What shown below is a reference form the text Analysis on Manifolds by James Munkres.
Theorem
Let be $Q$ be a rectangle in $\Bbb R^n$; let $f:Q\times[a,b]\rightarrow\Bbb R$ be a continuous function. Denote $f$ by $f(x,t)$ for $x\in Q$ and $t\in[a,b]$. Then the function $$ F(x):=\int_a^bf(x,t)dt $$ is continuous on $Q$. Furthermore if $\partial f/\partial x_j$ is continuous on $Q\times[a,b]$ then $$ \frac{\partial F}{\partial x_j}=\int_a^b\frac{\partial f}{\partial x_j}(x,t)dt $$
This formula is called Leibniz's integral rule.
Proof. We show that $F$ is continuous. The rectangle $Q\times[a,b]$ is compact; therefore $f$ is uniformly continuous on $Q\times[a,b]$. That is, given $\epsilon>0$ there is a $\delta>0$ such that $$ |f(x_1,t_1)-f(x_0,t_0)|<\epsilon\,\,\,\text{whenever}\,\,\,|(x_1,t_1)-(x_0,t_0)|<\delta $$ and thus $$ |F(x_1)-F(x_0)|\le\int_a^b|f(x_1,t)-f(x_0,t)|\le\epsilon(b-a) $$ when $|x_1-x_0|<\delta$.
Now we prove the formula. So in calculating the integral and derivatives involved in Leibniz's rule, only the variables $x_j$ and $t$ are involved; all others are held constant. there fore it suffices to prove the theorem in the case where $n=1$ and $Q$ is an interval $[c,d]$ in $\Bbb R$. Let us set, for $x\in [c,d]$, $$ G(x):=\int_a^bD_1f(x,t)dt $$ We wish to show that $F'(x)$ exists and equals $G(x)$. For this purpose we apply (of all thins) the Fubini theorem. WE are given that $D_1 f$ is continuous on $[c,d]\times[a,b]$. Then $$ \int_c^{x_0}G(x)=\int_c^{x_0}\int_a^bD_1f(x,t)=\int_a^b\int_c^{x_0}D_1f(x,t)=\int_a^b[f(x_0,t)-f(c,t)]=F(x_0)-F(c) $$ Then for $x\in[c,d]$ we have $$ \int_c^{x}G(\xi)=F(x)-F(c)$ $$ Since $G$ is continuous by step 1 we may apply the fundamental theorem of calculus once more to conclude that $$ G(x)=F'(x) $$
So I did not understand why it suffices to prove the theorem for the case $n=1$ and thus I ask to explain it more formally or to prove the theorem with not this restriction. In particular I try to make the following two attempt of proof.
MY PROOF ATTEMPT
So we observe that $$ \frac{F(x+h\hat e_i)-F(x)}h=\int_a^b\frac{f(x+h\hat e_i,t)-f(x,t)}hdt $$ for each $x\in Q$ so that the statement would follows showing that $$ \lim_{h\rightarrow 0}\int_a^b\frac{f(x+h\hat e_i,t)-f(x,t)}h=\\ \int_a^b\lim_{h\rightarrow 0}\frac{f(x+h\hat e_i,t)-f(x,t)}h $$ but unfortunately I do not able to do it. Anyway if $D_if(x,t)$ is continuous on $Q\times[a,b]$ then through the same argument of set $1$ for any $\epsilon>0$ there must exist $\delta>0$ such that $$ |D_if(y,t)-D_i(x,t)|<\frac{\epsilon}{b-a} $$ whether $|y-x|<\delta$ so that for any $x\in Q$ let be $C(x,\delta)$ a cubic neighborhood of radius of such $\delta$ centered at $x$. Now by the mean value theorem if $x\in\text{int}\,Q$ there must exist $\lambda\in(0,1)$ such that $$ \frac{f(x+h\hat e_i)-f(x,t)}h=Df(x+\lambda h\hat e_i)\cdot\hat e_i=D_if(x+\lambda h\hat e_i) $$ and thus $$ \Biggl|\frac{F(x+h\hat e_i)-F(x)}h-\int_a^bD_if(x,t)\Biggl|=\\\Biggl|\int_a^bD_if(x+\lambda h\hat e_i,t)-D_if(x,t)\Biggl|\le\int_a^b|D_if(x+\lambda h\hat e_i,t)-D_if(x,t)|\le\int_a^b\frac{\epsilon}{b-a}=\epsilon $$ so that the formula holds when $x\in\text{int}\, Q$ but what's happen if $x\in\text{bd}\,Q$? Indeed in this case the positive or negative limt $$ \lim_{h\rightarrow 0^{\pm}}\int_a^b\frac{f(x+h\hat e_i,t)-f(x,t)}h $$ exists so that I ask if the equality $$ \lim_{h\rightarrow 0}\int_a^b\frac{f(x+h\hat e_i,t)-f(x,t)}h=\\ \int_a^b\lim_{h\rightarrow 0}\frac{f(x+h\hat e_i,t)-f(x,t)}h $$ effectively holds. I point out that I did not study Lebesgue integration and measure theory so that I ask courteously to not use them.
So could someone help me, please?