Integrating the Derivative- Filling in the Gaps

This will be a short continuation of the last post. Note that in our discussions in the previous post, we assumed throughout that F was continuous. Let’s drop that assumption and take care of the case of jump discontinuities. Fortunately there aren’t too many: between any jump discontinuity we can squeeze in a distinct rational number, so a bounded increasing function can have at most countably many discontinuities.

Define F(x^-) and F(x^+) to be F except defined on the left and right sides of each jump discontinuity respectively so that for F increasing, F(x^-)\le F(x)\le F(x^+). Define the jump function J(x) to be the sum of all jumps \alpha_n F(x^+_n)-F(x^-_n) at and to the left of the point, where x_n are points of discontinuity. In other words, we are constructing an increasing step functions where the discontinuities are exactly the same points as in the original function. Note that if F is bounded, the sum of these jumps and thus J is absolutely and uniformly convergent.

Why did we construct this? Well by design, F-J is continuous and increasing. Basically, the motivation graphically is to “drop down” all the discontinuities so that we get one continuous curve F-J, and then we add the jumps separately in the form of J. We can do this because it turns out:

Result 3: J'(x) exists and equals zero almsot everywhere.

Proof: Define E to be the set of x for fixed \epsilon for which the limit superior of the differential \frac{J(x+h)-J(x)}{h}>\epsilon. We want to show that m(E)=0. Because the series of jumps \sum\alpha_n converges, we can find N for any \nu such that the sum past the Nth jump is less than \nu. Then the jump function J_0 corresponding to all jumps past the Nth jump changes by less than \nu from J_0(a) to J_0(b). But J differs from J_0 by a finite number of summands, so the set E' for which the limit superior of \frac{J_0(x+h)-J(x)}{h} exceeds \epsilon differs from E by finitely many points. Pick a compact subset of E so that when we take out these points, the resulting compact subset K\subset E' is at least half the measure of E. For each x\in K, we have a neighborhood (a_x,b_x) where J_0(b_x)-J_0(a_x)>\epsilon(b_x-a_x). By compactness, we can pick a finite subcover, and then we can apply our old covering argument to get a disjoint sub-collection for which \sum^n_{j=1}m(I_n)\ge m(K)/3.

So we have \nu > J_0(b)-J_0(a)\ge \sum J_0(b_k)-J_0(a_k)>\epsilon\sum(b_k-a_k)\ge \frac{\epsilon m(E)}{6}, and because \nu can be anything, we have proven m(E) to be zero.

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