\section{Big picture of spaces in differentiation and integration}
FIXME: also have this over in "real analysis";
should integrate...

Most general thing I can differentiate:
  distribution
Most general thing I can integrate:
  measure
  (or better: I can integrate a distribution against any smooth
   function; I can integrate a measure over any measureable subset,
   tautologically)

$$
\xymatrix{C^0 & C^1 \ar[l] & C^2 \ar[l] & \cdots \ar[l]
  & C^\infty \supset C^\omega \supset \cP}
$$
$\cP$ is polynomials; all the stuff on the right end is preserved by taking
derivatives (and integrals), and $C^\infty$ is the biggest subspace of $C^0$
preserved by taking derivatives (if you are not $C^\infty$, then your
derivative is even less $C^\infty$).

one way of looking at this is that one can differentiate everything qua
distribution, but often the result can be nicely represented (eg, if you
differentiate a $C^1$ function, you get an actual function out of it, etc.).

Integration short exact sequence:
$$\xymatrix{0 \ar[r] & C^k \ar[r]^{I_a} & C^{k+1} \ar[r] & \bR \ar[r] & 0}$$
where the maps are $f \mapsto \int_a^x f$
and $g \mapsto g(a)$ (evaluation at $a$).
...the natural way of talking about the image of $I_a$.

FIXME: Should one think of the distributions that are derivatives of $C^0$
functions as being ``$C^{-1}$'' in some sense? Do I need to think about
Sobolev spaces??

\subsection{Between $C^0$ and $C^1$}
One natural space between $C^0$ and $C^1$ is the space of everywhere
differentiable (but not \emph{continuously} differentiable) functions.
The derivatives of these are functions that satisfy the Intermediate Value
Theorem (so I'll call them IVT functions), but not every IVT function is a
derivative (consider two IVT functions that differ at a point, like
$\sin(1/x)$ extended two different ways at the origin; then these can't be
distinguished by integrals).

Note that IVT functions fail to be continuous because they don't take compact
to compact (see notes in section on continuity).

Every derivative in this way is Denjoy (Generalized Riemann) integrable, but
needn't be Lebesgue integrable (consider $x^2 \cos (1/x^2)$; this is not
absolutely continuous, so the derivative is not in $L^1$).

This makes sense; since $L^1$ is \emph{absolutely integrable} functions, we
should only get absolutely continuous functions as Lebesgue integrals.

\subsection{Integrating measures}
$BV$ bounded variation, i.e., difference of two non-decreasing functions;
$AC$ absolutely continuous, i.e., difference of two continuous monotone
functions
$SC$ means singular continuous measures/functions

$\cM \to^I BV$

examples of each of these:
cantor measure (each half of cantor set has measure $1/2$, etc.),
which integrates to give devil's staircase
  (not differentiable on set of measure $0$)

enumerate rationals on $(0,1)$ and put weights of $1/2^n$ on each;
then the integral is an increasing function, continuous at each irrational and
discontinuous at every rational (and should be differentiable at each
irrational too)

\subsection{Distributions and Measures}
A measure gives a distribution $\cM \to \cD$; I think this is injective, as
otherwise we'd need a measure that vanished on every $C^\infty$ function, and
I have trouble thinking of that...
(is the map $L^1 \to \cD$ injective at least?)
Hence we can in particular differentiate measures.

FIXME: can we integrate distributions? I don't think so, as that would require
evaluating the distribution on $\chi_{[0,x]}$, and this is a step, not a
$C^\infty$ function. One could try to approximate the step by smooth
functions, but evaluating the distribution on this shouldn't in general
converge. If it does, I bet the distribution's actually a measure??

\subspace{Topologies on these spaces}
Note that $C^0$ is complete in the $C^0$ (aka, uniform on compact sets)
topology, but $C^1$ isn't a closed/complete subspace: hence you use
the $C^1$ topology, etc. -- that's where these come from.

