Question about baby rudin theorem 5.12 corollary The corollary says if $ f $ is differentiable on $ [a,b] $ then $ f' $ cannot have any simple discontinuities on $[a,b] $. 
I just don't how to prove it.
I think it should be proved on both two cases of simple discontinuities(first type and second type of simple discontinuities).
Thanks in advance.


Let $g \colon (a,b) \to \mathbb{R}$ a function. If $g$ has a simple discontinuity at $c \in (a,b)$, then $g$ doesn't have the intermediate value property.

Let's look at the case of a jump discontinuity. Replacing $g$ with $-g$ if necessary, we can assume that
$$L := g(c^-) < R := g(c^+).$$
Let $\varepsilon = \frac{R-L}{3}$. By definition of the one-sided limits, there are $\delta^-, \delta^+ > 0$ such that
\begin{align}
c - \delta^- < x < c &\implies \lvert g(x) - L\rvert < \varepsilon\qquad\text{and} \\
c < x < c + \delta^+ &\implies \lvert g(x) - R\rvert < \varepsilon.
\end{align}
With $\delta = \min \: \{\delta^-,\delta^+\}$, we thus have
$$g(x) \in (L-\varepsilon, L+ \varepsilon) \cup (R-\varepsilon, R+\varepsilon)$$
for $0 < \lvert x-c\rvert < \delta$. By the choice of $\varepsilon$, we have $L + \varepsilon < R - \varepsilon$, hence there is
$$v \in [L+\varepsilon,R-\varepsilon] \setminus \{g(c)\},$$
and there is no $x \in [c-\delta/2, c+\delta/2]$ with $g(x) = v$, but of course
$$g(c-\delta/2) < L+\varepsilon \leqslant v \leqslant R-\varepsilon < g(c + \delta/2).$$
Thus $g$ doesn't have the intermediate value property.
The argument for a removable discontinuity - that means $g(c^-) = g(c^+) \neq g(c)$ - is proved quite similarly.
Since derivatives have the intermediate value property (theorem 5.12), it follows that derivatives cannot have simple discontinuities.
