Measure 0 of set of points where $f$ is discontinuous Let $f: [a,b] \rightarrow \mathbb{R}$ be an increasing function. Show that $\{x: f \text{ is discontinuous at $x$}\}$ has measure $0$. Hint: Show that $\{x: o(f,x) > \frac{1}{n}\}$ is finite for each integer $n$. Use the fact that given $f: [a,b] \rightarrow \mathbb{R}$ an increasing function, if $x_1, ..., x_n \in [a,b]$ are distinct, then $\sum_{i=1}^n o(f, x_i) \leq f(b) - f(a)$.
Here is my proof:
Suppose for contradiction that that there exists an $N \in \mathbb{N}$ such that for all $n \geq N$, $\{x: o(f,x) > \frac{1}{N}\}$ is infinite. Pick distinct $x_1, ..., x_k \in [a,b]$ where $k$ satisfies $k > N \cdot (f(b)-f(a))$. Then, $\sum_{i=1}^k o(f, x_i) > \sum_{i=1}^k \frac{1}{N} = \frac{k}{N} > \frac{N \cdot (f(b)-f(a))}{N} = f(b) - f(a)$. But this contradicts the fact given in the hint, so our assumption is false and the set must be finite for any $n \in \mathbb{N}$. Since the set is finite for each $n \in \mathbb{N}$, and the condition of the set is equivalent to $f$ being discontinuous at point $x$, so the set has measure $0$. Since the union of measure $0$ sets is also measure $0$, so the set where $f$ is discontinuous at $x$ has to have measure $0$.
I feel like the steps makes sense, but let me know if I'm missing something in the proof. Thanks!
 A: I assume that "$o(f,x)$" means the oscillation of $f$ around $x$. It seems that you understand the overall procedure of the proof, but your presentation could be improved.

Suppose for contradiction that that there exists an $N \in \mathbb{N}$ such that for all $n \geq N$, $\{x: o(f,x) > \frac{1}{N}\}$ is infinite.

You don't need the "for all $n\geq N$" bit. It is also a good idea to start explaining what you are doing, something like "We will follow the hint given in the question. Suppose[...]"

Pick distinct $x_1, ..., x_k \in [a,b]$ where $k$ satisfies $k > N \cdot (f(b)-f(a))$. Then, $\sum_{i=1}^k o(f, x_i) > \sum_{i=1}^k \frac{1}{N} = \frac{k}{N} > \frac{N \cdot (f(b)-f(a))}{N} = f(b) - f(a)$. But this contradicts the fact given in the hint, so our assumption is false and the set must be finite for any $n \in \mathbb{N}$.

This is ok, but you should avoid just referring to "the set". Give it a name! Why not start the whole proof with "Given a positive integer $n$, define $O_n=\left\{x:o(f,x)>\frac{1}{n}\right\}$"?

Since the set is finite for each $n \in \mathbb{N}$, and the condition of the set is equivalent to $f$ being discontinuous at point $x$, so the set has measure $0$.

Substituting every instance of "the set" by the name you gave it makes the presentation clearer. Also there is a small part which is not right:

the condition of the set is equivalent to $f$ being discontinuous at point $x$.

What you really want to say is the following:

Recall that the function $f$ is discontinuous at a point $x$ if and only if there exists some $n$ such that $o(f,x)>\frac{1}{n}$. This means that the set of points of discontinuity of $f$ is the union $\bigcup_{n=1}^\infty O_n$.

You should be careful with the difference: The way you are phrasing it, it seems that $O_n$ is the set of points of discontinuity of $f$, which is not true!

Since the union of measure $0$ sets is also measure $0$, so the set where $f$ is discontinuous at $x$ has to have measure $0$.

Not any union, but countable unions.

Overall you seem to get the idea for the proof, but you should be more careful in your explanations. Do not try to write things in a concise manner if it makes the explanation worse, specially if it allows for a wrong interpretation (and you would probably lose marks for it). Naming your objects is usually preferred. A careful marker would take issue
So here is my suggested improved version (with a few extra changes):

We will follow the hint given in the question.
Given a positive integer $n$, define $O_n=\left\{x:o(f,x)>\frac{1}{n}\right\}$. Suppose for contradiction that there exists an $N \in \mathbb{N}$ such that $O_N$ is infinite.
Pick distinct $x_1, ..., x_k \in [a,b]$ where $k$ satisfies $k > N \cdot (f(b)-f(a))$. Then, $\sum_{i=1}^k o(f, x_i) > \sum_{i=1}^k \frac{1}{N} = \frac{k}{N} > \frac{N \cdot (f(b)-f(a))}{N} = f(b) - f(a)$. But this contradicts the fact given in the hint, so our assumption is false and $O_N$ must be finite for any $N \in \mathbb{N}$.
Since $O_n$ is finite for each $n \in \mathbb{N}$, then it has measure $0$.
Recall that the function $f$ is discontinuous at a point $x$ if and only if there exists some $n$ such that $o(f,x)>\frac{1}{n}$. This means that the set of points of discontinuity of $f$ is the union $\bigcup_{n=1}^\infty O_n$. Since a countable union of measure $0$ sets also has measure $0$, the set of points of discontinuity of $f$ has measure $0$.

