Almost everywhere differentiability implies weak differentiability on the real line, without additional assumption of any integrability?

Question

Let $f : \mathbb{R} \to \mathbb{R}$ be a function differentiable (in the ordinary sense) almost everywhere on $\mathbb{R}$. Denote by $f' : \mathbb{R} \to \mathbb{R}$ be its (a.e defined) ordinary derivative.

Then,

1. Is $f'$ necessarily measurable?
2. Does $f$ have a distributional derivative and does it coincide (a.e.) with $f'$?

I looked at existing posts, but they seem to assume local integrability of $f'$ at least. And here, I am curious about perhaps the most general case.

Could anyone please clarify for me?

Here, I do NOT assume any integrability on $f$ or $f'$. In fact, measurability of $f'$ is not really sure to me..

Answer 1

1. The answer is yes; the difference quotients are measurable and converge a.e. to $f’$.

2. Typically, weak derivatives refer to locally integrable functions, whereas distributional derivatives are distributions. Any locally integrable function has a distributional derivative. In both cases they need to satisfy $\langle f’,\phi\rangle =-\langle f,\phi’\rangle$. For example, $f = \chi_{[0,\infty)}$ is differentiable a.e. and its distributional derivative is a Dirac delta at zero. But you can check that it’s not weakly differentiable, so that’s a counterexample to the claim that differentiability a.e. implies weak differentiability.