$I$ 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 $g$


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