# Showing that the $L^1$-limit of a sequence of uniformly bounded A.C. functions is A.C.

The problem is the following:

Let $\{f_k\}$ be an uniformly bounded sequence of absolutely continuous differentiable functions on $[0,1]$. Suppose that $f_k\to f$ in $L^1[0,1]$ and that $\{f'_k\}$ is Cauchy in $L^1[0,1]$. Prove that $f$ is absolutely continuous on $[0,1]$.

• There is a subsequence of $\{f_k\}$ that converges almost everywhere to $f$. But even if the convergence is uniform, I would have nothing;

• Since $\{f_k'\}$ is Cauchy and $L^1$ is complete, there is $g\in L^1$ such that $f'_k\to g$ in $L^1$. It seems that there is no way to guarantee that $f'\to g$ uniformly. So I don't know what to do with this.

• What to do with the uniform boundedness of $\{f_k\}$?

What is the starting point to solve this problem? Any hint will be really appreciated.

• What do you know about absolutely continuous functions? Is there a connection between the derivative and the function? If you have $L^1$ convergence, this tells you something about certain integrals. Is there a characterization of absolutely continuous functions using integrals? – PhoemueX Feb 21 '17 at 14:00
• @PhoemueX the FTC: (1) $f$ is A.C if and only if there exists $f'\in L^1$ a.e and $f(x)-f(0)=\int_0^xf'd\lambda$ for $x\in[0,1]$; (2) The total variation is A.C. too; – Filburt Feb 21 '17 at 14:03
• @PhoemueX (3) it maps sets of measure 0 to sets of measure 0 – Filburt Feb 21 '17 at 14:14
• Nice! Now, can you transfer property (1) from the $f_n$ to $f$? – PhoemueX Feb 21 '17 at 14:14
• @Filburt Why your second application of FTC is not correct? – GaC Feb 21 '17 at 15:46

You already know (upon switching to a subsequence, which I will assume in the following) that $f_n (x) \to f(x)$ almost everywhere. Furthermore, $f_n ' \to g$ in $L^1$. In particular, $f_n(x_0) \to f(x_0)$ for some $x_0$. But we have $$f_n(x_0) - f_n (0) = \int_0^{x_0} f_n '(t) dt \to \int_0^{x_0} g(t) dt.$$ Hence, $(f_n (x_0) - f_n (0))_n$ and $(f_n(x_0))_n$ are convergent, and hence so is $(f_n (0))_n$, say $f_n (0) \to y$. But this again yields $$f_n (x) = f_n (0) + \int_0^x f_n ' (t) dt \to y + \int_0^x g(t) dt.$$ Since $f_n (x) \to f(x)$ almost everywhere, this means $$f(x) = y + \int_0^x g(t) dt$$ almost everywhere. Now redefine $f$ so that $f(x) = y + \int_0^x g(t) dt$ everywhere and conclude that this redefined $f$ is absolutely continuous.
This redefinition can not be avoided, since the $L^1$ convergence $f_n \to f$ only determines $f$ almost everywhere.