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I am looking for a citable reference for the result on differentiation under the integral sign for integration against a measure.

The result states that if $R \subset \mathbb R$, $(X,\mathcal F, \mu)$ is a measure space, and $f: R \times X \rightarrow \mathbb{R}$ satisfies:

1) $x \mapsto f(t,x)$ is integrable for all (fixed) $t \in R$.

2) $t \mapsto f(t,x)$ is almost everywhere differentiable for (fixed) $x \in X$.

3) There exists integrable $g: X \rightarrow \mathbb R$ such that

$$ | \partial_t f(t,x) | \leq g(x) $$

Then the function

$$F(t) = \int_X f(t,x) \mu(dx) $$ is differentiable with derivative

$$F'(t) = \int_X \partial_t f(t,x) \mu(dx) $$

user101427
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  • What is E? Is it another subset of the reals? –  Nov 01 '13 at 12:08
  • @Eupraxis1981, sorry that was a typo: $E$ should have been $X$, and $X$ can be an arbitrary measure space. – user101427 Nov 01 '13 at 16:41
  • This is not exactly right as it stands. If $t \mapsto f(t,x)$ is only almost everywhere differentiable, then the most you could expect would be that $F$ is almost everywhere differentiable. (Suppose $X$ has only one point.) And I'm not completely convinced that is always true. If $t \mapsto f(t,x)$ is everywhere differentiable then you get your result. – Nate Eldredge Nov 01 '13 at 18:03
  • @nate: The differentation lemma only requires differentiability-a.e, not for all x. –  Nov 01 '13 at 18:41
  • @Eupraxis1981: So what about my example? Let's say $X$ is any probability space, $g(t)$ is a function which is differentiable almost everywhere but not everywhere, with $g'$ bounded, and $f(t,x)=g(t)$. Then $F(t)=g(t)$ as well, which is clearly not "differentiable" (everywhere). – Nate Eldredge Nov 01 '13 at 20:32
  • @NateEldredge: If the domain of X is a single point, then f(t,x) reduces to a single-variable problem $f'(t)$ defined on the real line, in which case it only needs to be almost everywhere differentiable in t, in the Radon-Nikodym sense, for the theorem to hold. –  Nov 01 '13 at 20:39

1 Answers1

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I found the answer to your question on p.142 of "Probability Theory: A Comprehensive Course, Ed. 2" by Klenke. Its called the Differentiation Lemma in that text. You've stated, almost word for word, the pre-conditions of that lemma. Here is a link that describes the conditions on exchanging integration and diffeentiation.

  • That's great. Cheers! I found the statement in a set of lecture notes, so I'm guessing that they were based on Klenke's book. – user101427 Nov 01 '13 at 17:52
  • I'd say Klenke is citing a common theorem. He certaintly did not come up with it...it's rather "old hat" measure theory. For example, no one cites the primary source of the Fundamental Theorem of Calculus. –  Nov 01 '13 at 17:54
  • Sure, I'm aware that this is a classical result... just couldn't find a reference that was reputable enough to cite! – user101427 Nov 01 '13 at 19:30
  • I see, my apologies. I misinterpreted what you meant by "based on Klenke's book". –  Nov 01 '13 at 19:33
  • I've looked up the Klenke version, which requires $f$ to be differentiable everywhere in $R$ for almost every $x \in X$ This is because we need mean value theorem. I'm not sure the conditions are the same as OPs – Mark Dec 20 '14 at 22:35