In Stein's real analysis book, we consider a bounded increasing function $F$ on $[a,b].$ Consequently, we know that the set of discontinuities of $F$ on this interval is countable.

Because $F$ is bounded, we concern ourselves with jump discontinuities. Define $F(x^-)$ to be the limit of $F(y)$ as $y\rightarrow x^-$ and $F(x^+)$ to be the limit from the right.

If $\{x_n\}$ denotes the set of points of discontinuity of $F$, let $\alpha_n=F(x_n^+)-F(x_n^-)$ be the oscillation and $F(x_n)=F(x_n^-)+\theta_n \alpha_n$ for some $\theta$ in $[0,1]$.

Now the book defines the following function: $j_n(x)=0$ if $x<x_n$, $j_n(x)=\theta_n$ if $x=x_n$, and $j_n(x) = 1$ if $x>x_n$. The jump function is now formulated as $J_F(x)=\sum_{n=1}^\infty \alpha_n j_n(x)$. Graphically, this is just a function that is constant on the intervals where $F(x)$ is continuous.

The book then goes on to prove the result that $J_F(x)$ is differentiable and vanishes almost everywhere by proving that the set of $x$ s.t. $\limsup_{h\rightarrow0} \frac{J_F(x+h)-J_F(x)}{h}>\epsilon$ has measure 0.

My question might be a bit naive: Since $J_F(x)$ is constant on the intervals where $F(x)$ is continuous, don't we already know that the derivative has to be zero? And because $J_F(x)$ is not continuous at $\{x_n\}$'s which is a countable set, then it is not differentiable at the $\{x_n\}$'s? And moreover, the $\{x_n\}$'s are singletons and so have meassure 0; consequently, the countable union also has measure zero? What am I overlooking here?


1 Answer 1


Most of what you said is good, but the problem lies in an implicit assumption in the first sentence:

"Since $J_F(x)$ is constant on the intervals where $F(x)$ is continuous..."

assumes that, at the very least, $F$ has any intervals on which it is continuous. Now, we know that countable sets can be dense everywhere, so this assumption is problematic. I am not sure whether it can be shown that, given boundedness, that some analogue of this statement can be made rigorous. But I do know that I've never seen a similar argument carried correctly to completion, so I suspect that there is a good counterexample lurking around.

(When I say "most of what you say is good" I mean that if you had some regularity on the discontinuities, like say that the ordering on $\mathbb{R}$ transferred to a well-ordering on the discontinuity set, then all of the things you mentioned are excellent arguments)


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