$f'(t)$ where $f(t)$ is not differentiable everywhere

Question

1. Can we talk of $f'(t)$ where $f(t)$ is not differentiable everywhere?

Apparently, 'if $f(t)$ and $f'(t)$ are integrable on $\mathbb R$, then $\int_{\mathbb R} f'(t) dt = 0$' is false but true if $f(t)$ is differentiable everywhere. I actually read the quote as saying that $f$ is differentiable everywhere otherwise we can't talk about $f'(t)$ in the first place. So actually $f(x) = |x|$ or $f(x) = \max\{0,x\}$ does have an '$f'(t)$' even though $f$ doesn't have a derivative? Or $f$ has a derivative which is $f'(t)$ but not defined on all $\mathbb R$ thus is not differentiable on all $\mathbb R$?

2. Can $f'(t)$ be integrable on $\mathbb R$ if there are $t$'s where $f'(t)$ is undefined? I have a feeling it's a different for Lebesgue integration.

Answer 1

The definition of a Lebesgue integral does not depened on the values of the integrand on a set of measure zero. Easy examples include any countable (posssibly infinte) set. That means even though $f'(t)$ is not defined at $t=0$ for $f(t)=|t|$, it makes sense to talk about (for example) the integral of $f'(t)$ from -1 to 1. No matter what value you substitute for $f'(0)$, the integral is always the same.

Answer 2

Sure $f(x)=|x|$ has a derivative - it just doesn't have a continuous derivative. Consider

$g(x) = \begin{cases} -1, & \text{if }x<0 \\ 0, & \text{if }x=0 \\ 1, & \text{if }x>0 \end{cases}$

then

$\int_a^b g(x) dx = |b|-|a|$

so $f'(x)=g(x)$. And we can set $g(0)$ equal to any value we like without changing the value of the integral.

Similar if $f(x) = \max(0,x)$ then

$g(x) = \begin{cases} 0, & \text{if }x \le 0 \\ 1, & \text{if }x>0 \end{cases}$

is a derivative of $f(x)$.