Here is a question I had as I read about absolutely continuous functions. If $f:[a,b] \rightarrow \mathbb{R}$ is continuous and real valued, $f'$
integrable on $[a, b],$ and $\int_a^b f' = f(b) – f(a)$, must $f$ be absolutely continuous? What if $f$ is monotone increasing as opposed to continuous?
Intuitively I would think yes to my first question and no to my second but I am not sure. My intuition say yes to the first question since there is a theorem in Royden that tells us that a function $f$ on a closed, bounded interval $[a, b]$ is
absolutely continuous on $[a, b]$ if and only if it is an indefinite integral over
$[a, b].$
Best Answer
About your intuition: The identity $f(b)-f(a)=\int_a^b f'(t)\,dt$ does not say that $f$ is an indefinite integral. For that you would require instead that $f(x)-f(a)=\int_a^x f'(t)\,dt$ for all $x$.
COMMENTS:
You don't specify but here is the everywhere case that perhaps you intended.
Suppose that $f:[a,b]\to \mathbb R$ is everywhere differentiable (so of course it is continuous). Then the following is well-known (one hopes):
If $f'$ is Riemann integrable then $f(x)-f(a)=\int_a^x f'(x)\,dx$ in the sense of the Riemann integral. [But a derivative does not have to be Riemann integrable.]
If $f'$ is Lebesgue integrable then $f(x)-f(a)=\int_a^x f'(x)\,dx$ in the sense of the Lebesgue integral. [But a derivative does not have to be Lebesgue integrable.]
In general $f'$ must be Denjoy-Perron integrable and $f(x)-f(a)=\int_a^x f'(x)\,dx$ in the sense of that integral.
Maybe you are interested instead in the case where $f:[a,b]\to \mathbb R$ is almost everywhere differentiable. There are some tidbits that one should know:
a. There are nonconstant continuous functions that have a derivative equal to zero almost everywhere.
b. Therefore two different continuous functions can have the same a.e. derivative.
c. Given any measurable function $g$ there is a continuous function $f$ so that $f'=g$ almost everywhere.
Ex. Take a continuous function $f$ so that $f(0)=0$, $f(\frac12)=1$ and $f(1)=0$ but with $f'(x)=0$ a.e.. Then $$ 0 = f(1)-f(0) = \int_0^1 f'(x)\,dx = 0 $$ in the sense of the Lebesgue integral and this meets your conditions for a function that is not absolutely continuous.
Your other question is about continuous monotone functions. If $f:[a,b]\to \mathbb R$ is a continuous, monotone nondecreasing function then (you will recall) the $f'$ exists a.e. and is Lebesgue integrable. Moreover $$ \int_a^b f'(x)\,dx \leq f(b)-f(a). \tag{1}$$ You might think that you get strict inequality when $f$ is not absolutely continuous and you get equality only in the case where $f$ is absolutely continuous.
There is a better version of (1) that makes this clearer. Instead of the inequality (1) we can write instead:
$$ f(b)-f(a) = \int_a^b f'(x)\,dx + \lambda_f(D_\infty). \tag{2}$$
Here $\lambda_f$ is the Lebesgue-Steiltjes measure associated with $f$ and $D_\infty$ is the set of points in $[a,b]$ where $f'(x)=+\infty$. The second term is zero if and only if $f$ is absolutely continuous. [This is a theorem of the Belgian mathematician de la Vallée Poussin.] So that is an answer to your second question.
[Dinner time so a bit rushed. If you need references I can find some.]