"minimal" conditions for Leibniz Rule for changing the integral and derivative I am trying to find "minimal" conditions for Leibniz Rule: $$\dfrac{d}{dx}\left(\displaystyle\int_{a}^{b}f(x,y)\;dy\right)=\displaystyle\int_{a}^{b}f_x(x,y)\;dy.$$
Let $$F(x)=\displaystyle\int_{a}^{b}f(x,y)\;dy,$$ which is a single variable function. 
Let us take the derivative of $F(x)$ using the definition of derivative: $$\dfrac{d}{dx}F(x)=\lim\limits_{h\to0}\dfrac{\displaystyle\int_{a}^{b}f(x+h,y)dy-\displaystyle\int_{a}^{b}f(x,y)dy}{h}$$
$$=\lim\limits_{h\to0}\dfrac{\displaystyle\int_{a}^{b}(f(x+h,y)-f(x,y))dy}{h}$$
$$=\lim\limits_{h\to0}\displaystyle\int_{a}^{b}\dfrac{f(x+h,y)-f(x,y)}{h}dy.$$
To rest of prove, I want to continue with definition of limit: 
Given $\epsilon>0$. I need to find $\delta >0$ such that if $$|h|<\delta,$$ then $$\left|\displaystyle\int_{a}^{b}\dfrac{f(x+h,y)-f(x,y)}{h}dy-\displaystyle\int_{a}^{b}f_x(x,y)dy\right|$$$$=\left|\displaystyle\int_{a}^{b}\left[\dfrac{f(x+h,y)-f(x,y)}{h}-f_x(x,y) \right]dy\right|<\epsilon.$$
I know that $$\left|\displaystyle\int_{a}^{b}\left[\dfrac{f(x+h,y)-f(x,y)}{h}-f_x(x,y) \right]dy\right|\le \displaystyle\int_{a}^{b}\left|\dfrac{f(x+h,y)-f(x,y)}{h}-f_x(x,y) \right|dy.$$ 
Therefore, I want to continue with this inequality. 
If I will find a $\delta>0$ for all $y\in[a,b]$ such that $$\left|\dfrac{f(x+h,y)-f(x,y)}{h}-f_x(x,y) \right|<\frac{\epsilon}{b-a}$$ whenever $|h|<\delta$, then the  proof will complete. 
First Attempt: (Uniformly Continuous) If $f_x(x,y)$ is continuous on $C\times[a,b]$ where $x\in C$ is a closed interval having more than one element, then it will be uniformly continuous and so we can find such $\delta>0$.
Second Attempt: If $f_x(x,y)$ is piece-wise continuous on $y$, we can separate the integral and make the pieces continuous and then apply the First Attempt to the pieces.
(I wrote that proof by myself. Am I correct with this proof; and are the Attempts correct?)
How can I make the conditions weaker? (Please don't say dominated convergence theorem directly, I am trying to understand proofs).
Thanks for help in advance.
 A: Towards the weakest requirements, provided that $f_x(x,y)$ exists, for every $(x,y)$, and $f_x(x,\cdot)$ is Riemann integrable over $[a,b]$.
Observe that
$$
\frac{1}{h}\bigg(\int_a^b f(x+h,y)\,dy-\int_a^b f(x,y)\,dy\,\bigg)-\int_a^b f_x(x,y)\,dy\\ =\frac{1}{h}\int_a^b\int_x^{x+h} \big(\,f_x(t,y)-f_x(x,y)\big)\,dt\,dy,
$$
and hence
$$
\bigg|\,\frac{1}{h}\bigg(\int_a^b f(x+h,y)\,dy-\int_a^b f(x,y)\,dy\,\bigg)-\int_a^b f_x(x,y)\,dy\,\bigg| \\ \le \frac{1}{h}\int_a^b\int_x^{x+h} \big|\,f_x(t,y)-f_x(x,y)\big|\,dt\,dy \\=
\frac{1}{h}\int_x^{x+h} \int_a^b \big|\,f_x(t,y)-f_x(x,y)\big|\,dt\,dy=
\frac{1}{h}\int_x^{x+h}\|\,f_x(t,\cdot)-f_x(x,\cdot)\|_{L^1[a,b]}\,dt
$$
where $\,\|g\|_{L^1[a,b]}=\int_a^b |g(x)|\,dx$.
So, if
$$
\lim_{h\to 0}\frac{1}{h}\int_x^{x+h}\|\,f_x(t,\cdot)-f_x(x,\cdot)\|_{L^1[a,b]}\,dt=0, \tag{1}
$$
then
$$
\frac{d}{dx}\int_a^b f(x,y)\,dy=\int_a^b f_x(x,y)\,dy.
$$
Note that, Condition $(1)$ is significantly weaker than the condition
$$
\lim_{h\to 0}\|\,f_x(x+h,\cdot)-f_x(x,\cdot)\|_{L^1[a,b]}=0,
$$
which in turn, is significantly weaker than the uniform continuity of $f_x$.
