FALSE: pointwise-integrable generator constants imply a locally Lipschitz WGAN loss

Set $a_n=4^{-n}$, $b_n={64}^{-n}$, $p_n=2^{-n}$, $t_n(\theta )=\max \{1-\mathrm{\mid }\theta -a_n\mathrm{\mid }\mathrm{/}{b}_n,0\}$ and $u_n(z)=\max \{1-4\mathrm{\mid }z-n\mathrm{\mid },0\}$. Put $g_{\theta }(z)={\sum }_{n\ge 1}{u}_n(z)t_n(\theta )$ and $\rho (A)={\sum }_{n\ge 1}{p}_n{\mathbf{1}}_A(n)$ for Borel $A$. The nonnegative-series formula is countably additive: for disjoint sets, equality holds first for finite partial sums and then by taking suprema of finite sums of nonnegative terms. By [F1] its total mass is one. Each $g_{\theta }$ is Borel because the sum is pointwise a limit of continuous partial sums.

On every bounded latent neighborhood only finitely many $u_n$ are nonzero, and the supports $[n-1\mathrm{/}4,n+1\mathrm{/}4]$ are disjoint. Each factor is bounded by one, $u_n$ is $4$-Lipschitz and $t_n$ is $b_n^{-1}$-Lipschitz. Thus each product is jointly Lipschitz with respect to $\mathrm{\mid }\theta -{\theta }^{\mathrm{\prime }}\mathrm{\mid }+\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }$, and the locally finite sum is jointly locally Lipschitz and takes values in $[0,1]$.

Write $I_n=[a_n-b_n,a_n+b_n]$. These intervals are pairwise disjoint: $a_n-a_{n+1}=3a_n\mathrm{/}4$, while $b_n+b_{n+1}=(65\mathrm{/}64)b_n\le (65\mathrm{/}1024)a_n<3a_n\mathrm{/}4$. The measurable function $L(\theta ,z)={\sum }_{n\ge 1}(4+b_n^{-1}){\mathbf{1}}_{I_n}(\theta ){\mathbf{1}}_{(n-1\mathrm{/}2,n+1\mathrm{/}2)}(z)$ supplies a local joint Lipschitz constant at every $(\theta ,z)$. If $\theta \mathrm{\notin }{I}_n$, a small parameter neighborhood misses $I_n$; if $z$ is outside all active latent supports, a small latent neighborhood makes the function zero. At support boundaries the displayed positive constant bounds the piecewise-linear product. At a fixed $\theta$ at most one $I_n$ is active, so $\int L(\theta ,z)d\rho (z)$ is either zero or the finite number $p_n(4+b_n^{-1})$. In particular it is zero at $\theta =0$.

Let $\iota (y)=(0,y)$ and let ${\gamma }_{\theta }={\iota }_{\mathrm{\#}}{P}_{\theta }$. The map $\iota$ is continuous in the product topology, so [F6] makes ${\gamma }_{\theta }$ a Borel probability; for every Borel $A\subseteq X$ its marginals satisfy ${\gamma }_{\theta }(A\times X)={\mathbf{1}}_{\{0\in A\}}={\delta }_0(A)$ and ${\gamma }_{\theta }(X\times A)=P_{\theta }(A)$, so it is a coupling. If $\gamma$ is any coupling and $C\subseteq X\times X$ is Borel, set $B_C={\iota }^{-1}(C)$, which is Borel by continuity. The first marginal gives $\gamma ((X\setminus \{0\})\times X)=0$, so $\gamma (C)=\gamma (C\cap (\{0\}\times X))=\gamma (\{0\}\times B_C)$. It also gives $\gamma ((X\setminus \{0\})\times B_C)=0$, and the second marginal gives $\gamma (X\times B_C)=P_{\theta }(B_C)$; hence $\gamma (\{0\}\times B_C)=P_{\theta }(B_C)={\gamma }_{\theta }(C)$. Thus this coupling is unique, and [F6] gives its cost as ${\int }_{X}yd{P}_{\theta }(y)=\int g_{\theta }d\rho$. For any nonnegative simple $s={\sum }_j{c}_j{\mathbf{1}}_{A_j}$, the definition of $\rho$ and the simple-integral formula give $\int sd\rho ={\sum }_n{p}_{n}s(n)$, since the finite sum over $j$ commutes with the nonnegative series. Choose simple $s_k\uparrow g_{\theta }$; [F2] and [F7] give $\int g_{\theta }d\rho ={\lim }_k{\sum }_n{p}_n{s}_k(n)={\sum }_n{p}_n{g}_{\theta }(n)$, where the last equality follows by taking suprema over finite partial sums and over $k$. Since $g_{\theta }(n)=t_n(\theta )$ and the parameter intervals in step 3.1 are disjoint, at most one term is nonzero. Therefore $V(\theta )={\sum }_{n\ge 1}{p}_n{t}_n(\theta )$. Consequently $V(0)=0$ and $V(a_n)=p_n$, so $V(a_n)\mathrm{/}{a}_n=2^n\to \mathrm{\infty }$. No finite local Lipschitz constant at zero is possible.

A common-neighborhood generator envelope at zero would provide a measurable full-measure set $E$ and an integrable $K$ such that the envelope inequality holds for every $z\in E$ and all parameters in one neighborhood $U$. Since $\rho (\{n\})=p_n>0$ and $\rho (E^c)=0$, every integer $n$ lies in $E$. For all sufficiently large $n$, $a_n\in U$, and comparison of parameters $0,a_n$ at $z=n$ gives $K(n)\ge \mathrm{\mid }{g}_{a_n}(n)-g_0(n)\mathrm{\mid }\mathrm{/}{a}_n=1\mathrm{/}{a}_n=4^n$. By [F7], $\int Kd\rho ={\sum }_n{p}_{n}K(n)\ge {\sum }_{n\ge N}{2}^n=+\mathrm{\infty }$, contradicting integrability. Thus the missing common-neighborhood premise is real. This witness is smooth off the countable set of triangular corners and zero, so it makes no claim to refute almost-everywhere differentiability.