$(c.1)$ (“Compactness” of Action constraint set)

Each action constraint set $S^{i}$ ($i\in\mathcal{N}$) is a closed and bounded subset of the action space $\mathbb{U}^{i}$ ($i\in\mathcal{N}$) which is itself a finite dimensional vector space. ### $(c.2)$ (Lower semicontinuity of $L$) $L(\xi,u^{1},\dots,u^{N})$ is a.s. jointly lsc in $(u^{1},\dots,u^{N})=:\mathbf{u}$ on $\mathbf{U}:=\mathbb{U}^{1}\times\dots \times \mathbb{U}^{N}$; ### $(c.3)$ (Measurement set is finite + each measurement has positive probability) Each measurement set $\mathbb{Y}^{i},i\in\mathcal{N}$ is finite, with no element receiving zero probability from the probability measure $P$; or equivalently, for each $i\in\mathcal{N}$, the partition set $\mathbf{Y}^{i}$ has a finite number of elements, with each element receiving positive probability from $P$; ### $(c.4)$ ($L$ has a minimum and $L$ is integrable) $L(\xi,u^{1},\dots,u^{N})$ is bounded from below, and $E_{\xi|y^{i}}[L(\xi;u^{1},\dots,u^{N})]$ is finite for every $y^{i}\in\mathbb{Y}^{i},u^{j}\in\mathbb{U}^{j},i,j\in\mathcal{N}$.

Remark

Note that under finite spaces $(c.1)$ implies our policy space $\mathbf{\Gamma}$ is compact which already gives us half of the Weierstrass Theorem.