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$I$ is an open interval

Using the following fact to show this:

${\{f_n\}}$ converges pointwise on $I$ to the function $f$, and ${\{f'_n}\}$ converges uniformly on $I$ to the function $g$

Attempt:

$|{f'_n}(x)-g(x)|<\epsilon/(2(b-a))$, and $|{f_n}(a)-f(a)|<\epsilon/2$

Integrate the former equation from a to b to cancel out (b-a) so first equation $<\epsilon/2$

Add the two and then $|{f_n}(x)-f(x)|<\epsilon$

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Integrating $|f'_n - g|$ does not yield $f_n - g$. Instead, try mean value theorem as follows: $$\begin{eqnarray} |f_n(x) - g(x)| &\leq& |(f_n - g)(x)-(f_n -g)(a) |+|f_n(a) -g(a)|\\&\leq& |x-a||f_n'(c)-g'(c)| + |f_n(a) -g(a)|\\&\leq& |b-a||f_n'(c)-g'(c)| + |f_n(a) -g(a)|. \end{eqnarray}$$

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