We can use integrals to make a set function 
out of any integrable function 
.
We call this set function the “indefinite integral” of , and it is defined for all measurable subsets 
. This isn’t quite the same indefinite integral that we’ve worked with before. In that case we only considered functions 
, picked a base-point 
, and defined a new function 
on the domain. In our new language, we’d write ![F(x)=\nu\left([a,x]\right)](http://s0.wp.com/latex.php?latex=F%28x%29%3D%5Cnu%5Cleft%28%5Ba%2Cx%5D%5Cright%29&bg=e6e6e6&fg=333333&s=0 "F(x)=\nu\left([a,x]\right)")
, so the two concepts are related but they’re not quite the same.
Anyhow, the indefinite integral is “absolutely continuous”. That is, for every 
there is a 
so that 
for all measurable 
with 
. Indeed, if 
is an upper bound for 
, then we can show that
And so if we make 
small enough we can keep 
small.
Further, an indefinite integral is countably additive. Indeed, if 
is a characteristic function then countable additivity of follows immediately from countable additivity of 
. And countable additivity for general simple functions is straightforward by writing each such function as a finite linear combination of characteristic functions.
With the exception of this last step, nothing we’ve said today depends on the function being simple, and so once we generalize our basic linearity and order properties we will immediately have absolutely continuous indefinite integrals.
May 27, 2010
Posted by John Armstrong |
Analysis, Measure Theory |
19 Comments
