Let $(\Omega,\mathcal F,\mu)$ be a measure space and let $f_x:\Omega\rightarrow\mathbb R$ be a measurable function for each $x\in\mathbb R$ such that $\frac{\partial}{\partial x}f_x(\omega)$ exists for (almost) all $\omega\in\Omega$ and for all $x\in\mathbb R$.
If there is an integrable function $g:\Omega\rightarrow\mathbb R$ such that $\left\vert \frac{\partial}{\partial x}f_x\right\vert\leq g$ (almost everywhere) for all $x\in\mathbb R$, then $$\frac{\mathrm d}{\mathrm dx}\int_\Omega f_x\,\mathrm d\mu = \int_\Omega \frac{\partial}{\partial x}f_x\,\mathrm d\mu\tag{$\star$}.$$ This is clear and I understand how the Dominated Convergence Theorem is utilized to prove this statement.
However, I have now come across two slightly different formulations of this theorem:
For some $C>0$ there is an integrable function $g:\Omega\rightarrow\mathbb R$ such that $\left\vert\frac{f_{x+h} - f_x}{h}\right\vert \leq g$ for each $\vert h\vert < C$. Then $(\star)$ holds.
For each (closed and bounded) neighborhood $N$ of $0$, there exists an integrable function $g_N:\Omega\rightarrow\mathbb R$ such that $\left\vert\frac{\partial}{\partial x}f_x\right\vert\leq g_N$ (almost everywhere) for all $x\in N$. Then $(\star)$ holds.
The first statement requires existence of a dominating function of the difference quotient only on some neighborhood of $0$. However, the usual statement of the theorem requires global boundedness of the partial derivative. Why is this enough? The second statement seems like a weaker version of the well-known statement since it requires the derivative to be bounded on each (closed) neighborhood of $0$ instead of global boundedness. But why is it enough to check for neighborhoods of $0$?