\times\nu) =
	\int_Y \left[ \int_X f d\mu \right] d\nu.
$$
\end{thm}

When can we write
$$
	\frac{d}{dt} \int_a^b \phi(x,t) \;dx =
	\int_a^b \frac{\D \phi(x,t)}{\Dt} \;dx?
$$

\begin{thm}[Differentiation under the integral sign: \cite{bib:rudin} p. 237]
\label{thm:diff_int}
Let\index{differentiation under the integral sign} $\phi(x,t): [a,b] \times
[c,d] \to \R$.  If the following are true:
\begin{itemize}
\item for all $t$, $\phi$ is integrable along $x$, and
\item there is an $s \in [c,d]$ such that $\D\phi/\Dt$ is continuous
	for all $x \in [a,b]$ and for all $t$ in some open interval around $s$,
\end{itemize}
then
$$
	\left[\frac{d}{dt} \int_a^b \phi(x,t) \;dx\right]_{t=s}
	=
	\int_a^b \left[\frac{\D \phi(x,t)}{\Dt}\right]_{t=s}
	\;dx.
$$
\end{thm}

Here is an alternate characterization which Assane has been using.

\begin{thm}[Differentiation under the integral sign]
Let\index{differentiation under the integral sign} $\phi(x,t): [a,b] \times
[c,d] \to \R$.  If the following are true:
\begin{itemize}
\item $\D\phi/\Dt$ exists almost everywhere,
\item $|\D\phi/\Dt| < F$ for some integrable $F$, and
\item $\phi(x,t)$ is integrable,
\end{itemize}
then
$$
	\left[\frac{d}{dt} \int_a^b \phi(x,t) \;dx\right]_{t=s}
	=
	\int_a^b \left[\frac{\D \phi(x,t)}{\Dt}\right]_{t=s}
	\;dx.
$$
\end{thm}

[xxx elaborate on the following heuristic: when we have an integral problem
with two parameters, e.g. $e^{-st}$ somewhere in the integrand, go ahead and
differentiate it, then move the derivative inside and hope the view improves.]

%% ----------------------------------------------------------------
\newpage
\section{Weighted integrals}

\begin{defn}
The \bfidx{Riemann integral} is familiar from calculus.

The \bfidx{Riemann-Stieltjes integral} is a  Riemann integral with a weight
function.  Example:  Let $f(x) = x$ and $w(x) = 1/(1+x^2)$.  Then integrate $f$
with $w$ as weight.  This is just
$$
	\int_{-\infty}^{\infty} f(x) w(x) dx.
$$
Notation:  Let $W(x)$ be an antiderivative of $w(x)$, namely
$$
	W(x) = \begin{cases}
		\lambda((0, x]),  & x \ge 0 \\
		-\lambda((x, 0]), & x < 0. \\
	\end{cases}
$$
Then we write
$$
	\int_{-\infty}^{\infty} f(x) dW(x)
	= \int_{-\infty}^{\infty} f(x) w(x) dx.
$$
\end{defn}

\begin{mnem}
How to remember this?  The usual (uniformly weighted) Riemann integral
uses $w(x) = 1$, with $W(x) = x$.  Then we think of
$$
	dW(x) = \frac{dW}{dx} dx = w\, dx = dx.
$$
This is just a mnemonic for the special case of continuous $w(x)$, though ---
the Riemann-Stieltjes construction works for $W(x)$ that aren't even
continuous.  [xxx something --- use an example --- of $W$ with steps.  Use
these to obtain point masses.]
\end{mnem}

\begin{defn}
The \bfidx{Lebesgue-Stieltjes integral}:  direct analogy?
\end{defn}

xxx say how to construct it for any non-decreasing (LSC?) $W$.

%% ----------------------------------------------------------------
\newpage
\section{Functional analysis}

\begin{defn}
Let $1\le p \le \infty$.  Let $(X,\mcM,\mu)$ be a measure space.
For a real-valued function $f:X \to \R$, we write
$$
	\|f\|_p = \left[ \int_X \left| f \right|^p d\mu \right]^{1/p}.
$$
for the \bfidx{norm} of $f$.
\end{defn}

\begin{defn}
Let $1\le p \le \infty$.  Let $Y$ denote the set of infinite complex-valued
(or complex-valued) sequences.
For a sequence $\vecx = (x_1, x_2, x_3, \ldots) \in Y$, we write
$$
	\|\vecx\|_p = \left[ \sum_{k=1}^\infty |x_k|^p \right]^{1/p}.
$$
for the \bfidx{norm} of $f$.
\end{defn}

