Show that if $g$ is continuous, then: \begin{align*} \frac{\partial}{\partial t} \int_{-L}^L g(x,t) \, dx &= \int_{-L}^L \frac{\partial g}{\partial t} \, dx \\ \end{align*}
My PDE textbook presents this as a given, and I'd like to see the proof.
By @cmk's advice, I can apply the difference quotient definition of the derivative to yield:
\begin{align*} \lim\limits_{h \to 0} \frac{1}{h} \left[ \left( \int_{-L}^L g(x,t+h) \, dx \right) - \left( \int_{-L}^L g(x,t) \, dx \right) \right] &= \int_{-L}^L \lim\limits_{h \to 0} \frac{1}{h} [g(x,t + h) - g(x,t)] \, dx \\ \end{align*}
I'm not sure how to go from there.