A function $f \in L_1(\mathbb{R})$ is said to be absolutely continuous if there exists $g \in L_1(\mathbb{R})$ such that: $$f(b)-f(a) = \int_a^b g(t) dt \tag{1}$$ for all $a, b \in \mathbb{R}$. The function $f$ is $L_1$ differentiable if there exists a $g' \in L_1(\mathbb{R})$ such that: $$\lim_{h \to 0} \int_\mathbb{R} \left \vert \frac{f(x+h)-f(x)}{h} - g'(x)\right\vert dx = 0. \tag{2}$$
Prove that if $f$ is absolutely continuous then $f$ is also $L_1$ differentiable.
My attempt
Let $f \in L_1(\mathbb{R})$ be absolutely continuous. Then there is a $g \in L_1(\mathbb{R})$ which satisfies equation $(1)$. We need to show there is a $g' \in L_1(\mathbb{R})$ which satisifes equation $(2)$. Almost certainly, we expect $g' = g$ to be the function that works. So using this and using the definition of absolutely continuity, plugging this into equation $(2)$ we get: \begin{align} \lim_{h \to 0} \int_\mathbb{R} \left \vert \frac{f(x+h)-f(x)}{h} - g'(x)\right\vert dx &= \lim_{h \to 0} \int_\mathbb{R} \left \vert \frac{1}{h} \int_x^{x+h} g(t)dt - g(x)\right\vert dx\\ &\leq \lim_{h \to 0} \int_\mathbb{R} \frac{1}{h} \int_x^{x+h} \vert g(t) - g(x) \vert dt dx \end{align} Now, if we can interchange the limit with the integral we can then use Lebesgue's theorem to say the function inside the integral is $0$ almost everywhere and thus the integral is $0$, completing the proof. However, I not able to figure out how we can interchange the limit with the integral. I don't see how monotone convergence theorem or how dominated convergence theorem can be applied here.
Could I please get some hints on how I can proceed from here?
The proof presented below is on the interval, but you can figure out how to generalize it to all of $\mathbb{R}$.