Take $G: \mathbb{R}^M\rightarrow \mathbb{R}$ with $G(a)\equiv \mathbb{E}_{\mathbb{P}}(\max_{k\in \{1,...,M\}}V_k+a_k)$ for any $a\equiv (a_1,...,a_M)\in \mathbb{R}^M$, where:
A1: $V\equiv(V_1,...,V_M)$ is an $M\times 1$ random vector with absolutely continuous distribution and support on an open subset $\mathcal{V}\subseteq \mathbb{R}^M$ with positive Lebesgue measure.
($\mathbb{E}_{\mathbb{P}}$ denotes expectation using probability measure $\mathbb{P}$).
I want to show that $G$ is lower semicontinuous.
My attempt:
1) Notice that, under A.1, for any $y\in \{1,...,M\}$ $$ \mathbb{P}(V_y+\alpha_{xy}\geq V_k+\alpha_{xk}\text{ }\forall k=1,...,M)= \frac{\partial \mathbb{E}_{\mathbb{P} } (\max_{k\in \mathcal{Y}} V_k+\alpha_{k})}{\partial \alpha_{y}} $$ See for example here for a proof. This shows that $G$ is differentiable at any $a\in \mathbb{R}^M$.
2) Differentiability implies continuity on $\mathbb{R}^M$.
3) Continuity implies lower semicontinuity on $\mathbb{R}^M$.
Correct?
The result is essentially the Leibniz formula with some attention to detail since the $\max$ function is not differentiable everywhere.
Let $\psi(t) = \max(t_1,...,t_n)$, and note that $\psi$ is Lipschitz with rank one and the set of points $E$ where $\psi$ is not differentiable has Lebesgue measure zero.
Assume that $E \|V\| < \infty$ so that the relevant expectations exist.
Let $\phi(x,a) = \psi(x+a)$ from which we get $|\phi(x_1,a_1) -\phi(x_2,a_2)| \le \|x_1-x_2+a_2-a_2\|$.
In particular, note that for any fixed $a$, we have $|{\phi(x+h,a) - \phi(x)}| \le \|h\|$ and that $E_a = \{ x | (x,a) \mapsto \phi(x,a) \text{ is not differentiable with respect to } a \}$ has Lebesgue measure zero.
Since $V$ has an absolutely continuous distribution, we see that $P\{ \omega | V(\omega) \in E_a \} = 0$. (This is what makes the following approach work.)
Then ${G(a+ t e_k) - G(a) \over t} = \int {\phi(V(\omega), a+t e_k)-\phi(V(\omega), a) \over t}dP(\omega)$, and from an application of the dominated convergence theorem we get ${\partial G(a) \over \partial a_k} = \int {\phi(V(\omega), a) \over \partial a_k}dP(\omega)$, and hence ${\partial G(a) \over \partial a} = \int {\phi(V(\omega), a) \over \partial a}dP(\omega)$.
It is straightforward to show that a differentiable function is continuous (Proving differentiable function is continuous.).
If $G$ is continuous at $a$ then it follows that if $a_k \to a$, then $\liminf_k G(a_k)= G(a)$ from which lower continuity follows.