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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:

  1. 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.

  2. 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 $0$?

Syd Amerikaner
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    1 is a stronger condition than the partial derivative being dominated. If all the difference quotients for various $h$ are dominated by the same $g$, then their limit as $h \to 0$ will be as well. – Paul Sinclair Oct 19 '23 at 21:34
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    2 is also a stronger condition, trivially so. $\Bbb R$ is a closed neighborhood of $0$, so if $2$ holds for all closed neighborhoods, it holds for $N = \Bbb R$, and you can just use $g_{\Bbb R}$ as the $g$ in the main result. – Paul Sinclair Oct 19 '23 at 21:37
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    2 is too trivial. I'm pretty sure you've misstated it. Perhaps it should be "for almost all $x_0\in \Bbb R$, there is some neighborhood $N$ of $x_0$ and integrable $g_N$ such that $\left|\frac{\partial}{\partial x}f_x\right|\le g_N$ for almost all $x \in N$. Then $(\star)$ holds." This follows because differentiation only depends upon the local behavior, not the global behavior. – Paul Sinclair Oct 19 '23 at 21:53
  • In fact, it didn't say closed neighborhood, but closed and bounded neighborhood ("for each $C>0$ and ${x\in\mathbb R : \vert x \vert \leq C}$"). But I think this does not change much. Regarding your concern, maybe there is a typo, or, since it's an exercise on applying the dominated convergence theorem, the technical details of almost everywhere were ignored. – Syd Amerikaner Oct 20 '23 at 04:01
  • What do you mean by "depends upon the local behavior, not the global"? And regarding 1.: the difference quotients are bounded for all $h$ that are less than some $C>0$. Or do you mean if we have that the difference quotients are bounded for all $h$ less than some $C>0$, we can use the function that dominates the difference quotients to bind the limit too? – Syd Amerikaner Oct 20 '23 at 04:05
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    You need to make that correction to 2 in your question, instead of leaving it only in the comments. Note that this holds for all $C > 0$. I.e., no matter how big $x$ is, there will be values of $C$ that are bigger, so for any $x$ there is an $N$ big enough to serve as a neighborhood for it. That is all that is necessary. What I meant by local behavior is that for any $x_0$, the derivatives at $x_0$ only depend on $f_x$ for $x$ near $x_0$. You could change functions $f_x$ to be radically different except in some small neighborhood of $x_0$, without affecting the derivatives at $x_0$ at all. – Paul Sinclair Oct 20 '23 at 04:51
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    So the fact that for each $x_0$, domination is only guaranteed in some neighborhood of $x_0$ still allows the main result to be proved at $x_0$. Since $x_0$ is arbitrary, that means the formula holds everywhere. As for 1, of course! If $\phi$ is a function with $|\phi(h)| \le M$ for all $0 < |h| < C$, then $\lim_{h \to 0} |\phi(h)| \le M$ as well. It is a simple result to prove. – Paul Sinclair Oct 20 '23 at 05:03
  • I see, thank you very much. – Syd Amerikaner Oct 20 '23 at 16:36

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