Approximate privacy admits purification of a joint measure with fixed history marginal

By [F1], the integrands defining $R,S$ are measurable. Countable additivity follows by applying monotone convergence to disjoint sections, and total mass is one. The same argument gives $\mu =(R+S)\mathrm{/}2=\lambda \otimes M$, where $M=(K+L)\mathrm{/}2$ is the explicit mixture probability kernel and the notation $\lambda \otimes M$ denotes the displayed iterated integral. Both $R,S$ have marginal $\lambda$. By [A1], choose a density of $R$ relative to $\mu$. The inequality $0\le R\le 2\mu$ forces this density into $[0,2]$ almost everywhere: a positive-measure level set violating a bound would contradict that measure inequality. Clip it into $[0,2]$ and call the result $p$; put $q=2-p$. Thus $p,q$ are finite nonnegative product-measurable densities of $R,S$ relative to $\mu$. The indicator identities extend to nonnegative integrands by [F2].

Write $a=\exp (\epsilon )\ge 1$ by [F4]. Let $E=\{p>aq\}$ and $F=\{p<q\mathrm{/}a\}$, both product-measurable. For every measurable history event $C$, apply the assumed event inequality to $E_h$ for $h\in C$, and integrate. The density identities give ${\int }_{C\times Y}(p-aq{)}_{+}d\mu =R(E\cap (C\times Y))-aS(E\cap (C\times Y))\le \delta \lambda (C)$. Similarly ${\int }_{C\times Y}(q\mathrm{/}a-p{)}_{+}d\mu \le (\delta \mathrm{/}a)\lambda (C)$. Define $\alpha (h)=\int (p-aq{)}_{+}(h,y)M(h,dy)$ and $\beta (h)=\int (q\mathrm{/}a-p{)}_{+}(h,y)M(h,dy)$. These are finite measurable functions. Localization over all $C$ implies $\alpha \le \delta$ and $\beta \le \delta \mathrm{/}a$ almost everywhere, since a violating positive-measure level set could be used for $C$. Likewise, the common history marginals imply $\int p(h,y)M(h,dy)=\int q(h,y)M(h,dy)=1$ for $\lambda$-almost every $h$, by equality of their integrals over every $C$. Let $G$ be the measurable conull set where all four scalar conditions hold. Only the two specified joint level sets and scalar marginal identities have been used; no simultaneous fiberwise density representation is asserted.

Put $c=\min \{\max \{p,q\mathrm{/}a\},aq\}$ and $m(h)=\int c(h,y)M(h,dy)$. On $G$, $m=1-\alpha +\beta$ and $q\mathrm{/}a\le c\le aq$. If $m<1$, set $w=c+\frac{1-m}{a-m}(aq-c)$; if $m>1$, set $w=c-\frac{m-1}{m-1\mathrm{/}a}(c-q\mathrm{/}a)$; if $m=1$, set $w=c$. Every used denominator is strictly positive and its fraction lies in $[0,1]$. These formulas give $q\mathrm{/}a\le w\le aq$ and $\int w(h,y)M(h,dy)=1$ on $G$. Off $G$, set $w=q$. All formulas are measurable by [F1] and [F3]; each division is confined to its stated measurable branch. In particular, when $a=1$, $c=q$ and $m=1$ on $G$, so no zero denominator is used.

Define $T(B)={\int }_{B}wd\mu$. By [F2] this is a measure. The conditional normalization on $G$ and the fact that $G^c$ is $\lambda$-null imply $T(C\times Y)=\lambda (C)$, so $T$ is a probability measure with the required marginal. Integrating $q\mathrm{/}a\le w\le aq$ gives $S\mathrm{/}a\le T\le aS$. On $G$, clipping changes $p$ by conditional absolute integral $\alpha +\beta$, and the correction changes $c$ in one direction by conditional absolute integral $\mathrm{\mid }1-m\mathrm{\mid }=\mathrm{\mid }\alpha -\beta \mathrm{\mid }$. Consequently $\int \mathrm{\mid }p-w\mathrm{\mid }d\mu \le 2\int \max \{\alpha ,\beta \}d\lambda \le 2\delta$. Because $p,w$ have equal total integral, the positive and negative parts of $p-w$ have the same integral; every event integral lies between these two extrema, attained by its sign sets. Thus $\mathrm{TV}(R,T)=\frac{1}{2}\int \mathrm{\mid }p-w\mathrm{\mid }d\mu \le \delta$. This proves all claims, including $\delta =0$ and $\epsilon =0$.