\begin{mnem}
This generalizes the familiar square-root-of-sum-of-squares computation in
the Euclidean distance formula.
\end{mnem}

\begin{defn}
Let $1\le p \le \infty$.  Let $(X,\mcM,\mu)$ be a measure space.  We write
\plainidx{$\mathcal{L}^p(X)$}
for the class of real-valued functions on $X$ such that
$$
	\|f\|_p < \infty.
$$
Define an equivalence relation whereby $f \sim g$ iff $f = g$ almost
everywhere.  We write \index{$L^p(X)$}
$$
	L^p(X) = \mcL^p(X) / \sim.
$$
\end{defn}

\begin{defn}
We write \plainidx{$\ell^p(\R)$} for the set of real-valued infinite sequences
$\vecx = (x_1, x_2, x_3, \ldots)$
such that
$$
	\|\vecx\|_p < \infty.
$$
\end{defn}

\begin{defn}
A \bfidx{Banach space} is a vector space which is normed and complete.
\end{defn}

\begin{mnem}
Q:  What's yellow, normed, and complete?  A:  A Bananach space.
\end{mnem}

\begin{defn}
A \bfidx{Hilbert space} is a Banach space which is also an inner product space.
\end{defn}

\begin{prop}[H\"older's inequality]
\label{prop:hoelder}
\index{H\"older's inequality}
\begin{eqnarray*}
	\|fg\|_1 \le \|f\|_p \, \|g\|_q
\end{eqnarray*}
with $1/p+1/q=1$.
\end{prop}

With $p=q=2$ we have the following:

\begin{cor}[Cauchy-Schwarz inequality]
\label{prop:cauchy_schwarz_for_functions}
\index{Cauchy-Schwarz inequality}
\begin{eqnarray*}
	\|fg\|_1 \le \|f\|_2 \, \|g\|_2.
\end{eqnarray*}
\end{cor}

\begin{rem}
\label{rem:hoelder_insertion}
To use H\"older, one needs a product of two functions.  Whenever we see an
integral inequality using $L^p$ norms, and if there is a single function $f$ on
the left-hand side, we can expect that we will need to \bfidx{insert} $g=1$, so
that $\|f\|_1$ becomes $\|f \cdot 1\|_1$.
\end{rem}

\begin{prop}
\label{prop:p_norm_1_norm}
$\|f\|_p^p = \|f^p\|_1$.
\begin{proof}
$$
	\|f\|_p^p = \left[\left( \int |f|^p \right)^{1/p} \right]^p
		= \int |f|^p = \int |f^p| = \|f^p\|_1.
$$
\end{proof}
\end{prop}

\begin{prop}[Minkowski inequality]
$$
	\|f+g\|_p \le \|f\|_p + \|g\|_p.
$$
\end{prop}

\begin{thm}[Projection theorem]
Let\index{Projection theorem} $H$ be a Hilbert space and let $M$ be a closed
subspace of $H$.  If $u \in H$ then there is a unique vector $v$ in $M$ closest
to $u$.  In particular, $u-v$ is perpendicular to $M$.
\end{thm}

\begin{defn}
A linear functional $\lambda$ is said to be \bfidx{bounded} if there is $M \in
\R$ such that for all $x \in H$, $\lambda(x) \le M \|x\|$.
\end{defn}

\begin{thm}[Riesz-Fr\'echet representation theorem]
In\index{Riesz-Fr\'echet representation theorem} a Hilbert space, all bounded
linear functionals $L$ come from a $u^*$.
\end{thm}

\begin{ex}
How could a linear functional failed to be bounded?  How ``linear'' could such
a functional be?  In finite dimensions, this isn't possible.  However, it is
possible in infinite dimensions it is.  Take a separable Hilbert space $H$ ---
it then has a countable basis $\{\vece_1, \vece_2, \ldots\}$.  Let
$\lambda(e_k) = k$, and extend linearly from the basis to all of $H$.  This is
clearly unbounded.
\end{ex}

\begin{rem}
This is a familiar fact from finite-dimensional vector spaces:  all linear
functionals may be represented by row vectors.
\end{rem}

\begin{defn}
Let $H$ be a Hilbert space.  Let $u_n$ be a sequence in $H$, and let $u \in H$.
Then $u_n$ converges to $u$ \bfidxtwo{weakly}{weak convergence} iff for all $v
\in H$,
$$
	\langle u_n, v \rangle \to \langle u, v \rangle.
$$
We say that $u_n$ converges to $u$ \bfidxtwo{strongly}{strong convergence}
if
$$
	\|u_n - u \| \to 0.
$$
\end{defn}

