# total variation measure vs. total variation of its distribution function

Let's say that $f:[0, \infty)\rightarrow \mathbb{R}$ for simplicity, although the question is also intended for a variety of other types of domains. (Probably interval is all that is required.)

Suppose that $f$ is of bounded variation on finite intervals, and cadlag. Then although it is not possible to find a signed measure $\mu$ for which $\mu((a, b])=f(b)-f(a)$, it is possible to do so uniquely on finite intervals within the domain of f of that type. (left open, right closed.) So let us say that we take $(a', b'] \subset [0, \infty)$ and we obtain the measure $\nu$ corresponding to f restricted to $(a', b']$ Is it true that $|\nu|((a', b'])=TV(f|_{[a,b]})$? Here TV means total variation, and the absolute value signs denote total variation norm of a measure. The sup implied on the LHS is clearly more inclusive than the one on the right, which allows only finite interval partitions. How do I go the other way?

• Once this is shown, then by summing the positive and negative variation we would extract as a corollary that the operation of taking total variation and passing between BV functions and signed measures (via distribution function) commutes. In order to deliver what is promised, we use the uniqueness property of the Jordan decomposition for both the function and measure side. Starting with the more demanding measure side, obtain the Jordan decomposition $\mu=\mu_+-\mu_-$. Observe that the minimum property of the measures is apparently stronger than for functions, and use uniqueness. – Jeff Jul 4 '15 at 19:21