TITLE: Showing uniform continuity
QUESTION [5 upvotes]: 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

REPLY [3 votes]: As you mentioned, $f$ is absolutely continuous, and showing this isn't really harder than showing uniform continuity directly.
If $g$ is integrable and $\varepsilon>0$ is given, there is a $\delta>0$ such that $m(A)<\delta$ implies $\int_A|g|<\varepsilon$.  To see this, you could for instance first take $h$ bounded by $M>0$ such that $\int_a^b|g-h|<\frac{\varepsilon}{2}$, and then take $\delta = \frac{\varepsilon}{2M}$.
Once you have this, you have $|\int_x^y g|<\varepsilon$ whenever $|x-y|<\delta$.  And as mentioned, this extends to showing absolute continuity.  Boundedness of $f'$ would imply the stronger condition of Lipschitz continuity.