\begin{ex}
An infinite orthonormal sequence converges weakly to zero, but not strongly.
\end{ex}

\begin{defn}
\bfidx{strong topology} \ldots.
\end{defn}

\begin{defn}
Let $E$ be a Banach space.  The \bfidx{weak topology} on $E$ is the coarsest
topology such that every element of $E^*$ is continuous.
\end{defn}

\begin{defn}
Let $E$ be a Banach space.  The \bfidx{weak* topology} on $E*$ is the coarsest
topology such that every element of $E^{**}$ is continuous.
\end{defn}

\begin{defn}
A sequence of functions $f_n$ is said to \bfidx{converge in measure}
if for all $\veps > 0$ there is $N$ such that for all $n \ge N$,
$$
	\mu\{x : |f(x)-f_n(x)| > \veps \} < \veps.
$$
\end{defn}

%% ----------------------------------------------------------------
\newpage
\section{Fourier analysis}

xxx needs elaboration

\begin{defn}
Write
$$
	\langle f, g \rangle = \int \overline{f} g.
$$
\end{defn}

\begin{rem}
Note that
$$
	\|f\|_2^2 = \langle f, f \rangle.
$$
\end{rem}

\begin{defn}
For $f(x) \in L^2$ (usually on $[0,1]$ or $\R$), and for basis functions
$b_k(x)$,
$$
	\hat{f}(k) = \langle f, b_k \rangle.
$$
\end{defn}

\begin{rem}
For $f(x) \in L^2([0,1])$, basis functions $b_k(x)$ may be taken to be
$e^{i2\pi k x}$ for $k \in \Z$.  For $f(x) \in L^2(\R)$, the $k$'s may take on
any real value.
\end{rem}

\begin{defn}
We write
$$
	f(x) * g(x) = \int f(x-y) g(y) dy.
$$
This is the \bfidx{convolution} of $f$ and $g$.
\end{defn}

\begin{prop}[Convolution for Fourier integrals]
Let\index{convolution} $\mathcal{F}$ denote the Fourier transform.  Then
$$
	\mathcal{F}(g * h) = \mathcal{F}(f) \cdot \mathcal{F}(g).
$$
That is,
$$
	\widehat{g*h} = \hat{g} \hat{h}.
$$
\end{prop}

\begin{rem}
The saying is that \emph{convolution in the time domain is multiplication
in the frequency domain}.
\end{rem}

\begin{prop}[Bessel's inequality, Parseval's identity, and Plancherel's
theorem] Let $H$ be a Hilbert space.  Let $f \in H$ and let $b_k$,
$k=1,2,3,\ldots$, be a sequence of orthonormal vectors in $H$.  Then
$$
	\sum_{k=1}^\infty |\langle f, b_k \rangle|^2 \le \|f\|^2.
$$
This is \bfidx{Bessel's inequality}.  If the $b_k$'s are furthermore
a basis for $H$, then we have \bfidx{Parseval's identity}:
$$
	\sum_{k=1}^\infty |\langle f, b_k \rangle|^2 = \|f\|^2.
$$
This is a special case of \bfidx{Plancherel's theorem}.  (This statement of
Plancherel is due to Assane Lo.  See also remark \ref{rem:faris_plancherel}
below.) Namely, for $f,g \in H$,
$$
	\sum_{k=1}^\infty \overline{\hat{f}(k)} \hat{g}(k) = \langle f, g \rangle.
$$
(Bessel and Parseval then follow for $f=g$, recalling that $\|f\|^2=\langle
f,f\rangle$ and $\hat{f}(k) = \langle f, b_k \rangle$.)
\end{prop}

\begin{rem}
The context of this remark is a closed interval of the real line (nominally,
the unit interval).  Then $f(x) \in L^2([0,1])$ and $\hat{f}(k) \in
\ell^2(2\pi\Z)$.
Using Fourier series we reconstruct $f$ as a linear combination of
the orthonormal basis, where the $k$th coefficient is the dot with the
$k$th basis vector.  That is,
$$
	f = \sum_{k=1}^\infty \langle f, b_k \rangle b_k.
	\qquad (*)
$$
Parseval's identity guarantees that this series converges when the $b_k$'s
form a basis for $H$.  Parseval's identity may be paraphrased by saying that
when we insert norms in $(*)$, we still have an equality.
\end{rem}

\begin{mnem}
Engineers routinely paraphrase Parseval's identity by saying that the power in
the time-domain spectrum is the same as the power in the frequency-domain
spectrum.  Briefly:  \emph{power in equals power out}.
\end{mnem}

\begin{rem}
\label{rem:faris_plancherel}
Faris and Wikipedia both state Plancherel's theorem by saying that if $f \in
L^1(\R,dx) \cap L^2(\R,dx)$, then $\hat{f} \in L_2(\R,dk/2\pi)$.  Here, we are
taking the Fourier transform from $\R$ to $\R$, so both $x$ and $k$ are allowed
to take all real values.
\end{rem}

\begin{prop}[Riemann-Lebesgue lemma]
If $f \in L^1([0,1])$, then $\hat{f}(k) \to 0$ as $k \to \infty$.
\end{prop}

