The Radon-Nikodym Chain Rule
Today we take the Radon-Nikodym derivative and prove that it satisfies an analogue of the chain rule.
If ,
, and
are totally
-finite signed measures so that
and
, then
-a.e. we have
By the linearity we showed last time, if this holds for the upper and lower variations of then it holds for
itself, and so we may assume that
is also a measure. We can further simplify by using Hahn decompositions with respect to both
and
, passing to subspaces on which each of our signed measures has a constant sign. We will from here on assume that
and
are (positive) measures, and the case where one (or the other, or both) has a constant negative sign has a similar proof.
Let’s also simplify things by writing
Since and
are both non-negative there is also no loss of generality in assuming that
and
are everywhere non-negative.
So, let be an increasing sequence of non-negative simple functions converging pointwise to
. Then monotone convergence tells us that
for every measurable . For every measurable set
we find that
and so for all the simple we conclude that
Passing to the limit, we find that
and so the product serves as the Radon-Nikodym derivative of
in terms of
, and it’s uniquely defined
-almost everywhere.
[…] Corollaries of the Chain Rule Today we’ll look at a couple corollaries of the Radon-Nikodym chain rule. […]
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Could you quickly clarify why there is no loss of generality in assuming that f and g are everywhere non-negative? Sorry for the frequent recent posts, and thank you for the great resource too!
Basically, if they’re not you can always decompose into positive and negative parts, use the result there, and put everything back together.
Could anyone tell me, were we exactly use the increasing sequences and why? I dont really see it 🙂
You mean where we use the fact that the sequence
is increasing? That’s a requirement of the monotone convergence theorem.
I know that theorem. I just wanted to know why we need it here, where it is used in the last part V(E)=… . I’m sorry, im really not used to this notations at all and don’t get the point of it by now
Okay, well the point is that any measurable function
can be approximated as the limit of an increasing sequence of measurable functions
, and the simple functions are basically constants times characteristic functions. So what we do is prove our result for the characteristic functions
. Then the fact that everything in sight is linear means that it holds for simple functions. And finally we can pass to the limit (using monotone convergence) and get the result for all measurable functions.
Ah, now I see it, thanks for your time an help 🙂
how do we know fg is lamda-integrable?
Because
is the Radon-Nikodym derivative of
with respect to
. To put it another way, you know that there has to be one, since
(because
is transitive), so we can call it
, and it’s uniquely defined (
-a.e.). But the product
satisfies the same condition as
: you can get the
-measure of any set
by integrating either
or
with respect to
over
. So they’re equal
-a.e. and
must be
integrable, and so it’s as good a representative for the Radon-Nikodym derivative as any.