The following result is established in this paper (Thm. 4 in appendix A.1). I'll reproduce the theorem and the beginning of the proof, which contains the step I don't understand.
Theorem: Let $\mathcal{P}$ be the set of all distributions over $(X_n, X_s, Y)$ that are absolutely continuous with respect to Lebesgue measure and such that the conditional $Y|X_s$ is the same for all distributions. Define $f_s(x_n, x_s) = \mathbb{E}(Y|X_s=x_s)$. Then, $$f_s \in \text{argmin}_{f \in \mathcal{C}^0}\; \sup_{\mathbb{P} \in \mathcal{P}} \mathbb{E}_{(X_s, X_s, Y) \sim \mathbb{P}}[(Y-f(X_n,X_s))^2].$$
Proof: Consider a function $f \in \mathcal{C}^0$ possibly different from $f_s$. We'll prove that, for each distribution $\mathbb{Q} \in \mathcal{P}$, we can construct a distribution $\mathbb{P} \in \mathcal{P}$ such that: $$\int (y-f(x_n,x_s))^2 \; d\mathbb{P} \geq \int (y-f_s(x_n,x_s))^2 \; d\mathbb{Q}$$
...
The remainder of the proof (that I've omitted) only proves that the claim in bold is true. I understand that part of the proof, but why is this claim sufficient to prove the theorem? In other words, I don't understand why the claim in bold is equivalent to (or implies) the result in the theorem.