Why it's not true that $\frac{d}{dt}\int f(t,x)dx=\int \frac{\partial }{\partial t}f(t,x)dx$ where $f(t,x)=g(x)sign(t-x)$? Let $g\in \mathcal C^\infty_c (\mathbb R)$ a test function. Set $$f(t,x)=g(x)\text{sgn}(t-x)$$ where $$\text{sgn}(x)=\begin{cases}1&x>0\\0&x=0\\-1&x<0\end{cases}.$$

I have a theorem that says : if 
1) $x\mapsto f(t,x)$ is $L^1$ for all $t$,
2) For a.e. $(t,x)\in \mathbb R^+\times \mathbb R$, $\frac{\partial }{\partial t}f(t,x)$ exist and $\left|\frac{\partial f}{\partial t}(t,x)\right|\leq h\in L^1$ for a certain function $h$, 
then
$$F(t)=\int_{\mathbb R}f(t,x)dx$$
  is derivable and $$F'(t)=\int_{\mathbb R}\frac{\partial }{\partial t}f(t,x)dx.$$

So, of course that $\frac{\partial f(t,x)}{\partial t}$ exist for almost all $t$ and almost all $x$ (since it's derivable on $\mathbb R^2\setminus \{(x,x)\mid x\in\mathbb R\}$ and the derivative is $0$). So, by the theorem, we should have $$F'(t)=0.$$
However, in my solution, they wrote $F'(t)=2f(t)$, and I really don't get how (well, I see that they used the fact that the derivative in distribution sense of the sign function is $2\delta _0$), but in somehow, this mean that the theorem before is not correct anymore because the derivative should be $0$. So what's wrong in the hypothesis here ?
 A: AFAIK there is another condition in the stated "theorem" that we require: that $f(\cdot,x)\in C^1(\Bbb R )$ for a.e. $x\in \Bbb R $, what clearly doesn't holds in your case because $g(x)\operatorname{sign}(\cdot-x)$ is discontinuous for each $x\in \Bbb R $ when $g\neq 0$. 
In view of this it seems that this condition can be weakened to assert that $\partial f(\cdot,x)$ must exists, as a function on it first argument, for a.e. $x\in \Bbb R $ and that it must be Lebesgue-integrable. Then probably in the stated theorem it must be changed the condition

For a.e. $(t,x)\in \mathbb R^+\times \mathbb R$, $\frac{\partial }{\partial t}f(t,x)$ exist

to

For a.e. $x\in \mathbb R$, $\partial f({\cdot},x)$ exist and is Lebesgue-integrable

In any case this condition doesn't hold either for your function because $\partial f(\cdot,x)$ doesn't exists for any chosen $x$ when $g\neq 0$.
Anyway we can check that $F'=2f$:
$$
\begin{align*}
F(t)&=\int_{\Bbb R }g(x)\operatorname{sign}(t-x)\,\mathrm d x\\
&=\int_{\Bbb R }g(t-x)\operatorname{sign}(x)\,\mathrm d x\\
&=\int_{0}^{\infty }g(t-x)\,\mathrm d x-\int_{-\infty }^0g(t-x)\,\mathrm d x\\
&=\int_{-\infty }^tg(w)\,\mathrm d w-\int_t^{\infty }g(w)\,\mathrm d w\\
&=2G(t)
\end{align*}
$$
with the change of variable $w:=t-x$ and where $G$ is some primitive of $g$ (such primitive exists because $g$ is a test function), so $F’=2g$ as expected. However is clear that $\int_{\Bbb R }\partial _t f(t,x)\,\mathrm d x=\int_{\Bbb R }0\,\mathrm d x=0$, or by
$$
\begin{align*}
\int_{\Bbb R }\partial_t[ g(x)\operatorname{sign}(t-x)]\,\mathrm d x&=\int_{\Bbb R }\partial _t[g(t-x)\operatorname{sign}(x)]\,\mathrm d x\\
&=\int_{0}^{\infty }-g'(t-x)\,\mathrm d x-\int_{-\infty }^0g'(t-x)\,\mathrm d x\\
&=\int_{-\infty }^\infty g'(w)\,\mathrm d w\\
&=0
\end{align*}
$$
A: It is not true because ${\partial\over\partial t}f(t,x)$ cannot be identified with a reasonable function of the two variables $t$ and $x$, let alone with a function to which Leibniz' formula can be applied. In fact we have
$$\phi(t):=\int_{\mathbb R} f(t,x)\>dx=\int_{-\infty}^\infty g(x)\,{\rm sgn}(t-x)\>dx=\int_{-\infty}^t g(x)\,1\>dx+\int_t^\infty g(x)(-1)\>dx$$
and therefore
$${d\over dt}\int_{\mathbb R} f(t,x)\>dx=\phi'(t)=2 g(t)\ .$$
