Differentiability under Integral Sign
Jump to navigation
Jump to search
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 $x \mapsto f \left({t, x}\right)$ 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:
- $\displaystyle \forall \tuple {t, x} \in \openint a b \times X: \size {\frac \partial {\partial t} \map f {t, x} } \le \map g x$
Then the mapping $h: \openint a b \to \R$ defined by:
- $\displaystyle \map h t := \int \map f {t, x} \map {\rd \mu} x$
is differentiable, and its derivative is:
- $\displaystyle D_t \map h t = \int \frac \partial {\partial t} \map f {t, x} \map {\rd \mu} x$
Proof
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $11.5$