proposition (Differentiating Under an Integral (non-negative)):
If |∂∂tf(x,t)|≤g(x)∈L1, then letting F(t)=∫f(x,t) dt, ∂∂tF(t):=lim
To justify passing the limit, let h_k \to 0 be any sequence and define \begin{align*} f_k(x, t) = \frac{f(x, t+h_k)-f(x, t)}{h_k} ,\end{align*} so f_k \overset{k\to\infty}\longrightarrow{\frac{\partial f}{\partial t}\,} pointwise.
Apply the MVT to f_k to get f_k(x, t) = f_k(\xi, t) for some \xi \in [0, h_k], and show that f_k(\xi, t) \in L_1.
#todo Examples
exercise (?):