Re: simple measure theory question

In article <funtf6$550$1@xxxxxxxxxxxxxxxx>,
mjhardy@xxxxxxx (Michael J Hardy) wrote:

Let (Omega, F, n) be a measure space and suppose
n(Omega) is finite. Let X:Omega ---> (0,infinity)
be a measurable function. For A in F let m(A)
= integral over A of X dn. Then X is the
Radon-Nikodym derivative dm/dn. Now for x >or= 0
let N(x) = n({omega : X(omega) > x}), and let
M(x) = [integral over {omega : X(omega) > x}] of X dn.
Then a question is: when does dM/dN = x ?

Always.

Let mu and nu denote the images of m and n under the
transformation X. For example,

mu(B) = m( X^-1}(B) )

when B is a Borel subset of the line. The absolute continuity
of m with respect to n implies the same for mu with
respect to nu, and the usual change of variables formula
for integrals then shows that the Radon-Nikodym derivative
of mu with respect to nu is the identity function.


Notice that N and M are non-increasing functions
of x, but may fail to be one-to-one. But M is
necessarily a strictly increasing function of N
(at least if M is always finite); it cannot fail
to be one-to-one. Since M is therefore a function
of N, one can ask whether it is differentiable and
what its derivative is.

The answer would appear to be that if there is
any interval (a,b) along which N (and therefore
also M) remains constant, then dM/dN does not
exist at that point, since the derivative from
the left could not be less than b and that from
the right could not be more than a; but at all
other values of N we would have dM/dN = x. The
proof would be that if N is a strictly decreasing
function of x, then x is a _continuous_ function
of N, and the difference quotient that approaches
dM/dN would be squeezed between x and x + Delta x,
and continuity would imply that Delta x ---> 0.

So "dm/dn = X" is easy, and "dM/dN = x" is a bit
more work.

My question is: is this a standard result found
in textbooks or other published sources (that
should be mentioned if one relies on this in
published work)? -- Mike Hardy

--
A.

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