Showing uniform continuity 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
 A: We assume that $\|f'\|_{L^1} = \int_{a}^{b} |f'|\,dt \lt \infty$. Let $A_{n} = \{x \,:\,|f'(x)| \leq n\}$. Put $g_{n} = [A_{n}] f'$, where $[A_n]$ denotes the characteristic function of $A_n$. Then we have $g_{n} \to f'$ almost everywhere, and, as Nate points out in his comment, dominated convergence implies that $\int_{a}^{b} |g_n - f'|\,dt \to 0$ as $n \to \infty$ (the integrand is bounded by the integrable function $2|f'|$).
Now, given $\varepsilon \gt 0$, choose $n$ so large that $\int_{a}^{b} |g_n - f'|\,dt \lt \varepsilon /2$. As $|g_{n}|$ is bounded by $n$, we have that $|\int_{x}^{y} g_{n}(t)\,dt| \leq n|y-x|$. Thus,
$$\left\vert \int_{x}^{y} f'(t)\,dt\right\vert \leq \left\vert\int_{x}^{y} |f'-g_n|\,dt\right\vert + \left\vert \int_{y}^{x} |g_n(t)|\,dt\right\vert \leq \varepsilon/2 + n \cdot |y-x|$$
and for $\delta = \frac{\varepsilon}{2n}$ we get for all $x,y$ with $|y-x| \lt \delta$
that 
$$|f(y) - f(x)| = \left\vert \int_{x}^{y} f'(t)\,dt \right\vert \leq \varepsilon/2 + \delta n \lt \varepsilon$$
which is the very definition of uniform continuity of $f$.
In fact, we get the even more general estimate that for $\mu(E) \lt \delta$ we have $\int_{E} |f'|\,dt \lt \varepsilon$. But that's exactly absolute continuity of $f$.
It is of course not true that an integrable first derivative is bounded in the sup-norm. For instance, for $f(x) = \sqrt{x}$ we have $f'(x) = \frac{1}{2\sqrt{x}}$ which is not bounded  but integrable on $[0,1]$.
A: 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.
