Let $a<b$ and call $f:[a,b]\to\mathbb C$ absolutely continuous if there is a $h\in L^2((a,b))$ with $$f(x)-f(a)=\int_a^xh(y)\:{\rm d}y\;\;\;\text{for all }x\in[a,b]\tag1.$$ Assume $(f_n)_{n\in\mathbb N}\subseteq L^2([a,b])$ is a sequence of absolutely continuous functions with $f_n(a)=f_n(b)=0$ and $f\in L^2([a,b])$ with $f_n\to f$ in $L^2([a,b])$. Does it follow that $f$ is absolutely continuous?
By assumption, there are $h_n\in L^2((a,b))$ with $f_n(x)=\int_a^xh_n(y)\:{\rm d}y$. Now, my guess is that we need to show that $(h_n)$ is Cauchy in $L^2((a,b))$ and then conclude by completeness. Are we able to do so?
Yes, you need to show that $(h_n)$ is a Cauchy sequence in $L^2$, and no, it's not possible to show that. For example say $(a,b)=(-1,1)$ and $h_n=n\chi_{(0,1/n)}$.
Or, to make it seem more face-palm obvious, note that the trigonometric polynomials are dense in $L^2((0,2\pi))$. (I said "seem": I don't think thiis really gives a simpler counterexampple, because we need to verify that the tps with $p(0)=0$ are dense, which is no easier than the triviality above. But I do think it makes it "clear" that the answer "must be" no...)