I am stuck on a remark in my lecture notes concerning switching the order of differentiation and limits of a sequence of functions. The below proof uses the fact that, under the condition that the convergence is uniform, the limit of $\int_a^x f_n(t) d t$ equals $\int_a^x f(t) d t$ and that a sequence of continuous functions converge to a continuous function.
Theorem. Let $\left(f_n\right)_1^{\infty}$ be a sequence of continuously differentiable functions in the interval $E$. Assume that the sequence is pointwise convergent to the function $f$ and that the differentiated sequence $\left(f_n^{\prime}\right)_1^{\infty}$ is uniformly convergent to the function $g$ in $E$. Then $f$ is continuously differentiable in $E$ and $f^{\prime}=g$.
Proof: Let $a$ be a point in $E$. We have that $$ \int_a^x f_n^{\prime}(t) d t=f_n(x)-f_n(a), \quad x \in E, $$ for $n=1,2, \ldots$. Now, letting $n \rightarrow \infty$, we get $$ \int_a^x g(t) d t=f(x)-f(a), \quad x \in E . $$ $g$ is continuous [since $\left(f_n^{\prime}\right)_1^{\infty}$ converges uniformly] and it follows from this that the function $f$ is differentiable in $E$ with $f^{\prime}(x)=g(x)$.
Remark. In the proof, the uniform convergence is used only in the interval $[a,x]$. This implies that the conclusion of the theorem is true also under the following weakend assumption: For each $x\in E$, there is an open interval $E_x\ni x$ such that $\left(f_n^{\prime}\right)_1^{\infty}$ is uniformly convergent in $E_x$.
The "open interval" in the remark confuses me. I have tried rewriting the proof with this assumption, but I do not know how it would start. Something like:
Let $a\in E$. Then there is an open interval $E_a$. Now how is this useful?
Any help is appreciated.