Differentiation under the integral sign for Lebesgue integrable derivative The problem is the following:

Let $a,b,c,d \in \mathbb R$ be given such that $a<b$ and $c<d$. 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!
 A: I have found a non-trivial counterexample:
Let $I = [0,1]$. I will construct a counterexample $f \in L^1(I\times I)$. Showing that additional assumptions along the lines of


*

*For every $t_0\in I$ there exists $g\in L^1(I)$ and a neighbourhood $U$ of $t_0$, s.t. for all $t\in U$ we have: 
$$|\partial_1f(t,x)| \le g(x) \qquad\text{almost everywhere}$$


cannot be dispensed with (if we want the claimed equality to hold for all $t$ rather than almost all $t$).
We define $f: I\times I\to \mathbb R_{\ge 0}$ by
$$f(t,x) = \sum_{n=1}^\infty \chi_{I_n}(x) \beta_n(t)$$
where


*

*$I_n = (2^{-n}, 2^{-n+1})$ and $\chi_{I_n}$ denotes the characteristic function of $I_n$,

*$\beta_n:I \to \mathbb R$ is differentiable with

*

*$0\le \beta_n(t) \le \dfrac{2^n}{n(n+1)}$, $0\le \beta_n'(t)$ for all $t\in I$,

*$\beta_n(t) = \begin{cases} 0 & 0\le t \le \dfrac{1}{2n} \\
\dfrac{2^n}{n(n+1)} & \dfrac 1n \le t \le 1 \end{cases}$



i.e. every $\beta_n$ is a monotonic differentiable function which is constant near $t=0$.
For a fixed $x$ at most one summand in the definition of $f(t,x)$ is not equal to zero, so there are no problems with convergence.
Thus we obtain $\partial_1f(t,x) = \sum_{n=1}^\infty \chi_{I_n}(x) \beta_n'(t)$ and - using nonnegativity of all summands
$$
\begin{align}
\int_0^1\int_0^1 |\partial_1f(t,x)|\, dt\, dx 
&=  \int_0^1\int_0^1 \left(\sum_{n = 1}^\infty \chi_{I_n}(x)\beta_n'(t)\right) \, dt\, dx\\
&= \sum_{n=1}^\infty |I_n| (\beta_n(1) - \beta_n(0)) \\
&= \sum_{n=1}^\infty 2^{-n} \frac{2^n}{n(n+1)} \\
&= \sum_{n=1}^\infty \left(\frac1n - \frac1{n+1}\right) \\
&= 1 < \infty
\end{align}
$$
Similarly a quick calculation shows
$$
\int_0^1\int_0^1 |f(t,x)|\, dt\, dx \le 1 < \infty
$$
Therefore $f, \partial_1f \in L^1(I\times I)$. But now we observe
$$
\begin{align}
\int_0^1 \frac{f(1/m, x)-f(0,x)}{1/m} \, dx 
&= m \int_0^1 \left(\sum_{n=1}^\infty \chi_{I_n}(x)[\beta_n(1/m) - \beta_n(0)]\right)\, dx \\
&= m\sum_{n=1}^\infty |I_n|\beta_n(1/m) \\
&\ge m \sum_{n=m}^\infty 2^{-n} \frac{2^n}{n(n+1)} \\
&= m\cdot \frac{1}{m} = 1
\end{align}
$$
for all $m\in \mathbb N$. In particular, with $\partial_1f(0,x) = \sum_n \chi_{I_n}(x) \underbrace{\beta_n'(0)}_{=0\; \forall n} = 0$ we see: 
$$\limsup_{t\to 0} \int_0^1 \frac{f(t, x)-f(0,x)}{t} \, dx \ge 1 \ne 0 = \int_0^1 \partial_1f(0,x) \, dx$$ 
Showing that $f$ is a counterexample as claimed.
A: You need $\partial_1f(t,x)$ be dominated by an integrable function, i.e., you need $|\partial_1f(t,x)|\le g(x)$ for every $x,t$ for you to take the limit $t'\to t$ inside the integral for every $t$.
Just $\partial_1f(t,x)$ being integrable over $[c,d]$ for every fixed $t\in [a,b]$ isn't sufficient.
Edit:
Let me clarify things here. The problem is that you can not straightaway write $$\frac{\mathrm d}{\mathrm dt}\int_c^d f(t,x) \, \mathrm dx 
= \lim_{t'\to t} \frac{1}{t'-t}\int_c^d f(t',x)-f(t,x) \, \mathrm dx $$ unless you have already shown that $\int_c^d f(t,x) \, \mathrm dx$ is differentiable w.r.to $t$.
So, let $t_n$ be a sequence converging to $t$ and let $$g_n(x)=\frac{f(t_n,x)-f(t,x)}{t_n-t}$$ Then $\lim g_n(x)=\partial_1 f(t,x)$ (given).
Then by mean value theorem $$|g_n(x)|\le \sup_{t\in [a,b]}|\partial_1 f(t,x)|\le g(x)$$ so DCT applies to $g_n(x)$.
Therefore $$\lim_{t_n\to t} \frac{1}{t_n-t}\int_c^d [f(t_n,x)-f(t,x)]\mathrm dx$$ exists and is equal to $$ \int_c^d \lim_{t_n\to t}\frac{f(t_n,x)-f(t,x)}{t_n-t}\mathrm dx=\int_c^d \partial_1f(t,x) \, \mathrm dx$$
I have just reproduced here what I once learnt from Folland's Real Analysis book.
