Hellinger invariance and comparison with library total variation

By [A1], [F3], and [F4], take finite-valued nonnegative densities $p=dP\mathrm{/}d\mu$ and $q=dQ\mathrm{/}d\mu$ for a common sigma-finite dominator $\mu$. Define $\rho (A):=P(A)+Q(A)$. For disjoint measurable $A_k$, countable additivity of $P,Q$ and regrouping nonnegative series give $\rho ({\bigcup }_k{A}_k)={\sum }_k\rho (A_k)$; also $\rho (\mathrm{\varnothing })=0$ and $\rho (\mathsf{X})=2$. Thus $\rho$ is a finite, hence sigma-finite, measure by [F6]. Additivity of the nonnegative integral [F10] gives $\rho (A)={\int }_A(p+q)d\mu$, with $p+q$ measurable by [F7].

Using $p,q$ against $\mu$, the set $B:=\{p\ge q\}$ is measurable by [F7,F8]. On $B$, $p=(p-q)+q$, so additivity [F10] gives $\delta :=P(B)-Q(B)={\int }_B(p-q)d\mu \ge 0$. On $B^c$, $q=(q-p)+p$ and the total masses are one, hence ${\int }_{B^c}(q-p)d\mu =Q(B^c)-P(B^c)=\delta$. Additivity then gives $\int \mathrm{\mid }p-q\mathrm{\mid }d\mu =2\delta$. For any measurable $A$, the difference on $A\cap B$ lies in $[0,\delta ]$ by monotonicity [F10], while the difference on $A\cap B^c$ lies in $[-\delta ,0]$; thus $\mathrm{\mid }P(A)-Q(A)\mathrm{\mid }\le \delta$, and $A=B$ attains $\delta$. Therefore [F2] gives $\parallel P-Q{\parallel }_{\mathrm{T}\mathrm{V}}=\delta =\frac{1}{2}\int \mathrm{\mid }p-q\mathrm{\mid }d\mu$.

On $\{p+q>0\}$ define $\bar{p}=p\mathrm{/}(p+q)$ and $\bar{q}=q\mathrm{/}(p+q)$, and set both to zero on $\{p+q=0\}$. For $r<0$, $\{\bar{p}>r\}$ is the whole space; for $r\ge 0$, it is $\{p+q>0\}\cap \{p-r(p+q)>0\}$. Thus [F7,F8] make $\bar{p}$ measurable, and the same argument applies to $\bar{q}$. Since $p=q=0$ where $p+q=0$, [F9] gives ${\int }_A\bar{p}d\rho ={\int }_A\bar{p}(p+q)d\mu =P(A)$ and ${\int }_A\bar{q}d\rho =Q(A)$ for every measurable $A$. Hence $\bar{p},\bar{q}$ are finite-valued density representatives of $P,Q$ against $\rho$.

Pointwise $\mathrm{\mid }p-q\mathrm{\mid }=\mathrm{\mid }\sqrt{p}-\sqrt{q}\mathrm{\mid }(\sqrt{p}+\sqrt{q})$. Hölder [F5] therefore gives $\int \mathrm{\mid }p-q\mathrm{\mid }d\mu \le h(P,Q)(\int (\sqrt{p}+\sqrt{q}{)}^{2}d\mu {)}^{1\mathrm{/}2}$. The squared second factor is $2+2\int \sqrt{pq}d\mu \le 4$, since [F5] applied to $\sqrt{p},\sqrt{q}$ bounds their product integral by $1$. Step 1.2 now gives $\parallel P-Q{\parallel }_{\mathrm{T}\mathrm{V}}\le h(P,Q)$.

Write $H_{\lambda }(P,Q{)}^2$ for the defining Hellinger integral computed from supplied densities against a common dominator $\lambda$. Applying [F1] and [F9] to the densities in step 2.1 gives $H_{\rho }(P,Q{)}^2=\int (\sqrt{\bar{p}}-\sqrt{\bar{q}}{)}^{2}d\rho ={\int }_{\{p+q>0\}}(\sqrt{p}-\sqrt{q}{)}^{2}d\mu =H_{\mu }(P,Q{)}^2$, because both original densities vanish where $p+q=0$. For any measure $\alpha$ and finite-valued nonnegative densities $f,g$ of the same finite measure $\mathrm{\Lambda }$ with respect to $\alpha$, they agree $\alpha$-almost everywhere: on $E_m:=\{f\ge g+1\mathrm{/}m\}$, equality of their density integrals gives $0={\int }_{E_m}(f-g)d\alpha \ge \alpha (E_m)\mathrm{/}m$, so $\alpha (E_m)=0$. Since $\{f>g\}={\bigcup }_{m\ge 1}{E}_m$, and the same argument with $f,g$ swapped handles $\{g>f\}$, this proves uniqueness almost everywhere. Thus changing supplied representatives against a fixed dominator does not change the Hellinger integral.

For any other common sigma-finite dominator $\nu$, repeat steps 1.1 and 2.1 using its densities to obtain density representatives ${\bar{p}}_{\nu },{\bar{q}}_{\nu }$ against the same finite measure $\rho =P+Q$. The uniqueness argument in step 3.1 gives ${\bar{p}}_{\nu }=\bar{p}$ and ${\bar{q}}_{\nu }=\bar{q}$ $\rho$-almost everywhere. The same density-change calculation as in step 3.1 therefore identifies the Hellinger integral against $\nu$ with the same canonical value against $\rho$, and hence with that against $\mu$.

Pointwise $(\sqrt{p}-\sqrt{q}{)}^2\le \mathrm{\mid }p-q\mathrm{\mid }$. Integrating by monotonicity [F10] and using [F1] and step 1.2 gives $h(P,Q{)}^2\le \int \mathrm{\mid }p-q\mathrm{\mid }d\mu =2\parallel P-Q{\parallel }_{\mathrm{T}\mathrm{V}}$.