Hellinger distance is a metric and controls L1 and total variation

Since $p,q$ are measurable, [F8] makes $\sqrt{p},\sqrt{q}$ measurable and [F7] makes their difference and absolute value measurable. Their unit integrals put both roots in $L^2(\mu )$. The triangle inequality in [F2] gives symmetry, nonnegativity, and the triangle inequality for $h$. If $h(P,Q)=0$, [F5] gives $\sqrt{p}=\sqrt{q}$ almost everywhere, hence $p=q$ almost everywhere and $P=Q$. Conversely, put $g=p-q\in L^1(\mu )$ by [F6]. If $P=Q$, then for every measurable $A$, $\int g{\mathbf{1}}_{A}d\mu =\int p{\mathbf{1}}_{A}d\mu -\int q{\mathbf{1}}_{A}d\mu =P(A)-Q(A)=0$ by [F6]. For each integer $m\ge 1$, the set $E_m:=\{g\ge 1\mathrm{/}m\}$ is measurable by [F8]. Define $u_m:=g{\mathbf{1}}_{E_m}$; it is nonnegative and measurable by [F7], and linearity gives $\int u_{m}d\mu =0$. By [F5], $u_m=0$ almost everywhere; since $u_m\ge 1\mathrm{/}m$ on $E_m$, $\mu (E_m)=0$. Applying the same argument to $(q-p){\mathbf{1}}_{\{q-p\ge 1\mathrm{/}m\}}$ shows $\mu (\{g\le -1\mathrm{/}m\})=0$. Since $\{g\mathrm{\ne }0\}$ is the countable union of these null sets, $p=q$ almost everywhere and $h(P,Q)=0$. Thus $h$ is a metric on these laws.

Put $g:=p-q$ and $A:=\{x:g(x)\ge 0\}$, which is measurable by [F7, F8]. Since $\int gd\mu =0$, the positive and negative parts have equal integral, so ${\int }_{A}gd\mu =\frac{1}{2}\int \mathrm{\mid }g\mathrm{\mid }d\mu$. For every measurable $B$, positivity of $g$ on $A$ and nonpositivity on $A^c$, together with monotonicity [F9], give ${\int }_{A^c}g\le {\int }_{B}g\le {\int }_{A}g$; equality in the absolute supremum is attained at $A$ (or $A^c$). Thus [F4] gives $\parallel P-Q{\parallel }_{\mathrm{T}\mathrm{V}}=\frac{1}{2}{d}_1(P,Q)$.

The functions in the pointwise inequality are measurable by [F7, F8]. If $p\ge q$, then $(\sqrt{p}-\sqrt{q}{)}^2\le p-q$, because $0\le (\sqrt{p}-\sqrt{q})\mathrm{/}(\sqrt{p}+\sqrt{q})\le 1$ when the denominator is positive; the case $q\ge p$ is symmetric, and when both vanish the inequality is zero on both sides. Monotonicity [F9] and [F1] give $h(P,Q{)}^2\le d_1(P,Q)$.

By [F7, F8], the displayed products and sums of square roots are measurable. Factor $\mathrm{\mid }p-q\mathrm{\mid }=\mathrm{\mid }\sqrt{p}-\sqrt{q}\mathrm{\mid }(\sqrt{p}+\sqrt{q})$. By [F3], $d_1(P,Q)\le h(P,Q)\parallel \sqrt{p}+\sqrt{q}{\parallel }_2$. Also $\parallel \sqrt{p}+\sqrt{q}{\parallel }_2^2=2+2\int \sqrt{pq}d\mu \le 4$, again by [F3] and the unit $L^2$ norms. Hence $d_1(P,Q)\le 2h(P,Q)$, and step 1.2 gives $\parallel P-Q{\parallel }_{\mathrm{T}\mathrm{V}}\le h(P,Q)$.