If I choose my $X=\{(x_{1},x_{2},..,x_{n})\in \mathbb{R}^n| \ x_{1}+x_{2}+...+x_{n}=1\}$, and $\int_{\Omega}(f(x,\omega))dP(\omega)=\int_{\Omega}\bigg[max\bigg(\sum_{i=1}^{n}x_{i}(t-r_{i}),0\bigg)\bigg]^\alpha dP(\omega)$ where $\alpha \geq 1$, $t$ be a constant and $r_{i}'s$ are random variables.
Will the differentiation(with respect to $x_{i}$) under integration theorem work here? Because my X is not open subset of $\mathbb{R}^n$. Theorem says for $f:X\times\Omega \longrightarrow \mathbb{R} $, X should be a open subset of $\mathbb{R}^n$. But I have seen that people used this theorem for this case. What is the reason behind this ?