# 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 $N$th jump is less than $\nu$. Then the jump function $J_0$ corresponding to all jumps past the $N$th 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.