# 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 $x \mapsto f \left({t, x}\right)$ 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:

$\displaystyle D_t \map h t = \int \frac \partial {\partial t} \map f {t, x} \map {\rd \mu} x$