Romanian Mathematical Olympiad

a) Let us consider the functions $f(x) = x$, for all $x \in \mathbb{R}$, and
$$g(x) = \begin{cases} x, & x \in \mathbb{Q} \\ x+1, & x \in \mathbb{R} \setminus \mathbb{Q} \end{cases}.$$
The function $g$ is discontinuous at any real point. Let $a < b < c$ be three arbitrary real numbers. For any sequence $(x_n)_{n \ge 1}$ of rational numbers converging to $b$, we have $\lim_{n \to \infty} g(x_n) = \lim_{n \to \infty} x_n = b \in (a, c) = (f(a), f(c))$.

b) Consider $b \in \mathbb{R}$ a continuity point of $g$. We will prove $g(b) = f(b)$ by contradiction. If $g(b) < f(b)$ then, based on the continuity of $f$ at the point $b$, there is $a < b$ such that $f(a) > g(b)$. Then, for any sequence $(x_n)_{n \ge 1}$ converging to $b$, we have $\lim_{n \to \infty} g(x_n) = g(b) < f(a)$, contrary to the hypothesis. If $g(b) > f(b)$ then, by using again the continuity of $f$ at $b$, we can find $c > b$ such that $f(c) < g(b)$. For any sequence $(x_n)_{n \ge 1}$ converging to $b$, we have $\lim_{n \to \infty} g(x_n) = g(b) > f(c)$, contrary to the hypothesis. So $g(b) = f(b)$. We conclude that $g(x) = f(x)$ at any continuity point $x$ of $g$.

Let $x$ be an arbitrary real number. The monotone function $g$ has lateral limits at $x$. The set of discontinuity points of $g$ is at most countable. So, for any $n \in \mathbb{N}^*$, there are two continuity points of $g$: $u_n \in (x - 1/n, x)$ and $v_n \in (x, x + 1/n)$. Then $\lim_{t \searrow x} g(t) = \lim_{n \to \infty} g(u_n) = \lim_{n \to \infty} f(u_n) = f(x)$ and $\lim_{t \nearrow x} g(t) = \lim_{n \to \infty} g(v_n) = \lim_{n \to \infty} f(v_n) = f(x)$. Thus, $\lim_{t \searrow x} g(t) = \lim_{t \nearrow x} g(t) = f(x)$. From the monotony of $g$, we get $g(x) = f(x)$.