Convex function on closed interval: boundary points? $I=[a,b]$, let the function $f:I\rightarrow\mathbb{R}$ be convex.
(1) Is it possible to prove the existence of the  limits:
$$\lim_{x\rightarrow a^+}f(x) \ \ \ \ \ \lim_{x\rightarrow b^-}f(x)$$
If not, can we find a continuous function $g:I\rightarrow\mathbb{R} $ such that $g(x)=f(x)\ \forall x \in int(I)$?
I know that convexity implies that $f$ is continuous on the interior $int(I)$ of $I$ but I do not have any information on what happens at the boundary points $a$ and $b$.
(2) Moreover, how can I use the fact that the $left-derivative$ $f'_{-}$ of a convex function is $increasing$ in order to prove that each element of the following series is negative?
$$\sum_{j=0}^{y-1}\int_{\frac{2\pi j}{yk}}^{\frac{2\pi (j+1)}{ky}}f'_{-}(\frac{z}{k}) \frac{sin(z)}{k}dz$$
 A: More detail than you ever wanted to know :-). 
I am taking the interval to
be $[0,1]$ to simplify some formulae. I am taking for granted the fact that a
convex function is continuous on the relative interior of its domain. (This
is straightforward to establish, but not the focus here.)
I need to establish two results first:
The first result shows that a function with a closed epigraph is
lower semi continuous (lsc.). This is actually an equivalence, but we only need
one direction here. Convexity is not used here.
Suppose $C \subset \mathbb{R}^{n+1}$ is closed and has the property that if
$(x,t) \in C$ then $(x,s) \in C$ for all $s \ge t$ and we define
$c:\mathbb{R}^n \to [-\infty, \infty]$ by
$c(x) = \inf \{ t | (x,t) \in C \}$.
Then $\operatorname{epi} c = C$. Furthermore, $c$ is
lower semi continuous on $A= \{ x | (x,t) \in C \text{ for some } t \in \mathbb{R}^n \}$. Note that $A$ is not necessarily closed and $c(x) = +\infty$ for
$x \notin A$..
To see this, suppose $(x,t) \in C$, then
$c(x) \le t$ and so $(x,t) \in \operatorname{epi} c$. Now suppose
$(x,t) \in \operatorname{epi} c$, then $\phi(x) \le t$. Hence there are $t_n \downarrow \phi(x)$ with
$(x, t_n) \in C$. Since $C$ is closed, we see that $(x,\phi(x)) \in C$ and so
$(x,t) \in C$.
Now suppose $x_n \in A$ and $x_n \to x$. Let $t_n = \phi(x_n)$ and
$t = \liminf_n \phi(x_n)$.
Suppose $t=+\infty$. Then we have trivially that $\phi(x) \le t$.
Suppose $t \in \mathbb{R}$, then
since $(x_n,t_n) \in C$, we see that $(x, t) \in C$, and so
$\phi(x) \le t$.
Finally, suppose $t=-\infty$. Then for any $M$ there is some subsequence
such that $(x_n,M) \in C$ and so it follows that $(x,M) \in C$ and so
$\phi(x) \le M$. Hence $\phi(x)=-\infty = t$.
The second result shows that under some mild conditions, a convex function
is bounded below on a bounded, convex set.
Now suppose $A \subset \mathbb{R}^n$ is a bounded, non empty, convex set and $f:A \to [-\infty, \infty]$ is convex. In addition, suppose there is some $x \in \operatorname{ri} A$ such that $f(x) \in \mathbb{R}$.
Then $f$ is bounded below.
There is no loss of generality in assuming $x \in A^\circ$. Since $f$ is convex,
it is continuous on $A^\circ$ and there is some $\epsilon>0$ such that
$\overline{B}(x,\epsilon) \subset A^\circ$. Since $\overline{B}(x,\epsilon)$ is
compact, $\underline{f} = \min_{y \in \overline{B}(x,\epsilon)} f(y)$ is finite.
Now let $y \in A \setminus \overline{B}(x,\epsilon)$, then
$\underline{f} \le f(x+\epsilon {y-x \over \|y-x\|}) \le f(x) + \epsilon {1 \over \|y-x\|}(f(y)-f(x))$. Rearranging gives
$f(y) \ge f(x) + {\|y-x\| \over \epsilon } ( \underline{f}- f(x))$, and since
$A$ is bounded, we see that $f$ is bounded below.
Now suppose $f$ is as given in the question, then we see that $f$ is bounded
below and if we let $E=\operatorname{epi} f$
and define
$\phi(x) = \inf \{t | (x,t) \in \overline{E} \}$, then $\phi$ is also
bounded below and is lsc. Since $\overline{E}$ is convex, it follows that
$\phi$ is convex too. (The function $\phi$ is known in convex
analysis as the closure of $f$.)
We see that $\phi(x) \le f(x)$ and if $x \in (0,1)$, then $\phi(x)=f(x)$.
To see this, note that there is some $\epsilon>0$ such that $\overline{B}(x,\epsilon) \subset (0,1)$. Let $L=\overline{B}(x,\epsilon) \times \mathbb{R}$, then we have $L \cap \overline{E} = L \cap E$.
One direction follows immediately, so suppose $(y,t) \in L \cap \overline{E}$.
There are $(y_n,t_n) \in E$ such that $(y_n,t_n) \to (y,t)$ and hence by
continuity of $f$, we have $f(y) \le t$, and so $(y,t) \in E$. Hence
$\phi(x)=f(x)$.
If we can show that $\phi$ is also upper semi continuous (usc.) on $[0,1]$, then
$\phi$ is continuous and the result follows.
To see this, note that if $x \in [{1 \over 2} , 1]$ then
$x = (2x-1) + (1-(2x-1)) {1 \over 2}$ and so
$\phi(x) \le (2x-1)\phi(1) + (1-(2x-1)) \phi({1 \over 2})$. Hence
if $x_n \uparrow 1$, then $\limsup_n \phi(x_n) \le \phi(1)$ and so
$\phi$ is lsc. at $x=1$. A similar analysis shows that the same holds
true for $x=0$, and so $\phi$ is continuous. (This result can be
generalised somewhat if the domain can be 'approximated' at the boundary by a suitable
collection of simplices. See, for example, Theorem 10.4 in Rockafellar, "Convex Analysis".)
The above can be generalised slightly to allow $f$ to be $+\infty$ at $0,1$ (the
limits may be $+\infty$, of course, for example, with $x \mapsto {1 \over x}$).

Simpler answer: Here is a proof that works for an interval on the real line.
The function $f$ is not necessarily differentiable everywhere, but we do
have that the slope (secant) is non decreasing (see here).
Consider $x > { 1\over 2}$ and let $d(x) = {f(x)-f({1 \over 2}) \over x-{1 \over 2}}$, and note that $d$ is non decreasing, bounded above by $d(1)$ and hence we have the limit $d^* = \lim_{x \to 1} d(x)$. Since
$f(x) = f({1 \over 2}) + d(x) (x- {1 \over 2})$, we see that
$\lim_{x \to 1} f(x) = f({1 \over 2}) + {1 \over 2}d^*$.
The same sort of analysis applies to the other limit (or consider
$x \mapsto f(1-x)$).
Original answer: (As Julian pointed out, this doesn't answer the question asked.)
The function $f=1_{\{0,1\}}$ is convex but not continuous at the boundaries.
