A moving uniform support contributes a missing score boundary term

The density is normalized because ${\int }_0^{\theta }{\theta }^{-1}dx=1$. By [F3] and [F4], the constant payoff has $J(\theta )={\int }_0^{\theta }{\theta }^{-1}dx=1$, hence $J^{\mathrm{\prime }}(\theta )=0$. In the representation $X_{\theta }=\theta U$, $H(X_{\theta })=1$ for every $U$, so its pathwise derivative is zero by [F2].

On $0<x<\theta$, $p_{\theta }(x)=1\mathrm{/}\theta$, so [F1] gives $S_{\theta }(x)=-1\mathrm{/}\theta$ almost surely under $P_{\theta }$. Since $H=1$, the score product is the constant $-1\mathrm{/}\theta$ almost surely, and [F5] gives ${\mathbb{E}}_{\theta }[H(X_{\theta })S_{\theta }(X_{\theta })]=-1\mathrm{/}\theta$. This differs from $J^{\mathrm{\prime }}(\theta )=0$, so the fixed-support score identity does not apply.

The integral is $J(\theta )=\theta \cdot {\theta }^{-1}$. Differentiating this product gives the interior density-derivative contribution ${\int }_0^{\theta }H(x){\mathrm{\partial }}_{\theta }({\theta }^{-1})dx=-1\mathrm{/}\theta$ and the moving-endpoint contribution $H(\theta )p_{\theta }({\theta }^{-})=1\mathrm{/}\theta$, where $p_{\theta }({\theta }^{-})$ is the left limit of the density. Their sum is zero, exactly $J^{\mathrm{\prime }}(\theta )$ from step 1.1; the interior score term alone omits this boundary correction.