I've been trying to show that a function $f$ on the real interval $[a,b]$ which satisfies
$ f(x)=f(a)+\int_a^xf'(s)\,ds\qquad\text{($f'$ defined almost everywhere)} $
must be uniformly continuous on $[a,b]$.
Since the condition above is equivalent to absolute continuity I know that I could show what I need by means of the proof that absolute continuity - from its fundamental definition - implies uniform continuity: I have seen a proof of that. However, I would like to show the above without involving another form of continuity in the process.
I know that - since what I have stated is also equivalent to there existing any integrable function in place of f' -the proof should not involve the properties of the derivative. However, in establishing a bound I get only as far as
\left|f(y)-f(x)\right|=\left|\int_x^yf'(s)ds\right|\leq\int_x^y\left|f'(s)\right|\,ds
Is it possible to show that an integrable first derivative (or indeed any integrable function) is bounded in sup norm?
Thank you.
Marko