74th Romanian Mathematical Olympiad

a) Let us consider the functions $f(x)=x$, for all $x\in \mathbb{R}$, and $$g(x)=\{\begin{array}{ll}x,& x\in \mathbb{Q}\\ x+1,& x\in \mathbb{R}\setminus \mathbb{Q}\end{array}.$$ 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 $n\in {\mathbb{N}}^{\ast }$, 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 $g(x)=f(x)$.