It suffices to assume $f’_n-g\overset{L^1}{\to}0$.

Consider large $N$ such that $||f’_n-g||_{L^1}<\infty~(\forall n\ge N)$, put $h_n=f_n-f_N$ and $h=f-f_N$. We have that $h_n\overset{unif}{\to}h$ and $h’_n \overset{L^1}{\to} g-f’_N$.

Since $h’_n$ exists everywhere and is $L^1~(\forall n\ge N)$ the Fundamental Theorem of Calculus holds:

$\frac{h(x)-h(y)}{x-y}=\lim_{n\to\infty}\frac{h_n(x)-h_n(y)}{x-y}\\~~~~~~~~~~~~~~=\lim_{n\to\infty}\frac{1}{x-y}\int^x_y h’_ndt\\~~~~~~~~~~~~~~=\frac{1}{x-y}\int^x_y g-f’_Ndt$

Since $g-f’_N\in L^1$ we have $h’=g-f’_N$ at every Lebesgue point of $g-f’_N$ hence almost everywhere. Finally since $f’_N$ exists everywhere hence whenever $h’=(f-f_N)’$ exists $f’$ must also exist

$\implies f’=(f’-f’_N)+f’_N=h’+f’_N=g$ a.e.

It suffices to assume $f’_n-g\overset{L^1}{\to}0$.

Consider large $N$ such that $||f’_n-g||_{L^1}<\infty~(\forall n\ge N)$, put $h_n=f_n-f_N$ and $h=f-f_N$. We have that $h_n\overset{unif}{\to}h$ and $h’_n \overset{L^1}{\to} g-f’_N$.

Since $h’_n$ exists everywhere and is $L^1~(\forall n\ge N)$ the Fundamental Theorem of Calculus holds:

$\frac{h(x)-h(y)}{x-y}=\lim_{n\to\infty}\frac{h_n(x)-h_n(y)}{x-y}\\~~~~~~~~~~~~~~=\lim_{n\to\infty}\frac{1}{x-y}\int^x_y h’_ndt\\~~~~~~~~~~~~~~=\frac{1}{x-y}\int^x_y g-f’_Ndt$

Since $g-f’_N\in L^1$ we have $h’=g-f’_N$ at every Lebesgue point of $g-f’_N$ hence almost everywhere. Finally since $f’_N$ exists everywhere hence whenever $h’=(f-f_N)’$ exists $f’$ must also exist, and so

$$f’=(f’-f’_N)+f’_N=h’+f’_N=g$$ almost everywhere, as desired.

It suffices to assume $f’_n-g\overset{L^1}{\to}0$.

Consider large $N$ such that $||f’_n-g||_{L^1}<\infty~(\forall n\ge N)$, put $h_n=f_n-f_N$ and $h=f-f_N$. We have that $h_n\overset{unif}{\to}h$ and $h’_n \overset{L^1}{\to} g-f’_N$.

Since $h’_n$ exists everywhere and is $L^1~(\forall n\ge N)$ FTC holds:

$\frac{h(x)-h(y)}{x-y}=\lim_{n\to\infty}\frac{h_n(x)-h_n(y)}{x-y}\\~~~~~~~~~~~~~~=\lim_{n\to\infty}\frac{1}{x-y}\int^x_y h’_ndt\\~~~~~~~~~~~~~~=\frac{1}{x-y}\int^x_y g-f’_Ndt$

Since $g-f’_N\in L^1$ we have $h’=g-f’_N$ at every lebesgue point of $g-f’_N$ hence almost everywhere. Finally since $f’_N$ exists everywhere hence whenever $h’=(f-f_N)’$ exists $f’$ must also exist

$\implies f’=(f’-f’_N)+f’_N=h’+f’_N=g$ a.e.

It suffices to assume $f’_n-g\overset{L^1}{\to}0$.

Consider large $N$ such that $||f’_n-g||_{L^1}<\infty~(\forall n\ge N)$, put $h_n=f_n-f_N$ and $h=f-f_N$. We have that $h_n\overset{unif}{\to}h$ and $h’_n \overset{L^1}{\to} g-f’_N$.

Since $h’_n$ exists everywhere and is $L^1~(\forall n\ge N)$ the Fundamental Theorem of Calculus holds:

$\frac{h(x)-h(y)}{x-y}=\lim_{n\to\infty}\frac{h_n(x)-h_n(y)}{x-y}\\~~~~~~~~~~~~~~=\lim_{n\to\infty}\frac{1}{x-y}\int^x_y h’_ndt\\~~~~~~~~~~~~~~=\frac{1}{x-y}\int^x_y g-f’_Ndt$

Since $g-f’_N\in L^1$ we have $h’=g-f’_N$ at every Lebesgue point of $g-f’_N$ hence almost everywhere. Finally since $f’_N$ exists everywhere hence whenever $h’=(f-f_N)’$ exists $f’$ must also exist

$\implies f’=(f’-f’_N)+f’_N=h’+f’_N=g$ a.e.