Differentiability under Integral Sign

Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $\openint a b$ be a non-empty open interval.

Let $f: \openint a b \times X \to \R$ be a mapping satisfying:

$(1): \quad$ For all $t \in \openint a b$, the mapping $x \mapsto \map f {t, x}$ is $\mu$-integrable

$(2): \quad$ For all $x \in X$, the mapping $t \mapsto \map f {t, x}$ is differentiable

$(3): \quad$ There exists a $\mu$-integrable $g: X \to \R$ such that:

$\ds \forall \tuple {t, x} \in \openint a b \times X: \size {\frac \partial {\partial t} \map f {t, x} } \le \map g x$

Then the mapping $h: \openint a b \to \R$ defined by:

$\ds \map h t := \int \map f {t, x} \map {\rd \mu} x$

is differentiable, and its derivative is:

$\ds \dfrac \d {\d t} \map h t = \int \frac \partial {\partial t} \map f {t, x} \map {\rd \mu} x$

Proof

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Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $11.5$