Solutions

a) Let $A=\{f(x):x<a\}$. Since $A$ is nonempty and is bounded above by $f(a)$, it has a least upper bound; let $L=\sup A$.

Pick any $\varepsilon>0$. Since $L$ is the least upper bound of $A$ and $f$ is non-decreasing, there exists $x_0<a$ such that $L-f(x_0)<\varepsilon$ and consequently $L-\varepsilon<f(x_0)$. Let $\delta=a-x_0$. For all $y$, we have

$$\begin{align*} 0<a-y<a-x_0=\delta &\implies x_0<y<a\\ &\implies L-\varepsilon<f(x_0)\le f(y)\le L\\ &\implies -\varepsilon < f(y)-L \le 0 < \varepsilon. \end{align*}$$

Hence,

$$\lim_{y\to a^-} f(y)=L.$$

The proof that

$$\lim_{y\to a^+} f(y)$$

exists is analogous.

b) Consider

$$L^- = \lim_{x\to a^-} f(x)$$

and

$$L^+ = \lim_{x\to a^+} f(x).$$

Both limits exist from part a), and there are two possibilities that follow. If these limits do not equal each other, then $f$ does not have a removable discontinuity by definition. On the other hand, if these two limits do equal each other, then they also equal $\lim_{x\to a} f(x)$. In fact, we have

$$\lim_{x\to a} f(x)=L^-\le f(a)\le L^+=\lim_{x\to a} f(x).$$

So $\lim_{x\to a} f(x)=f(a)$ and $f$ is continuous, in which case it cannot have a removable discontinuity.

c) Consider proving the contrapositive. Suppose $f$ is not continuous. Then from b), these limits are not equal to each other, otherwise $\lim_{x\to a}f(x)=f(a)$, which contradicts $f$ not being continuous. Since $f$ is nondecreasing, it ensues that

$$\sup\{f(x):x<a\}=\lim_{x\to a^-}f(x)<\lim_{x\to a^+}f(x)=\inf\{f(x):x>a\}.$$

Hence, $f$ does not take on all values between the limit from below and from above; that is, there is a gap.