Is the Hellinger distance invariant under the choice of the dominating measure?

Question

Let $\mu_1,\mu_2$ be two probability measures on $(\Omega,\Sigma)$, and $\nu$ be a $\sigma$-finite measure on $(\Omega,\Sigma)$, such that $\mu_1,\mu_2$ are absolutely continuous with respect to $\nu$. By the Radon-Nikodym Theorem, there exist measurable functions $\frac{d\mu_1}{d\nu},\frac{d\mu_2}{d\nu}$ mapping $\Omega\rightarrow[0,\infty)$, such that for all $A\in \Sigma$, $$ \int_A \frac{d\mu_1}{d\nu} d\nu = \int_A d\mu_1,\quad \int_A \frac{d\mu_2}{d\nu} d\nu = \int_A d\mu_2 $$ In a textbook I'm studying there is an exercise which asks to prove that for any two $\sigma$-finite measures $\nu_1,\nu_2$, with respect to which $\mu_1,\mu_2$ are absolutely continuous, the following holds: $$ \int\left(\sqrt{\frac{d\mu_1}{d\nu_1}}-\sqrt{\frac{d\mu_2}{d\nu_1}}\right)^2d\nu_1=\int\left(\sqrt{\frac{d\mu_1}{d\nu_2}}-\sqrt{\frac{d\mu_2}{d\nu_2}}\right)^2d\nu_2 $$ In other words, the Hellinger distance is invariant under changes of the dominating measure. A straightforward step towards showing this would be to expand the two expressions. After some cancelling, the statment reads: $$ \int \sqrt{\frac{d\mu_1}{d\nu_1}\frac{d\mu_2}{d\nu_1}}d\nu_1=\int \sqrt{\frac{d\mu_1}{d\nu_2}\frac{d\mu_2}{d\nu_2}}d\nu_2 $$ This is where I get stuck. Symbolically, it seems to make sense, as one might expect something like: $$ \sqrt{\frac{d\mu_1}{d\nu_1}\frac{d\mu_2}{d\nu_1}}d\nu_1=\sqrt{d\mu_1d\mu_2}=\sqrt{\frac{d\mu_1}{d\nu_2}\frac{d\mu_2}{d\nu_2}}d\nu_2 $$ But why is this true? Is it even generally true? How can it be shown?

Answer 1

This is a suggestion:

Consider $\nu=\frac12(\nu_1+\nu_2)$. Then $\nu_j\ll\nu$ and so \begin{align} \int\sqrt{\frac{d\mu_1}{d\nu_j}\frac{d\mu_2}{d\nu_j}}\,d\nu_j&=\int\sqrt{\frac{d\mu_1}{d\nu_j}\frac{d\mu_2}{d\nu_j}}\,\frac{d\nu_j}{d\nu}\,d\nu\\ &=\int\sqrt{\frac{d\mu_1}{d\nu_j}\frac{d\nu_j}{d\nu} \frac{d\mu_2}{d\nu_j}\frac{d\nu_j}{d\nu}}\,d\nu\\ &=\int\sqrt{\frac{d\mu_1}{d\nu}\frac{d\mu_2}{d\nu}}\,d\nu \end{align} for $j=1,2$.

One needs to check that if for $\sigma$ finite measures $P$, $Q$ and $R$, if $P\ll Q$ and $Q\ll R$, then $P\ll R$ and $\frac{dP}{dR}=\frac{dP}{dQ}\,\frac{dQ}{dR}$. This is just another manifestation of Radon-Nikodym's theorem.