Romanian Mathematical Olympiad - District Round

Suppose, without any loss, that $f$ is an increasing function.

a. Let $x_0 \in \mathbb{R}$. The set $\{f(x) \mid x < x_0\}$ is upper bounded by $f(x_0)$, because $f$ is increasing. Set $L = \sup\{f(x) \mid x < x_0\}$. We claim that $L = f(x_0 - 0)$.
To this end, let $\varepsilon > 0$ and notice that there exists $a < x_0$ such that $f(a) > L - \varepsilon$. Since $f$ is increasing, we have $|f(x) - L| = L - f(x) < \varepsilon$ for any $x \in (a, x_0)$, hence $L = f(x_0 - 0)$.
Similarly, $f(x_0 + 0) = \inf\{f(x) \mid x > x_0\}$.

b. Let $x_0 \in \mathbb{R}$ and $t, s, a, b \in \mathbb{R}$ such that $t < a < x_0 < s < b$. Then $f(t) \le f(a) \le f(x_0) \le f(s) \le f(b)$ and furthermore $g(a) = \lim_{t \searrow a} f(t) \le f(x_0)$ and $g(b) = \lim_{s \nearrow b} f(s) \ge f(x_0)$, that is $g(a) \le f(x_0) \le g(b)$.
Recall that $g$ is continuous to get $g(x_0) = \lim_{a \searrow x_0} g(a) = \lim_{b \nearrow x_0} g(b)$, hence $g(x_0) \ge f(x_0) \ge g(x_0)$ or $g(x_0) = f(x_0)$. Consequently $f = g$ and the claim follows.