Let $(X,\Omega)$ be a measurable space and let $\mu, \nu$ be two $\sigma$-finite measures on $(X,\Omega)$. Suppose $\nu \ll \mu$ and let $\phi$ be the Radon-Nikodym derivative of $\nu$ with respect to $\mu$ $(\phi = d\mu/d\nu)$. Define $V:L^2(\nu)\rightarrow L^2(\mu)$ by $Vf=\sqrt \phi f$. Show that $V$ is a well-defined linear isometry and $V$ is an isomorphism if and only if $\mu \ll \nu$ (that is, if and only if the measures are mutually absolutely continuous.)

This is exercise 5.8 in chapter 1 of Conway's A Course in Functional Analysis, second edition. It can be found on page $23$.


It is clear that $V$ is well-defined on equivalence classes and linear.

We have

$$\langle f,g \rangle_\nu = \int_X f\overline g \ d\nu = \int_X f\overline g \ \phi \ d \mu = \int_X (f\sqrt \phi)\overline {(g\sqrt \phi)} \ \ d \mu =\langle Vf, Vg\rangle_\mu.$$

This shows the operator is an isometry.

If $\mu$ and $\nu$ are mutually absolutely continuous, then define $U$ to be the analogous operator $L^2(\mu)\rightarrow L^2(\nu)$. By the same kind of manipulation as above, we see that $U\circ V$ and $V\circ U$ are the identity operators.

For the converse, assume that $V$ is an isomorphism, so that it surjective. To show that $\mu$ is absolutely continuous with respect to $\nu$, we must show that $\mu(A)=0$ for every set $A$ where $\nu(A)=0$. Suppose this is not the case, and that $\mu(B)\neq 0$ while $\nu(B)=0$ for some set $B$. Let $f$ be the function that is $1$ on $B$ and $0$ elsewhere. Then this function is not in the range of $V$, a contradiction.

  • $\begingroup$ The last paragraph is a bit casual. If you have a better explanation, feel free to post another answer. $\endgroup$
    – Potato
    Aug 3 '13 at 22:52

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