Maximal domain of function and its derivative if both have Lebesgue > 0

Question

Consider $f(x) = \ln(x), \sqrt{x}, |x|$ and $\lfloor x \rfloor$.

Based on these, I have the following conjecture. Please prove/disprove.

The maximal domain of $f$ and its derivative differ only by a set of Lebesgue measure zero

if

the maximal domain of $f'(x)$ has positive Lebesgue measure.

Now, the Weierstass function's maximal domain is $\mathbb R$, and its derivative's maximal domain is $\emptyset$, which is zero, so it's okay that the difference, which is $\mathbb R$, does not have Lebesgue measure zero.

To disprove I think of $f(x)$ which has maximal domain $\mathbb R$ and whose derivative is $f'(x)$ has maximal domain $[0,1]$. Could we construct some kind of piecewise weierstrass where $f$ is weierstrass outside $[0,1]$ and then some smooth function like $x^2$ inside $[0,1]$?

Answer 1

Let $$\phi(t)=\begin{cases}t^{2}&t>0\\0&t\le 0\end{cases} $$ Note that $\phi'$ exists and is still continuous.

Given two differentiable functions $f$ and $g$, we can consider $$ h(x)=\frac{\phi(x-1)f(x)+\phi(1-x)g(x)}{\phi(x-1)+\phi(1-x)}$$ Note that the denominator is always positive. Then $h(x)=f(x)$ and $h'(x)=f'(x)$ for $x>1$, whereas $h(x)=g(x)$ and $h'(x)=g'(x)$ for $x<-1$.

Answer 2

This is false, even if $f$ is continuous. (Your construction is "backwards".)

Let $W$ be a nowhere differentiable continuous function (with domain containing $[0,1]$), and set $$ f(x) = \begin{cases} W(0) ,& x \leq 0, \\ W(x) ,& 0 \leq x \leq 1, \\ W(1),& 1 \leq x \end{cases} \text{.} $$

Then the maximal domain of $f'$ is $(-\infty, 1] \cup [1,\infty)$, which has positive Lebesgue measure. However, the maximal domain of $f$ is $\mathbb{R}$ which differs from the previous set by $(0,1)$ having Lebesgue measure $1$.