Question about proof of interchanging integral and derivative using uniform convergence. Theorem: Assume that

*

*$\varphi(x,t):[a,b]\times[c,d]\to \Bbb R$

*$\varphi(x,t)\in R[a,b]$ ( $\varphi(x,t)$ is Riemann integrable on $[a,b]$ )$\ \ \forall t\in [c,d]$. i.e. $$\int_a^b\varphi(x,t)dx$$ exists $\ \forall t\in [c,d]$.

*Let $c\lt s\lt d $ and $\varepsilon \gt 0$, there exists $\delta \gt 0 $ s.t. $$\lvert \; \frac{\partial\varphi}{\partial t}(x,t) - \frac{\partial\varphi}{\partial t}(x,s)\;\rvert \lt \epsilon$$ for all $x\in [a,b]$, $t\in (s-\delta,s+\delta)$.

Define $$f(t)=\int_a^b\varphi(x,t)dx$$ , $t\in[c,d]$ .
Then $$\frac{\partial\varphi}{\partial t}(x,s)\in R[a,b]$$ and $$f'(s)=\int_a^b\frac{\partial\varphi}{\partial t}(x,s)dx$$
My attempt: Define $$\Psi(x,t)=\frac{\varphi(x,t)-\varphi(x,s)}{t-s}$$ for $0\lt\lvert t-s \rvert\lt\delta$.
By MVT, $\exists u$ between $s$ and $t$ s.t. $$\Psi(x,t)=\frac{\partial\varphi}{\partial t}(x,u)$$
So $$\lvert \Psi(x,t)-\frac{\partial\varphi}{\partial t}(x,s)  \rvert=\lvert\frac{\partial\varphi}{\partial t}(x,u)-\frac{\partial\varphi}{\partial t}(x,s)  \rvert\lt\varepsilon\ \forall x\in [a,b] \;\; \color{blue}{(\star)}$$ by assumption 3.
$\color{darkred}{\textrm{My question}}$: By $\color{blue}{(\star)}$, can I conclude that $$\Psi(x,t)\to\frac{\partial\varphi}{\partial t}(x,s) \text{ uniformly on } [a,b] \text{ as } t\to s  \;\;\color{darkorange}{(\star)}\text{ ? }$$ be the limit function.
I have learned uniform convergence for sequence of functions like $\{ f_n\}\to f$ uniformly means for any $\varepsilon \gt 0$, there exist $N\in \Bbb N$ s.t.  for any $n\gt N$ $\lvert f_n(x)-f(x) \rvert\lt\varepsilon\ \forall x\in X$ ( if $f$ is defined on $X$ ). How can I find such $N$ in $\color{darkorange}{(\star)}$ ?
Suppose $$\varepsilon=\frac1n$$ $(\forall n\in\Bbb N)\ \exists\delta_n$ in assumption 3 s.t. $$t\in (s-\delta_n,s+\delta_n)$$
which means $t$ getting closer and closer to $s$ as $n$ getting greater. i.e. $t\to s$ as $n\to\infty$
Can I let $$\Psi(x,t)=f_n(x)$$ in $\color{darkorange}{(\star)}$ ,where $n$ depends on $\delta_n$ related to $t$.
Let $$\frac{\partial\varphi}{\partial t}(x,s)=f(x)$$
Then $$f_n(x)\to f(x)\text{ uniformly on }[a,b]\text{ as }t\to s$$
Is my proof correct? Thanks for helping.
 A: Of course $\Psi(x,t)\to\frac{\partial\varphi}{\partial t}(x,s)$ as $t \to s$ uniformly on $[a,b]$. You have shown that there exists $u$ between $t$ and $s$ such that
$$\left|\Psi(x,t)-\frac{\partial\varphi}{\partial t}(x,s)  \right|=\left|\frac{\partial\varphi}{\partial t}(x,u)-\frac{\partial\varphi}{\partial t}(x,s)  \right|$$
Since $|u-s| \leqslant |t-s|$ it follows from condition (3) that there exists $\delta$ such that if $|t-s| < \delta$, then for all $x \in [a,b]$ we have
$$\left|\Psi(x,t)-\frac{\partial\varphi}{\partial t}(x,s)  \right|=\left|\frac{\partial\varphi}{\partial t}(x,u)-\frac{\partial\varphi}{\partial t}(x,s)  \right|< \varepsilon$$
Proving that interchanging the derivative and integral is valid is straightforward and requires no further complication with regard to uniformly convergent sequences.  Clearly, when $|t-s| < \delta$ we have
$$\left|\frac{f(t)-f(s)}{t-s}- \int_a^b \frac{\partial\varphi}{\partial t}(x,s)  \, dx\right| \leqslant \int_a^b \left|\frac{\partial\varphi}{\partial t}(x,u)-\frac{\partial\varphi}{\partial t}(x,s)  \right| \leqslant \varepsilon (b-a),$$
and it follows that
$$f'(s) = \lim_{t \to s} \frac{f(t)-f(s)}{t-s} = \int_a^b \frac{\partial\varphi}{\partial t}(x,s)  \, dx$$
All of this depends on the fact that $x \mapsto \frac{\partial \varphi}{\partial t}(x,t)$ is Riemann integrable on $[a,b]$ which you are also asked to prove. I leave that for you to consider.
