If $g$ is the $L^2$-derivative of a function $f\in L^2$, then integrating $g$ gives $f$ If $f(x), g(x) \in L^2(\mathbb{R})$ and 
$\lim\limits_{h\to\infty}\int_{\mathbb{R}}|f_h(x)-g(x)|^2dx=0$,
where  $f_h(x):=\frac{f(x+h)-f(x)}{h}$  for any $h\neq 0$,
show that $f(x)=\int_{[0,x]}g(t)dt+C$  for some constant C. 
 A: Since
$$
\lim_{h\to0}\int_0^x\left|\frac{f(t+h)-f(t)}{h}-g(t)\right|^2\,\mathrm{d}t=0\tag{1}
$$
Hölder's Inequality says
$$
\lim_{h\to0}\int_0^x\left|\frac{f(t+h)-f(t)}{h}-g(t)\right|\,\mathrm{d}t=0\tag{2}
$$
There is no way to derive the continuity of $f$ since changing $f$ on a set of measure $0$ will not affect $(1)$. However, the Lebesgue Differentiation Theorem says that
$$
\bar{f}(x)=\lim_{h\to0}\frac1h\int_x^{x+h}f(t)\,\mathrm{d}t\tag{3}
$$
exists and equals $f(x)$ almost everywhere. Suppose that $\bar{f}\!(a)$ exists. Then
$$
\begin{align}
\int_0^xg(t)\,\mathrm{d}t
&=\int_0^ag(t)\,\mathrm{d}t+\int_a^xg(t)\,\mathrm{d}t\\
&=\int_0^ag(t)\,\mathrm{d}t+\lim_{h\to0}\int_a^x\frac{f(t+h)-f(t)}{h}\,\mathrm{d}t\\
&=\int_0^ag(t)\,\mathrm{d}t
+\lim_{h\to0}\frac1h\int_x^{x+h}f(t)\,\mathrm{d}t
-\lim_{h\to0}\frac1h\int_a^{a+h}f(t)\,\mathrm{d}t\\
&=\int_0^ag(t)\,\mathrm{d}t+\bar{f}\!(x)-\bar{f}\!(a)\tag{4}
\end{align}
$$
Equation $(4)$ implies that if $\bar{f}\!(a)$ exists, then $\bar{f}\!(x)$ exists for all $x$, and
$$
\bar{f}\!(x)=\int_0^xg(t)\,\mathrm{d}t+\underbrace{\bar{f}\!(a)-\int_0^ag(t)\,\mathrm{d}t}_C\tag{5}
$$
Equation $(5)$ implies that $\bar{f}\!(x)$ is absolutely continuous.

Therefore, the result is true for $\bar{f}$, which is absolutely continuous and equal to $f$ almost everywhere.

