Extend absolutely continuous function to be bounded and continuous on $[0,1]$ So I am working on the following problem and I am not really sure how to go about approaching it. I always seem to get stuck when trying to start a problem involving the extension of a function to the boundary for some reason and so any tips, hints, comments or even similar problems would be greatly appreciated.

Suppose that $f:(0,1)\rightarrow(-\infty,\infty)$ is abolutely continuous on each closed subinterval of $(0,1).$ If $f'\in L^p(0,1)$ for some $1\leq p\leq \infty,$ prove that $f$ can be extended to be bounded and continuous on $[0,1].$

My attempt at a solution is as follows. So since we are on a finite measure space we automatically have that $$f\in L^q \ \forall q\leq p,$$ but most importantly $q=1.$ We also know that $f\in \textrm{AC}(0,1)$ guarantees $f'$ exists $[m]$ a.e.. Now we have by the Fundamental Theorem of Calculus that for any $x\in(0,1).$
$$f(x)=\int_a^x f'(y) dy + f(a),$$ where $a\in(0,1).$
Now this means
$$|f(x)|=\left| \int_a^x f'(y) dy + f(a)\right|\leq ||f'||_1+|f(a)|<\infty.$$
I can't see how this gives me what I want though, because I still don't know anything about the endpoints.
Thanks in advance for any assistance. 
 A: You already know
$$
f(x)= \int_a^x f'(t)\, dt + f(a)
$$
for all $x,a \in (0,1)$.
Fix some $a$. Now, the right hand side of the above formula also makes sense for $x\in [0,1]$ (why) and defines a continuous function (why), so you are done.
A: I will first show that $\lim_{x\downarrow0}f(x)$ exists. To see this, let $(x_n)_{n\in\mathbb N}$ be a sequence in $(0,1)$ converging to $0$. Let $a\in(0,1)$ be such that $x_n<a$ for all $n\in\mathbb N$. As you already observed, absolute continuity implies that $$f(x_n)=f(a)-\int_{x_n}^af'(t)\,\mathrm dt=f(a)-\int\mathbb 1_{(x_n,a)}(t)\times f'(t)\,\mathrm dt\quad\forall n\in\mathbb N.\tag{$\clubsuit$}$$ Now, the sequence of functions $(\mathbb 1_{(x_n,a)}\times f')_{n\in\mathbb N}$ converges pointwise the function $\mathbb 1_{(0,a)}\times f'$, and the sequence is also uniformly dominated (in absolute value) by the integrable function $f'$. Therefore, Lebesgue's dominated convergence theorem implies that $$\lim_{n\to\infty}\left\{\int_{x_n}^af'(t)\,\mathrm dt\right\}$$ exists and is equal to $$\int_0^af'(t)\,\mathrm dt,$$ which is well-defined because $f'\in L^1(0,1)$ (as you already observed). Together with ($\clubsuit$), this implies also that $\lim_{n\to\infty}f(x_n)$ exists, as claimed.
In a similar manner, it can be shown that $\lim_{x\uparrow 1}f(x)$ exists as well. Therefore,
\begin{align*}
F(x)\equiv\begin{cases}\lim_{t\downarrow0}f(t)&\text{if $x=0$,}\\\phantom{\lim_{t\downarrow0}\,}f(x)&\text{if $x\in(0,1)$,}\\\lim_{t\uparrow 1}f(t)&\text{if $x=1$}\end{cases}
\end{align*}
is a real-valued continuous function on $[0,1]$ extending $f$. Also, since $[0,1]$ is compact, $F$ is bounded.
