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By being "infinitely steep" on a set I mean that for each point $x$ in the set we have $\sup\limits_{\delta>0}\text{ }\inf\limits_{y\in(x-\delta,\text{ }x+\delta)\backslash\{x\}}\frac{f(y)-f(x)}{y-x}=\infty$. As a remark, the Cantor function is infinitely steep on the Cantor-Set which is uncountable, but of course has Lebesgue measure $0$.

Jonathan Hole
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1 Answers1

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The Denjoy–Young–Saks theorem gives the stronger result that the left and right lower derivatives can only be $\infty$ on a null set: defining $$D_-f(x):=\liminf_{y\to x^-} \frac{f(y)-f(x)}{y-x}$$ $$D_+f(x):=\liminf_{y\to x^+} \frac{f(y)-f(x)}{y-x}$$ then $\{x\mid D_-f(x)=\infty\}$ and $\{x\mid D_+f(x)=\infty\}$ are null sets. The set of $x$ such that $\liminf_{y\to x} \frac{f(y)-f(x)}{y-x}=\min(D_+f(x),D_-f(x))=\infty$ is therefore a null set.

Here is an cut-down argument for your specific question. For each integer $n\geq 1$ define $$C_n=\{x\mid f(\lambda)\leq f(\mu)\text{ for all }\lambda\in(x-1/n,x)\text{ and }\mu\in(x,x+1/n)\}.$$ Each $C_n$ is measurable because it's a closed set. For any real $q,$ the restriction of $f$ to the set $D_{n,q}$ of density points of $C_n\cap (q,q+1/n)$ is non-decreasing: between two points $\lambda<\mu$ in $D_{n,q}$ there's another density point, which forces $f(\lambda)\leq f(\mu).$ This restriction can therefore be extended to a non-decreasing function $g_{n,q}$ on the open interval $(\inf D_{n,q},\sup D_{n,q})$: define $g_{n,q}(x)=\sup\{f(y)\mid y\in (-\infty,x]\cap D_{n,q}\}.$ Since $g_{n,q}$ is non-decreasing it is a.e. differentiable. Whenever $g_{n,q}'(x)$ exists and $x\in D_{n,q}$ we have $$\liminf_{y\to x} \frac{f(y)-f(x)}{y-x}\leq \lim_{\substack{y\to x\\y\in D_{n,q}}}\frac{f(y)-f(x)}{y-x}=g_{n,q}'(x)<\infty\tag{1}$$

Let $S$ be the set of points with $\liminf_{y\to x} \frac{f(y)-f(x)}{y-x}=\infty.$ Since (1) holds for a.e. $x\in C_n\cap(q,q+1/n),$ each set $S\cap C_n\cap (q,q+1/n)$ is a null set. So $$S=\bigcup_{n=1}^\infty\bigcup_{q\in\mathbb Q} S\cap C_n\cap (q,q+1/n)$$ is a null set.

Dap
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  • Thanks for the answer! In showing that the restriction of f to $D_{n, q}$ is non-decreasing you only really need that between $\lambda, \mu \in D_{n,q}$ there is a point in $C_n$, right? Or am I missing something? Also, what is the definition of density point (you are using)? It seems to me that you only need points $x\in D_{n,q}$ to have the property that any neighbourhood of them contains points to the left and to the right of $x$ that belong to $C_n$. Or do you really need the stronger notion of density points as in Lebesgue’s density theorem? – Jonathan Hole Jan 20 '20 at 18:53
  • For more precise results on which sets can be "sets of infinite differentiability" for various types of functions, see the bottom third of p. 39 to the top 4 lines of p. 40 of Bruckner/Leonard's 1966 survey paper Derivatives and the references I provided in my answer to Characterization of sets of differentiability. Feel free to incorporate any of this into your answer, as I don't have time for the next few days to write anything new. – Dave L. Renfro Jan 20 '20 at 19:33
  • @JonathanHole: it might be overkill, but yes I do mean points where the lower Lebesgue density is $1,$ see for example https://www.encyclopediaofmath.org/index.php/Density_of_a_set. It's not quite enough to just have a point in $C_n$ between any distinct points in $D_{n,q}$ - this argument also needs that $D_{n,q}$ is dense-in-itself (every point of $D_{n,q}$ is in the closure of $D_{n,q}\setminus{x}$) to be able to get the equation (1) relating a lower derivative of $f$ to the derivative of $g_{n,q}.$ Using density points seems like a quick way to take care of these sort of issues. – Dap Jan 21 '20 at 14:53