D$ and $D\in{\mathscr{L}}_{\lambda}$. The hypothesis implies that $C\subset\overline{D}$ and $D\subset\overline{H_\lambda^+}$. As $\overline{H_\lambda^+}$ is closed, this immediately gives $C\subset\overline{H_\lambda^+}$, hence $C\in{\mathscr{L}}_{\lambda}$.

To show (ii) let $D_{1},D_{2}, \ldots, D_{p}$ be the chambers of ${\mathscr{T}}$ that are larger than $C$ with respect to the partial order $\leq$. If one of them is contained in $\overline{H_\lambda^+}$, then we are done. Otherwise, for every $i\in\{1,2,\ldots,p\}$, we can find a point $x_i\in D_i$ such that $(\lambda,x_i)<0$. Let $\mathcal{H}^{\prime}$ be the subarrangement of $\mathcal{H}$ where we remove all the hyperplanes of $\mathcal{H}$ that contain the cone $C$. In the arrangement $\mathcal{H}^{\prime}$ all the points $x_{i}$ are contained in the same chamber $C^{\prime}$. In particular, the convex hull $P$ of the points $x_{1}, x_{2}, \ldots, x_{p}$ is contained in $C^{\prime}$. The convex hull $P$ intersects the linear span of the cone $C$ in a point $x$. Since all the points $x_{i}$ are in the open half-space $H_\lambda^-$, so is the point $x$, that is, $(\lambda,x) < 0$. By inserting the hyperplanes of $\mathcal{H}$ that contain the cone $C$ back in the arrangement $\mathcal{H}^{\prime}$, we subdivide the region $C^{\prime}$ into regions $C_{1}, C_{2}, \ldots, C_{p}$. But the point $x$ belongs to the closure of each region $C_{i}$, thus the 