%\begin{rem}
%This is equivalent to saying that the integral of a highly oscillatory function
%is small?  [xxx needs a proof.]
%\end{rem}

%% ----------------------------------------------------------------
\newpage
\section{Radon-Nikodym derivatives}

\begin{defn}
Let $(X, \mcA, \mu)$ and $(X, \mcA, \nu)$ be measure spaces (on the same set
--- same $\sigma$-algebra too?).  Assume that $\mu$ and $\nu$ are
$\sigma$-finite.  Also assume that $\nu$ is signed and $\mu$ is positive.
The measure $\nu$ is said to be an
\bfidx{absolutely continuous measure} with respect to $\mu$,
written
$$
	\nu \ll \mu,
$$
if
$$
	\mu(E) = 0 \implies \nu(E) = 0.
$$
\end{defn}

\begin{ex}
\label{ex:radnyk}
Let $\nu$ be Lebesgue measure on $X=[-1,1]$ (with the Borel $\sigma$-algebra)
and let $\mu$ be $\mu + \delta_0$.  Let $E \subseteq X$.  If $0 \in E$, then
$\mu(E) = 1$.  If $0 \not\in E$, then $\mu(E) \nu(E)$.  Thus $\nu \ll \mu$.
\end{ex}

\begin{defn}
We say $h$ is the \bfidx{Radon-Nikodym derivative} of $\nu$ with respect to
$\mu$, written
$$
	h = \frac{\nu}{\mu},
$$
if for all measurable subsets $E$ of $X$,
$$
	\mu(E) = \int_E d\nu = \int_E h \; d\mu.
$$
\end{defn}

\begin{thm}[Radon-Nikodym derivatives exist]
If $\nu \ll \mu$ as above, then a Radon-Nikodym derivative exists.
\end{thm}

\begin{ex}
Let $\mu$ and $\nu$ be as in example \ref{ex:radnyk}.  Then take $h(x) =
d\nu/d\mu$ to be 1 if $x \ne 0$ and $0$ if $x = 0$.  That is, $h$ is designed
to pluck out 0, the one spot at which $\mu$ and $\nu$ differ.
\end{ex}

%% ----------------------------------------------------------------
\newpage
\section{Radon measures}

\bfidx{Radon measures}

\bfidx{Riesz representation} for Radon measures.

Pair Radon measure $\mu$ with continuous function $f$: $\langle \mu,f \rangle$.

\emph{Why} Radon measures?

%% ================================================================
\newpage
\section{Farisisms}

\begin{defn}
A \bfidx{lattice} is poset $P$ such that for all $a,b \in P$, $a\wedge b$ and
$a \vee b$ are in $P$.
\end{defn}

\begin{defn}
A \bfidx{Stone vector lattice} is:
\begin{itemize}
\item A vector space $L$ of real-valued functions (i.e. for $f,g \in L$ and
	$a \in \R$, $f+g$ and $af$ are in $L$),
\item which is a lattice ($f \wedge g$ is pointwise sup and $f \vee g$
	is pointwise),
\item with the \emphidx{Stone property}, namely, for all $f \in L$,
	$f \wedge 1 \in L$ (``locally constant'').
\end{itemize}
\end{defn}

\begin{defn}
A \bfidx{$\sigma$-ring} of real-valued functions on a set is a
Stone vector lattice which is closed under pointwise convergence.
\end{defn}

\begin{defn}
A \bfidx{$\sigma$-algebra} of real-valued functions on a set is a $\sigma$-ring
which also contains constant functions.  (By linearity, if $1 \in L$, then all
constant functions are in $L$.)
\end{defn}

\begin{defn}
An \bfidx{elementary integral} is linear, order-preserving, and satisfies
monotone convergence.  [xxx state this precisely.]
\end{defn}

\begin{mnem}
L-O-M --- eLOMentary integral --- linear, order-preserving, monotone
convergence.
\end{mnem}

\begin{thm}[Dini's theorem]
Let\index{Dini's theorem} $K$ be a compact topological space.  Let $f_n$ be a
sequence of continuous real-valued functions on $K$ such that $f_n \to 0$.
Then $f_n \to 0$ uniformly.
\end{thm}

%%% ================================================================
%\newpage
%\section{Skills}
%
%\bfidx{Standard machine} \ldots
%
%Some integrals to know (this list will grow):
%$$
%	\int \frac{du}{1+u^2}
%$$
%$$
%	\int \frac{du}{\sqrt{1+u^2}}
%$$

%% ================================================================
\newpage
\section{Problem solutions}

%% ----------------------------------------------------------------
\subsection{HW 15.1}

HW 15.1 --- two steps.

