${\mathcal{D}}_h^{\lambda, {\mathbb{H}_{d}}}$. Conversely, the same calculation implies that given a sample $Y$ from ${\mathcal{D}}_h^{\lambda, {\mathbb{H}_{d}}}$, and a vector $Z \in {B_d(\mathbf{0}, \frac 1 4)}$ sampled proportional to $\exp(-s\lambda \|\vz\|)$, $Y + Z$ is a sample from $\widehat{{{\mathcal{D}}_f^{\lambda}}}$. Noting that $Z$ is a sample from a efficiently sample-able log-concave distribution completes the equivalence.

We next argue that ${{\mathcal{D}}_f^{\lambda}}$ and $\widehat{{{\mathcal{D}}_f^{\lambda}}}$ are $\exp(-\Omega(d))$-close as distributions. Since $\widehat{{{\mathcal{D}}_f^{\lambda}}}$ is a conditioning of ${{\mathcal{D}}_f^{\lambda}}$, this is equivalent to showing that nearly all of the mass of ${{\mathcal{D}}_f^{\lambda}}$ lies in $\cup_{\vy \in {\mathbb{H}_{d}}} {B_d(\vy, \frac 1 4)}$. We write $$\begin{aligned}
  Z_{\lambda}^f &= \int_{{B_d(0, R)}} \exp(-\lambda f(\vx)) \;{\mathrm{d}}\vx\nonumber\\
  &\geq \sum_{\vy \in {\mathbb{H}_{d}}} \int_{{B_d(\vy, \frac 1 4)}} \exp(-\lambda f(\vx)) \;{\mathrm{d}}\vx \nonumber\\
  &= \sum_{\vy \in {\mathbb{H}_{d}}} \exp(-\lambda h(\vy)) \cdot \int_{{B_d(\vy, \frac 1 4)}} \exp(-{s}\lambda g(\vx)) \;{\mathrm{d}}\vx\nonumber\\\end{aligned}$$ Let $\widehat{Z_{\lambda}^f}$ denote this last expression. We will argue that $Z_{\lambda}^f \leq (1+\exp(-\Omega(d)))\widehat{Z_{\lambda}^f}$. We write $Z_{\lambda}^f$ as $$\begin{aligned}
  \int_{{B_d(0, R)}} \exp(-\lambda f(\vx)) \;{\mathrm{d}}\vx
  &\leq \sum_{\vy \in {\mathbb{H}_{d}}} \int_{{B_d(\vy, \frac 1 2)}} \exp(-\lambda f(\vx)) \;{\mathrm{d}}\vx + \int_{{B_d(0, R)}} \exp(-\lambda M) \;{\mathrm{d}}\vx \\
    &\leq \underbrace{\sum_{\vy \in {\mathbb{H}_{d}}} \exp(-\lambda h(\vy))\cdot \int_{{{B_d(\vy, \frac 1 2)}}} \exp(-{s}\lambda g(\vx)) \;{\mathrm{d}}\vx}_{\text{(A)}} + \underbrace{\int_{{B_d(0, R)}} \exp(-\lambda M) \;{\mathrm{d}}\vx}_{\text{(B)}} \\\end{aligned}$$ A simple calculation, formalized as Lemma \[lem:incomplete\_gamma\] in Appendix \[app:incomplete\_gamma\] shows 