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.


1 Answer 1


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.



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