%% ----------------------------------------------------------------
\subsection{A02.6A}

A02.6A --- a beautiful example of many things.

%% ----------------------------------------------------------------
\subsection{A02.4B}

Let $f$ be a continuously differentiable real-valued function on $[0,1]$
with $f(0) = 0$.  Prove the inequalities:
\begin{eqnarray*}
	(i)  && |f(x)| \le x^{1/q} \|f'\|_p, \qquad x \in (0,1) \qqand\\
	(ii) && \|f\|_p^p \le \frac{1}{p} \|f'\|_p^p\\
\end{eqnarray*}
where $1<p<\infty$, $\|\cdot\|_p$ is the $L^p(0,1)$-norm, and
$1/p+1/q=1$.

\begin{answer}
With regard to the free-association principle mentioned in section
\ref{sec:principles}, several things should immediately come to mind.  First,
when we see $f$ and $f'$ together, especially with the $f(0)=0$ item, we should
think of the corollary to the FTC (corollary \ref{cor:f_and_f_prime})
$$
	\int_0^x f'(t)dt = f(x).
$$
Second, integral inequalities with $L^p$ norms in them should lead us to
H\"older's inequality (proposition \ref{prop:hoelder}).  Since there is only a
single function $f$ involved, we may need to insert a 1 at some point, as
mentioned in remark \ref{rem:hoelder_insertion}.  Lastly, the toothpaste
theorem (see mnemonic \ref{mnem:toothpaste}) is used in many integral problems.

So, let's start:
\begin{eqnarray*}
	f(x) &=& \int_0^x f'(t) dt \\
	|f(x)| &=& \left|\int_0^x f'(t) dt\right|
		\le \int_0^x \left|f'(t)\right| dt
		= \int_0^x \left|1 \cdot f'(t)\right| dt \\
	&\le&
		\left[ \int_0^x 1^q \,dt \right]^{1/q}
		\left[ \int_0^x |f'(t)|^p \,dt \right]^{1/p}
		\qquad \textrm{(we knew we needed a $p$)} \\
	&\le&
		x^{1/q}
		\left[ \int_0^1 |f'(t)|^p \,dt \right]^{1/p}
		\qquad \textrm{(approximate to get the $p$-norm we need)} \\
	&=& x^{1/q} \|f'\|_p.
\end{eqnarray*}

%For part (ii), we could start at the same spot and follow our noses.  Or, we
%can use the fact that $\|f\|_p^p = \|f^p\|_1$ (proposition
%\ref{prop:p_norm_1_norm}), recognizing that the $|f(x)|$ from part (i) may
%play a role in computing $\|f\|_p^p$.

For part (ii), we could start at the same spot as we did in part (i) and follow
our noses.  Or, since we need to get the $p$-norm of $f$ on the left, and we
have $|f|$, we can raise the left-hand side of the part (a) result to the $p$th
power and integrate:
\begin{eqnarray*}
	|f(x)| &\le& x^{1/q} \|f'\|_p \\
	\int_0^1 |f(x)|^p \,dx &\le& \int_0^1 x^{p/q} \|f'\|_p^p \,dx.
\end{eqnarray*}
What is $p/q$?  Since $1/p+1/q=1$, we have
$$
	p/q = p-1.
$$
Also, $\|f'\|_p^p$ is a constant and so comes out of the right-hand integral.
We're left with
$$
	\|f\|_p^p \le \int_0^1 x^{p-1} dx \; \|f'\|_p^p
		= \left[ \frac{x^p}{p}\right]_0^1 \|f'\|_p^p 
		= \frac{1}{p} \|f'\|_p^p
$$
as desired.
\end{answer}

%% ----------------------------------------------------------------
% Folland
% Rudin
% 3rd?

% little book ...

% Krantz
% Garrity

%% ----------------------------------------------------------------
% skills
% defns
% thms

%% ----------------------------------------------------------------
% solns

% mcL^1:  no +-infty values.
% abs cvg => cvg

% xxx rev tri ineq
