Suppose $\{g_k(x)\}$ is absolutely continuous on $[a,b]$, and for all $k$,
$$|g_k^{\prime}(x)| \le F(x) \text{ a.e.}$$
where $F(x) \in L^1[a,b]$. If
$$ \lim_{k \rightarrow \infty} g_k(x) = g(x) \text{, and } \lim_{k \rightarrow \infty} g_k^{\prime}(x) = f(x) \text{ a.e.} $$
prove that $$g^{\prime}(x) = f(x) \text{ a.e.}$$
Alright, I'm tired of going in circles on this problem. I know, since $\{g_k(x)\} $ is absolutely continuous on $[a,b]$ that
$$g_k(x) = \int_a^x g_k^\prime(t) \, dt + g_k(a)$$
What I'm trying to do is let $k \rightarrow \infty$ and then use the dominated convergence theorem to show that $g^{\prime}(x) = f(x) \text{ a.e.}$. Can I simply take limits on both sides of the equality...then
$$\lim_{k \rightarrow \infty} g_k(x) = \lim_{k \rightarrow \infty} \left( \int_a^x g_k^\prime(t) \, dt + g_k(a) \right) \Rightarrow$$
$$\lim_{k \rightarrow \infty} g_k(x) = \lim_{k \rightarrow \infty} \left( \int_a^x g_k^\prime(t) \, dt \right) + \lim_{k \rightarrow \infty} g_k(a) \Rightarrow$$
$$\lim_{k \rightarrow \infty} g_k(x) = \int_a^x \lim_{k \rightarrow \infty} g_k^\prime(t) \, dt + \lim_{k \rightarrow \infty} g_k(a) \Rightarrow$$
Then using the hypothesis: $$g(x) = \int_a^x f(x) dt + g(a)$$
But where to go from here? Do I just differentiate both sides and use the dominated convergence theorem or can I just claim the conclusion here via the dominated convergence theorem?
All help is very much appreciated.
Rough Reasoning:
Define the measure $\mu$ on $[a,b]$ as $\mu((c,d))=g(d)-g(c)$, then $\mu((c,d))=\displaystyle\int_{c}^{d}f(x)dx$ as what the reasoning has written, then $\dfrac{d\mu}{dx}=f(x)$ a.e.
However, we know that $\dfrac{d\mu}{dx}=g'(x)$ a.e. by an application of Lebesgue Differentiation Theorem.
Another Reasoning?
We have $\lim_{\delta\rightarrow 0^{+}}\dfrac{1}{|(c-\delta,c+\delta)|}\displaystyle\int_{c-\delta}^{c+\delta}f(x)dx=f(c)$ for a.e. $c$ in $[a,b]$. But $\dfrac{g(c+\delta)-g(c-\delta)}{2\delta}=\dfrac{1}{|(c-\delta,c+\delta)|}\displaystyle\int_{c-\delta}^{c+\delta}f(x)dx$ and $g$ is absolutely continuous on $[a,b]$, so $g$ is differentiable a.e. so $g'(c)=f(c)$ a.e. in $[a,b]$.