# Distributional derivative of absolutely continuous function

In $$\textit{Rudin, Functional Analysis, p. 148}$$, the example 6.14 says that if $$\Omega \subset \mathbb{R}$$ is an interval and $$f$$ is a function of bounded variation which is left continuous at every point, then $$D\Lambda_f = \Lambda_\mu,$$ where $$\mu([a,b]) = f(b) - f(a)$$, $$\Lambda_f$$ and $$\Lambda_\mu$$ are the distributions asociated to $$f$$ and $$\mu$$, and $$D$$ denote the distributional derivative. The proof of this fact is pretty easy using Fubini's theorem. In addition, it says that $$D\Lambda_f = \Lambda_{f'}$$ if and only if $$f$$ is an absolutely continuous function, but there's no proof for this fact. How can we prove it?

If we assume $$\Lambda_{f'} = \Lambda_\mu$$, the proof is easy. Just take $$(a_k,b_k)$$ disjoints intervals on $$\Omega$$ and note $$\sum_k f(b_k) - f(a_k) = \sum_k \mu( (a_k,b_k)) = \sum_k \int_{a_k}^{b_k} f'(t)dt = \int_{\bigcup_k (a_k,b_k)} f'(t)dt.$$ Now, use the absolute continuity of the Lebesgue integral, and find $$\delta > 0$$ such that $$\sum_k b_k - a_k < \delta$$ implies that our integral is $$< \varepsilon.$$ This proves that $$f$$ es absolutely continuous.

For the other implication, I think we must use Radom-Nikodym theorem, but I can't figure it out.

Anyone can help me? Thank you very much.

You are to show the following for any $$\phi \in C_c^\infty(\mathbb{R})$$: $$\langle D\Lambda_f, \phi \rangle = -\langle \Lambda_f, \phi' \rangle = -\int f(x) \, \phi'(x) \, dx \stackrel{*}{=} \int f'(x) \, \phi(x) \, dx = \langle \Lambda_{f'}, \phi \rangle$$
The integration by parts (the equality marked with a star) is valid if and only if $$f$$ is absolutely continuous. The other equalities are just by definitions.