Let $X$ be an open subset of $\mathbb{R}^{n}$ and $\Omega$ be a measure space. Suppose that the function $f:X\times\Omega \longrightarrow \mathbb{R} $ satisfies the following theorem conditions:
- $f(x,\omega)$ is a measurable function of $\omega$ for each $x\in X$.
- For almost all $\omega \in \Omega$, the derivative $\frac{\partial f(x,\omega)}{\partial x_{i}}$ exists for all $x \in X.$
- There is an integral function $\Theta:\Omega\longleftrightarrow \mathbb{R}$ such that $|\frac{\partial f(x,\omega)}{\partial x_{i}}|\leq \Theta$ for all $x \in X.$ Then $$\frac{\partial }{\partial x_{i}}\int_{\Omega}^{}f(x,\omega)dP(\omega)=\int_{\Omega}^{}\frac{\partial}{\partial x_{i}}f(x,\omega)dP(\omega).$$
I have got a problem where other conditions satisfy except the openness. Can I invoke the theorem where $X$ is closed subset of $\mathbb{R}^n.$
If I choose my $X=\{(x_{1},x_{2},..,x_{n})\in \mathbb{R}^n| \ x_{1}+x_{2}+...+x_{n}=1\}$, and $f(x,\omega)=\bigg[max\bigg(\sum_{i=1}^{n}x_{i}(t-r_{i}),0\bigg)\bigg]^\alpha$ where $\alpha \geq 1$, $t$ be a constant and $r_{i}'s$ are random variables, Will the theorem work now.