I essentially need to prove Leibniz's Rule for differentiation under an integral: given a continuous function $f(x,y)$ and a continuous partial $\frac{\partial f}{\partial y}$ prove that if $G(y) = \int_a^b f(t,y) \, dt$ then $G'(y) = \int_a^b \frac{\partial f}{\partial y}(t,y) dt$.
There are lots of proofs for this online, but I can't quite understand why they require the partial derivative to be continuous. As far as I can see, I need to prove that
$$\lim_{h \to 0} \frac{G(y+h)-G(y)}{h} = \lim_{h \to 0} \int_a^b \frac{f(t,y+h)-f(t,y)}{h} dt = \int_a^b \frac{\partial f}{\partial y}(t,y) dt$$ which, using the epsilon-delta definition of the limit means that I want to show that for any fixed $\epsilon$ I can find a $\delta > 0$ such that as long as $h < \delta$ $$\left| \int_a^b \frac{f(t,y+h)-f(t,y)}{h} - \frac{\partial f}{\partial y}(t,y) dt \right| < \epsilon$$ At this point most of the proofs I've seen turn to the Mean Value Theorem, and then make use of the continuity assumption, but doesn't the existence of $\frac{\partial f}{\partial y}$ mean that $$\lim_{h \to 0} \frac{f(t,y+h)-f(t,y)}{h} = \frac{\partial f}{\partial y}$$ which means it's relatively easy to select a relevant $\delta$ without having to turn to continuity? Sorry if that was overly verbose, and thanks for any help!!
In general, you cannot interchange the limit and the integral. That is, in general $$ \lim_{h \to 0} \frac{f(t,y+h)-f(t,y)}{h} = \frac{\partial f}{\partial y},\quad a\le t\le b,\tag1 $$ does not imply $$ \lim_{h \to 0} \int_a^b\frac{f(t,y+h)-f(t,y)}{h}\,dt=\int_a^b \frac{\partial f}{\partial y}\,dt.\tag2 $$ When you choose $\delta$ in (1), it will depend on $t$.