This will be a short continuation of the last post. Note that in our discussions in the previous post, we assumed throughout that 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 and to be except defined on the left and right sides of each jump discontinuity respectively so that for increasing, . Define the jump function to be the sum of all jumps at and to the left of the point, where 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 is bounded, the sum of these jumps and thus is absolutely and uniformly convergent.
Why did we construct this? Well by design, is continuous and increasing. Basically, the motivation graphically is to “drop down” all the discontinuities so that we get one continuous curve , and then we add the jumps separately in the form of . We can do this because it turns out:
Result 3: exists and equals zero almsot everywhere.
Proof: Define to be the set of for fixed for which the limit superior of the differential . We want to show that . Because the series of jumps converges, we can find for any such that the sum past the th jump is less than . Then the jump function corresponding to all jumps past the th jump changes by less than from to . But differs from by a finite number of summands, so the set for which the limit superior of exceeds differs from by finitely many points. Pick a compact subset of so that when we take out these points, the resulting compact subset is at least half the measure of . For each , we have a neighborhood where . 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 .
So we have , and because can be anything, we have proven to be zero.