Measure Theory/Bounded Variation

Bounded Variation
In Lesson 0 of this section, we already introduced functions of bounded variation as an extension of monotone functions. We also showed that every function of bounded variation has an up-down decomposition.

This was all in the hope of proving the integral of the derivative equation, under nice conditions for a given function. We can now state that, at least one nice condition, is for a function to be of bounded variation.

Show that any function, $$f:[a,b]\to\Bbb R$$, of bounded variation is differentiable a.e.

This should be a quick proof from the observation that f has an up-down decomposition, each component function being monotone and therefore differentiable a.e. Therefore the set of points at which either is not differentiable has measure zero, and so on.

This ensures that the f' exists in the equation


 * $$\int_{(a,b)}f' = f(b)-f(a)$$

We'll also wish for a guarantee that f' is integrable, so that we may be assured $$\int_{(a,b)}f'$$ exists. This will occupy our concerns in this lesson.

BV Integrable f'
Let $$f:[a,b]\to\Bbb R$$ be a function of bounded variation, and prove that f' is integrable on [a,b], and


 * $$\int_{(a,b)}f' \le f(b)-f(a)$$

Hint: First show this for any monotonically increasing function.

Consider the set of points at which the derivative exists, and "fill the gaps" by defining g to be equal to the one of the Dini derivatives of f. Therefore $$f=g$$ a.e. and g has a finite Dini derivative everywhere on [a,b]. You can then show that $$\int_{(a,b)} g'$$ exists and infer that $$\int_{(a,b)}g'=\int_{(a,b)}f'$$ at the end.

Now to show that $$\int_{(a,b)}g'$$ exists, use the Dini derivative which you know exists to infer that the sequence of functions,


 * $$g_k(x) = \frac{g(x+1/k)-g(x)}{1/k}$$

converges pointwise to $$g'(x)$$ and $$0\le g'$$.

Then apply Fatou's to infer the desired inequality.

Note that in fact, we not only have the existence of $$\int_{(a,b)}f'$$, but also half of the equality that we're seeking!

Can we prove the reverse inequality? We answer this cliff-hanger in the next lesson.