User:Addemf/sandbox/Lebesgue Differentiation (version 3.0)/Absolute Continuity

Anti-Derivatives
If you're reading this then, for years, you probably have been familiar with the following fact. If f is differentiable on $$\Bbb R$$, then $$\int_a^x f' \ dt = f(x)-f(a)$$ where a may be any point.

Because the length-measure integral and the Riemann integral agree on any closed and bounded interval, the same is true when the area function $$\int_{(a,x)}f'$$ is given by length-measure integration.

In the previous lessons, we found that a function of bounded variation is differentiable a.e.  Is it true that a function of bounded variation also satisfies this relationship?

It is easy to find an example where it doesn't. The "unit step function"


 * $$u = \mathbf 1_{[0,\infty)}$$

has derivative equal to 0 everywhere except at $$x=0$$. Therefore $$\int_{(-1,x)}u' = 0$$ for every x. Since this visibly disagrees with being equal to u, we can see that this is a counter-example.

Perhaps the issue is discontinuity? Unfortunately, it turns out that even with the assumption that f is continuous and of bounded variation, this is not enough to ensure the equation $$\int_{(a,x)}f' = f(x)$$.

The example is again the Cantor-Lebesgue function, which is continuous and of bounded variation. As before, we skip the definition of the function and the demonstration that it has all the properties that we claim. Any interested reader can investigate more independently.

But it turns out that this is a function which is uniformly continuous, monotonic (and therefore of bounded variation), has derivative equal to 0 a.e., and yet is not identical to the zero function. Hence this function is not equal to the integral of its derivative.

Absolute Continuity
The above demonstrates that what we are seeking is some property that f might have, which


 * is weaker than differentiability everywhere, and yet
 * is stronger than uniform continuity plus bounded variation.

Let us look somewhat directly at the goal. Let us say that G is the area function, $$G(x)=\int_{(a,x)}f'(x)$$. Our goal is to find conditions under which we could have that $$G(x)=f(x)-f(a)$$.