Lebesgue Integral
Definition
\[\mathbb{M}(S \rightarrow \mathbb{C}) := \{ f: S \rightarrow \mathbb{C} \mid f \text{ is measurable} \} .\] \[L^{1}(\mu) := \{ f \in \mathbb{M}(S \rightarrow \mathbb{C}) \mid \abs{f} \text{ is integrable} \} .\]■
Theorem
- $(S, \mathcal{A}, \mu)$,
- measure space
- $I := (a, b) \subseteq \mathbb{R}$,
- \(\{f_{t}\} \subseteq \mathbb{M}(S \rightarrow \mathbb{C}) \cap L^{1}(\mu)\),
For all $(x, t) \in S \times I$, $ \frac{\partial f_{t}}{\partial t}(x)$ exists. If there exists an open interval $J \subseteq I$ and $g \in L^{1}(\mu)$ such that
\[\sup_{t \in J} \abs{ \frac{\partial f_{t}}{\partial t} } \le g \quad \mu \text{-a.e.}\]Then $F$ is differentiable over $J$ and for all $t \in J$
\[F^{\prime}(t) = \int \frac{\partial f_{t}}{\partial t} \ d \mu .\]proof
It is sufficnet that $f$
$\Box$