A function $ f: [a,b] \rightarrow \mathbb{R} $ strictly increasing is continous in [a,b]? I am trying to do the exercise number 5 of section 2.6 of the book: A user friendly introduction to lebesgue measure and integration by Gail S. Nelson it is on page 103 and says: Suppose $ f:[a,b] \rightarrow \mathbb{R}  $ is a strictly increasing function. Prove that $f$ is a measurable function.
At the beginning I though that $f$ must be continuous in $[a,b]$. But instead
I divided  my reasoning by cases: if $\alpha < f(a) $ so $f$ is measurable because $ \{x\in [a,b]: \alpha < f(x)\} $ is $[a,b]$ that is measurable. Now if $f(a)\le \alpha < f(b) $ then $ \{x\in [a,b]: \alpha < f(x)\} $ is $ (f^{-1}(\alpha),b] $ that is also measurable. Finally if $ f(b) \le\alpha $ then $ \{x\in [a,b]: \alpha < f(x)\} $ is $\emptyset$ that is measurable. 
Is that correct? 
 A: It is not true that a strictly increasing function is continuous, however it is true that an increasing function is continuous almost everywhere (in fact, the set of discontinuity points must be countable). For instance, consider the function $f : [0,1] \to \mathbb{R}$ given by
$$
f(x) := \begin{cases}
x & \text{if } 0 \leq x \leq \frac{1}{2},\\
2x & \text{if } \frac{1}{2} < x \leq 1.
\end{cases}
$$
As for the measurability of an increasing function, the proof you've given could be written in a more formal manner. To establish measurability, it will be enough to show that the set $$\{x \in [a,b] : f(x) > \alpha\}$$ is measurable for each $\alpha  \in \mathbb{R}$. 
To this end, we fix $\alpha \in \mathbb{R}$ and let $$E_\alpha := \{ x \in [a,b] : f(x) > \alpha\}.$$ If $E_\alpha = \emptyset$, there is nothing to show and we are done. Otherwise, by setting $s := \inf{E_\alpha}$ we find that $E_\alpha$ is an interval. Indeed, we have either $E_\alpha = (s,b]$ or $ E_\alpha = [s,b]$. In any case, $E_\alpha$ is measurable.

Edit: As requested, let's check that $E_\alpha$ must indeed be an interval.  By definition of $s$, we must have $f(s) \geq \alpha$. Hence, there are exactly two possibilities:


*

*Case $f(s) = \alpha$. In this case, we claim that $E_\alpha = (s,b]$. First, let $x \in E_\alpha$ so that $f(x) > \alpha = f(s)$. Since $f$ is increasing, we must have $x > s$ whence $x \in (s,b]$. That is, $E_\alpha \subseteq (s,b]$. Conversely, let $x \in (s,b]$ be given. Because $f$ is increasing, we see that $f(x) \geq f(s) = \alpha$. If $f(x) = \alpha$, then $x$ would be a lower-bound for $E_\alpha$ strictly larger than $s$ -- a contradiction. It follows that $f(x) > \alpha$, i.e. $(s,b] \subseteq E_\alpha$. To summarize, we have $E_\alpha = (s,b]$.

*Case $f(s) > \alpha$ so that, especially, $s \in E_\alpha$. Because $f$ is increasing, we have that $[s,b] \subseteq E_\alpha$. Conversely, for every $x \in E_\alpha$, the definition of $s$ tells us that $x \geq s$. Hence, $E_\alpha \subseteq [s,b]$. To summarize, $E_\alpha = [s,b]$.
So, in either case, $E_\alpha$ is an interval (and thus measurable).
