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Let $\mu_1,\mu_2$ be two Borel probability measures on $l^2(\mathbb{N})$ and let $g:\mathbb{R}^\infty\times l^2(\mathbb{N}) \to \mathbb{R}$ be a jointly measurable mapping. Lastly let $\rho$ denote a Borel probability measure on $\mathbb{R}^\infty$ and let $\langle \cdot , \cdot \rangle$ denote the Euclidean metric on $\mathbb{R}^k$, and $u\in l^2(\mathbb{N})$ have the decomposition $u=(u_{\leq k},u_{>k})=((u_1,...,u_k),(u_{k+1},u_{k+1},...))$ for some $k\in \mathbb{N}$.

I know that for $\rho$-almost all $w\in \mathbb{R}^\infty$, $\lambda^k$-almost all $v\in \mathbb{R}^k$, $\lambda$-almost all $s\in \mathbb{R}$, it holds that $$ \mu_1(u\in l^2(\mathbb{N}): \langle u_{\leq k}, v\rangle +g(w,u_{>k}) \leq s ) =\mu_2(u\in l^2(\mathbb{N}): \langle u_{\leq k}, v\rangle +g(w,u_{>k}) \leq s ) . $$ By cadlag properties of $s\mapsto LHS,RHS$ (they are cdf's on $\mathbb{R}$) this can be strengthened to all $s\in \mathbb{R}$, easily.

Problem: It should be possible (stated in an article in The Annals of Probability, but I can not figure out how to prove it), to also strengthen the statement in regards to the $v\in \mathbb{R}^k$.

That is, to show that for $\rho$-almost all $w\in \mathbb{R}^\infty$, all $v\in \mathbb{R}^k$ and all $s\in \mathbb{R}$, it holds that $$ \mu_1(u\in l^2(\mathbb{N}): \langle u_{\leq k}, v\rangle +g(w,u_{>k}) \leq s ) =\mu_2(u\in l^2(\mathbb{N}): \langle u_{\leq k}, v\rangle +g(w,u_{>k}) \leq s ) . $$

Any help will be much appreciated.

1 Answers 1

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For fixed $u$, $v\mapsto X_\nu (u):=\langle u_{\le k},\nu\rangle+\langle w,u_{>k}\rangle$ is continuous. Fix a good $w$ so that for $\lambda^k$ a.e. $\nu$ (say, for $\nu\in G$ with $\lambda^k(G^c)=0$) you have $$ \mu_1(u:X_\nu (u)\le s)=\mu_2(u:X_\nu (u)\le s),\qquad\forall s. $$ Given any $\nu\in\Bbb R^k$, choose $\nu_n\in G$ with $\lim_n\nu_n=\nu$. The two cdfs $\mu_i(u:X_{\nu_n} (u)\le s)$, $i=1,2$, then converge weakly; you conclude that the two sides of the above display agree, first for all $s$ that are continuity points for both limit cdfs, then for all $s$ by right continuity.