The problem is the following:
Let $a,b,c,d \in \mathbb R$ be given such that $a and $c
. Suppose $f: [a,b]\times [c,d] \to \mathbb R$ is a function such that $\partial_1 f: [a,b]\times [c,d]\to\mathbb R$ exists and is (Lebesgue-)integrable. Show that for $t\in [a,b]$ we have
$\frac{\mathrm d}{\mathrm dt}\int_c^d f(t,x) \, \mathrm dx = \int_c^d \partial_1 f(t,x)\, \mathrm dx$
I have tried the following: By the fundamental theorem of calculus, we have f(t',x) - f(t,x) = \int_{t}^{t'} \partial_1f(s,x)\, \mathrm ds. It follows that
\begin{align} \int_c^d f(t',x)-f(t,x) \, \mathrm dx &= \int_c^d \int_{t}^{t'} \partial_1f(s,x)\, \mathrm ds \, \mathrm dx \\ &= \int_{t}^{t'} \int_c^d \partial_1f(s,x)\, \mathrm dx \, \mathrm ds \end{align}
so that for almost all $t$, we have
\begin{align} \frac{\mathrm d}{\mathrm dt}\int_c^d f(t,x) \, \mathrm dx &= \lim_{t'\to t} \frac{1}{t'-t}\int_c^d f(t',x)-f(t,x) \, \mathrm dx \\ &= \lim_{t'\to t} \frac{1}{t'-t}\int_{t}^{t'} \int_c^d \partial_1f(s,x)\, \mathrm dx \, \mathrm ds \\ &= \int_c^d \partial_1f(t,x)\, \mathrm dx \end{align}
since $t \mapsto \int_c^d \partial_1f(t,x)\, \mathrm dx$ is an $L^1$ function (hence almost every point $t$ is a Lebesgue-point).
Unfortunately equality for almost every $t$ is not good enough in this case.
I have been thinking about this for quite a while now, but I cannot figure out the reason for why every point $t\in [a,b]$ has to be a Lebesgue point of
$t\mapsto \int_c^d \partial_1f(t,x)\, \mathrm dx$
I have also been thinking about possible counterexamples, but couldn't really come up with one. (At least not if one defines $\frac{\mathrm d}{\mathrm dt} \infty = 0$ for the constant function $\infty$...)
Some help would be appreciated, thanks!