## Difference of differentiation under integral sign between Lebesgue and Riemann

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 $$\frac{d}{dx} \left( \int_A f(x,t) dt \right)\bigg|_{x=x_0} = \int \frac{\partial{f}}{\partial{x}}(x_0,t) dt.$$

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 senses, the only gain of interpreting the integral as Lebesgue one is gettig rid of the continuity condition of the partial derivative. Is it right or are there other advantages of Lebesgue integral over Riemann integral for this problem?

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 Yes, of course it is true, since if the integrand is continuous, then the Lebesgue integral reduces to an (improper) Riemann integral. A quick proof is via the FTC and Fubini's theorem; and the the latter is less natural for Riemann integrals. – Pietro Majer Nov 28 at 17:03