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Here is a consequence of Lebesgue dominated convergence theorem on differentiation under integral sign.

Function $f(x, t)$ is differentiable at $x_0$ for almost all $t \in A$, and $t \to f(x, t)$ is integrable. Moreover, there exist an integrable function $g(t)$ such that $|(\partial{f}/\partial{x})(x,t)| \le g(t)$. Then we have $$\frac{\mathrm{d}}{\mathrm{d}x} \left( \int_A f(x,t) ~ \mathrm{d}t \right)\bigg|_{x=x_0} = \int \frac{\partial{f}}{\partial{x}}(x_0,t) ~ \mathrm{d}t. $$

Of course, here the word "integrable" means "Lebesgue integrable". But, if we read the word "integrable" as "improper Riemann integrable" and add an assumption that the partial derivative is continuous, then I think the statement is still true. Weierstrass M-test for integrals guarantees uniform convergence of the improper integral and we can interchange the differentiaion and integration. If $f$ is integrable in both of (improper) Riemann and Lebesgue sense, the only gain of interpreting the integral as Lebesgue one is gettig rid of the continuity condition of the partial derivative. Is it right?

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In your right hand side is it supposed to be df/dx(x,t)dt evaluated at x = x0 and not df/dx(x0,t)? –  Michael Smith Dec 5 '12 at 5:07

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