Rubin's integral Stark conjecture

Let $(L/K,S,T)$ be a basic triple, let $G=\operatorname{Gal}(L/K)$, and let $r\geq 0$ be an integer. Let $U_{S,T}$ be the group of $S$-units of $L$ that are congruent to $1$ at all places above $T$, let $X$ be the kernel of the augmentation map on the free abelian group generated by the places of $L$ above $S$, and let $\lambda^{(r)}$ be the map induced by the regulator $\lambda$ on $r$th exterior powers. Write $\Omega_{S,T,r}$ for the subgroup of $\bigwedge_0^r U_{S,T}$ supported on characters $\chi$ with order of vanishing $r(\chi)=r$, and let $e_[r]$ be the sum of the corresponding character idempotents. Assume $(\mathbf{St}1)$–$(\mathbf{St}4)$: $S$ contains the ramified places and, in characteristic zero, the infinite places; $S$ contains at least $r$ places splitting completely in $L/K$; $S$ contains at least $r+1$ places; and $U_{S,T}$ is $\mathbb{Z}$-torsion free. Rubin's conjecture. Under these hypotheses,

$$e_[r]\theta_{L/K,S,T}\bigwedge_{\mathbb{Z}[G],\mathrm{tf}}^r X\subseteq\lambda^{(r)}(\Omega_{S,T,r}).$$

This is Rubin's integral refinement of Stark's conjecture, asserting that the relevant Stickelberger element and the integral exterior power of the place module are generated by regulators of suitable $S$-units. The statement is presented as a conjecture attributed to Rubin and is not resolved in the supplied text.