Nonnegativity and equality for density Jensen–Shannon divergence

Define $\psi (0)=1$ and $\psi (t)=t\ln t-t+1$ for $t>0$. If $0<t<1$, then $\ln t=-{\int }_t^{1}du\mathrm{/}u>-(1-t)\mathrm{/}t=1-1\mathrm{/}t$, since $1\mathrm{/}u<1\mathrm{/}t$ for $t<u\le 1$. If $t>1$, then $\ln t={\int }_1^{t}du\mathrm{/}u>(t-1)\mathrm{/}t=1-1\mathrm{/}t$, since $1\mathrm{/}u>1\mathrm{/}t$ for $1\le u<t$. Thus $\psi (t)>0$ for $t\mathrm{\ne }1$ and $\psi (1)=0$. Also $-1\le \varphi (t)\le 0$ for $0<t\le 1$, by the first inequality and $\ln t\le 0$; and $0\le \varphi (t)\le t(t-1)\le 2$ for $1\le t\le 2$, since $\ln t={\int }_1^{t}du\mathrm{/}u\le t-1$. At $t=0$, $\varphi (0)=0$ and $\psi (0)=1$. Hence $\psi \ge 0$ with equality exactly at $1$, $\mathrm{\mid }\varphi (t)\mathrm{\mid }\le 2$, and $\psi (t)\le 5$ on $[0,2]$.

By [F1], [F7] and [F8], $\int md\mu =\frac{1}{2}\int (p+q)d\mu =1$. By [F2], $0\le r,s\le 2$. The ratio maps $R(a,b)=2a\mathrm{/}(a+b)$ and $S(a,b)=2b\mathrm{/}(a+b)$ on $a+b>0$, extended by $0$ at $(0,0)$, are Borel because they are continuous on $\{a+b>0\}$ and separately defined on the Borel singleton $\{(0,0)\}$. The functions $\varphi ,\psi$ are continuous on $[0,2]$, including at $0$. Thus coordinatewise measurability [F10], composition with these Borel maps [F11], and arithmetic closure [F9] give measurable $m,r,s,\varphi (r),\varphi (s),\psi (r),\psi (s)$; product closure [F12] then gives measurability of $m\varphi (r)$, $m\varphi (s)$, and $\frac{1}{2}m(\psi (r)+\psi (s))$. Step 1.1 gives $\mathrm{\mid }m\varphi (r)\mathrm{\mid },\mathrm{\mid }m\varphi (s)\mathrm{\mid }\le 2m$ and $0\le \frac{1}{2}m(\psi (r)+\psi (s))\le 5m$, so they are integrable. On $m>0$, $r+s=2$ and $\varphi (t)=\psi (t)+t-1$, so $\varphi (r)+\varphi (s)=\psi (r)+\psi (s)\ge 0$; when $m=0$ both products by $m$ vanish. Linearity in [F6] gives ${\mathrm{JS}}_{\mu }(P,Q)=\int \frac{1}{2}m(\psi (r)+\psi (s))d\mu \ge 0$.

If ${\mathrm{JS}}_{\mu }(P,Q)=0$, the nonnegative measurable function $h:=\frac{1}{2}m(\psi (r)+\psi (s))$ has integral zero by step 2.1. By [F5], $h=0$ $\mu$-almost everywhere. On $m>0$, both $\psi (r)$ and $\psi (s)$ are nonnegative, so they must each be zero almost everywhere there; step 1.1 gives $r=s=1$ there, hence $p=q$. On $m=0$, nonnegativity and $m=(p+q)\mathrm{/}2$ give $p=q=0$. Thus $p=q$ $\mu$-almost everywhere. For each $E\in \mathcal{A}$, [F1] and [F13] give $P(E)-Q(E)=\int p{\chi }_{E}d\mu -\int q{\chi }_{E}d\mu$. These restrictions are integrable because they are bounded by $p$ and $q$, whose integrals are $1$; [F8] and [F6] rewrite their difference as $\int u_{E}d\mu$ for $u_E:=p{\chi }_E-q{\chi }_E$. The positive and negative parts of $u_E$ are measurable by [F9, F12] and vanish almost everywhere, so [F5] makes both nonnegative integrals zero. By [F14], $\int u_{E}d\mu =0$, hence $P(E)=Q(E)$.

Conversely, suppose $P=Q$. For positive integers $n,k$, put $E_{n,k}:=\{p-q\ge 1\mathrm{/}n\}\cap \{p+q\ge 1\mathrm{/}k\}$, which is measurable by [F9]. Since $\{p>q\}={\bigcup }_{n,k\ge 1}{E}_{n,k}$, positive measure of $\{p>q\}$ would make some $E:=E_{n,k}$ have positive measure. On $E$, $(1\mathrm{/}k){\chi }_E\le (p+q){\chi }_E\le p+q$, so [F8], [F13], [F15] and [F16] give $\mu (E)\mathrm{/}k\le {\int }_E(p+q)d\mu \le \int (p+q)d\mu =2$ by [F1], [F7] and step 2.1; hence $\mu (E)<\mathrm{\infty }$. Put $u:=(p-q){\chi }_E$. By [F9, F12], $u$ is nonnegative measurable, and $(1\mathrm{/}n){\chi }_E\le u$, so [F8], [F15] and [F16] give $\int ud\mu \ge \mu (E)\mathrm{/}n>0$. Also $p{\chi }_E$ and $q{\chi }_E$ are integrable since they are bounded by $p$ and $q$; [F1], [F13] and [F6] therefore give $\int ud\mu =P(E)-Q(E)=0$, a contradiction. The same argument with $p,q$ exchanged rules out $\{q>p\}$, hence $p=q$ $\mu$-almost everywhere. Then $m\varphi (r)$ and $m\varphi (s)$ vanish almost everywhere: on $m>0$, $r=s=1$ almost everywhere, and on $m=0$ both products are zero. Their positive and negative parts are measurable by [F9, F12] and vanish almost everywhere, so [F5] makes all four nonnegative integrals zero. Step 2.1 gives their integrability, and [F14] and [F2] therefore yield ${\mathrm{JS}}_{\mu }(P,Q)=0$.

Steps 3.1 and 3.2 prove that ${\mathrm{JS}}_{\mu }(P,Q)$ is zero exactly when $P=Q$; step 2.1 proves it is finite and nonnegative. These are all claims in the Statement.