Here is my proof for (e): for $x\in int(I)$, let $s^+ (x, t) = \frac{f (x + t) - f (x)}{t} ; s^- (x, t) = \frac{f (x) - f (x - t)}{t}$, then $f^+ (x) = \lim_{t \rightarrow 0} s^+ (x, t)$.
Since $s^+ (x, t) = s^- (x + t, t) \leqslant f^+ (x + t)$, it holds that $f^+ (x) = \lim_{t \rightarrow 0} s^+ (x, t) \leqslant \lim_{t \rightarrow 0} f^+ (x + t)$.
For the inverse inequality, note that : $f^+ (x + t) = \lim_{\beta \rightarrow 0} s^+ (x + t, \beta) \leqslant s^+ (x + t, t) = 2 s^+ (x + 2 t) - s^+ (x + t)$.
Thus $\lim_{t \rightarrow 0} f^+ (x + t) \leqslant \lim_{t \rightarrow 0} (2 s^+ (x + 2 t) - s^+ (x + t)) = f^+ (x)$.
Combine above two inequalities, we get $f^+$ is right continues in $int(I)$.
But it seems that my proof automatically extends to the boundaries when $a\in I$, however i know it's not true.
A case is that for some continues convex function $f$ on $[a,b]$,let $g(x)=f(x)$ when $x \in (a,b]$ and $g(a)=f(a)+1$. Then $g$ is convex but $g^+(a)=-\infty$.
What`wrong with my proof? Did i miss anything?