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$\begin{aligned} G_k &:= \frac{C^+(k) - C^-(k)}{c_s(k)} \\ &= \frac{C(x(k) + \sum_{s' \in S} c_{s'}(k) \tilde{\Delta}_{s'}(k)) - C(x(k))}{c_s(k)} \\ &= \frac{C(x(k) + c_s(k) \hat{\Delta}(k)) - C(x(k))}{c_s(k)}, \end{aligned}$

where

$\hat{\Delta}(k) = \sum_{s' \in S} \frac{c_{s'}(k)}{c_s(k)} \tilde{\Delta}_{s'}(k)$

Since (i) $(\frac{c_s(k)}{c_{s'}(k)})^2 \to 1$ from A4, which implies $\hat{\Delta}(k)$ is bounded, and (ii) $C(\cdot)$ is convex and continuous, by Lemma 1 of [17] $\forall \epsilon > 0, \exists K_2 < \infty$ such that $\forall k \ge K_2$, w.p. 1

$|\hat{C}'(x(k); \hat{\Delta}(k)) - G_k| < \epsilon$

where $\hat{C}'(x(k); \hat{\Delta}(k))$ is the one-sided directional derivative of $\hat{C}$ at $x(k)$ with respect to vector $\hat{\Delta}(k)$ as defined in [17]. We know that there exists a subgradient $sg(x(k)) \in \partial C(x(k))$ such that

$\begin{aligned} \hat{C}'(x(k); \hat{\Delta}(k)) &= sg^T(x(k))\hat{\Delta}(k) \\ &= sg^T(x(k)) \sum_{s' \in S} \frac{c_{s'}(k)}{c_s(k)} \tilde{\Delta}_{s'}(k) \end{aligned}$

As $\frac{c_{s'}(k)}{c_s(k)} \to 1$, there exists a finite $K_3 \ge K_2$ such that $\forall \epsilon > 0$ and $k > K_3$, with probability one

$\left| \frac{C^{+}(k) - C^{-}(k)}{c_{s}(k)} - s g^{T}(x(k)) \sum_{s' \in S} \tilde{\Delta}_{s'}(k) \right| < \epsilon . \quad (18)$

As a consequence, for $k > \max{K_1, K_3}$,

$\begin{align*} E[\hat{g}_{s,i}(k)|x(k)] &= \frac{N_s}{N_s-1} E\left[\frac{C^+(k)-C^-(k)+\mu_s^+(k)-\mu_s^-(k)}{c_s(k)\Delta_{s,i}(k)}|x(k)\right] \\ &= \frac{N_s}{N_s-1} E\left[\frac{E[C^+(k)-C^-(k)|\Delta(k)]}{c_s(k)\Delta_{s,i}(k)}|x(k)\right] \\ &= \frac{N_s}{N_s-1} E\left[\frac{sg^T(x(k))\sum_{s'\in S}\tilde{\Delta}_{s'}(k)}{\Delta_{s,i}(k)}|x(k)\right] + \delta(k) \\ &= \frac{N_s}{N_s-1}\left(\frac{N_s-1}{N_s}sg_i(x(k))\right) + \delta(k) \\ &= sg_i(x(k)) + \delta(k) \end{align*}$