kernel and image of $D$ and $I$:

note that there is a map $\cL \to L^1$, considering each integrable function
as a measure -- but this has kernel the functions vanishing outside a set of
measure $0$.

Integral:
depends on your domain

kernel --------
domain = integrable functions (denote by $\cL$)
kernel is functions that vanish outside a set of measure zero,
i.e., exactly what we mod out by to get $L^1$ or talk about a function defined
``almost everywhere''

domain = measures
no kernel (we got rid of it by talking about $L^1$ instead of $\cL$)


image depends on your domain;
$$I\from L^1 \oplus SC\cM \oplus D \to BV=(AC \oplus SC \oplus D\cM)$$
SC = singular continuous (left I mean measures, right I mean functions)
D = point masses (er, on right should be BV step functions)

EZ to understand/break up:
any BV function is difference of two monotone;
fix up the discontinuities and get a continuous BV function;
  (this picks off the point masses)
then look at $\int f'$; this is your Abs Cont part;
the rest in the singular continuous part

derivative:
kernel:
constant functions (EZ by Rolle's theorem)

image:
urk!!!
image contains all $C^0$ functions (integrate then differentiate), but also
contains stuff like $\sin(1/x)$, which satisfy the IVT but aren't continuous
(look at comments elsewher).
Further, you can't get all IVT functions (as noted elsewhere: change the value
at zero of $\sin(1/x)$ and it's still IVT, but integral can't detect the
difference)

