IMC 2021 · Day 1 · P4

Hint: Start from a segment in place of $\mathbb{R}$ and use its compactness. Or recall the cool things called “the Lebesgue characterization theorem” and “the Baire characterization theorem”.

Since $g$ depends also on $\varepsilon$, let us use the notation $g(x, \varepsilon)$. Considering only $\varepsilon = 1/n$ for positive integer $n$ will suffice to reach our conclusions, hence we may use $\min\{ g(x, 1/m) \mid m \le n \}$ in place of $g(x, 1/n)$ and thus assume $g(x, \varepsilon)$ decreasing in $\varepsilon$.

For any $x \in \mathbb{R}$, choose $\delta_n(x) = \min\{ 1/n, g(x, 1/n) \}$. Of the $\delta_n(x)$-neighborhoods of all $x$ select (using local compactness of the reals) an inclusion-minimal locally finite covering $\{U_i\}$. From its inclusion-minimality it follows that we may enumerate $U_i$ with $i \in \mathbb{Z}$ so that $U_i \cap U_j \ne \emptyset$ only when $|i - j| \le 1$ and the enumeration goes from left to right on the real line. For an assumed $n$, let $x_i$ be the center of $U_i$ and $\delta_i = \delta_n(x_i)$, so that $U_i = (x_i - \delta_i, x_i + \delta_i)$ and $\delta_i < 1/n$ for all $i$.

Now define a continuous $f_n : \mathbb{R} \to \mathbb{R}$ so that it equals $f(x_i)$ in $U_i \setminus (U_{i-1} \cup U_{i+1})$, and so that $f_n$ changes continuously between $f(x_{i-1})$ and $f(x_i)$ in the intersection $U_{i-1} \cap U_i$.

Now we show that $f_n \to f$ pointwise. Fix a point $x$ and $\varepsilon = 1/m > 0$, and choose $n > \max\{ 1 / g(x, \varepsilon), m \}$.

Examine the construction of $f_n$ for any such $n$. Observe that $g(x, \varepsilon) > 1/n > \delta_i$ and $1/n < 1/m$. There are two cases:

• $x$ belongs to the unique $U_i$. Then using the monotonicity of $g(x, \varepsilon)$ in $\varepsilon$ we have $$|x_i - x| < \delta_i \le \min\Bigl\{ g\Bigl( x_i, \frac1n \Bigr), g(x, \varepsilon) \Bigr\} \le \min\{ g(x_i, \varepsilon), g(x, \varepsilon) \}.$$ Hence $$|f(x) - f_n(x)| = |f(x) - f(x_i)| < \varepsilon.$$
• $x$ belongs to $U_{i-1} \cap U_i$. Similar to the previous case, $$|f(x) - f(x_{i-1})|, \ |f(x) - f(x_i)| < \varepsilon.$$ Since $f_n(x)$ is between $f_n(x_{i-1}) = f(x_{i-1})$ and $f_n(x_i) = f(x_i)$ by construction, we have $$|f(x) - f_n(x)| < \varepsilon.$$

We have that $|f(x) - f_n(x)| < \varepsilon$ holds for sufficiently large $n$, which proves the pointwise convergence.

This solution uses the Baire characterization theorem: A function $f : \mathbb{R} \to \mathbb{R}$ is a pointwise limit of continuous functions if and only if its restriction to every non-empty closed subset of $\mathbb{R}$ has a point of continuity.

Assume the contrary in view of the above theorem: $A \subseteq \mathbb{R}$ is a non-empty closed set and $f$ has no point of continuity in $A$. Let's think that $f$ is defined only on $A$.

Then for all $x \in A$ there exist rationals $p < q$ for which $\limsup_x f > q$, $\liminf_x f < p$. Apply the Baire category theorem: If a complete metric space $A$ is a countable union of sets then some of the sets is dense in a positive radius metric ball of $A$. It follows that there exist $p$ and $q$, which serve for a subset $B \subset A$ which is dense on a certain ball (in the induced metric of the real line) $A_1 \subset A$. It yields that both sets $Q = f^{-1}(q, \infty)$ and $P = f^{-1}(-\infty, p)$ are dense in $A_1$.

Choose $\varepsilon = (q - p)/10$ and find $k$ for which the set $S = \{ x : g(x) > 1/k \}$ is also dense on a certain ball $A_2 \subset A_1$. Partition $S$ into subsets where $f(x) > (p+q)/2$ and $f(x) \leqslant (p+q)/2$, one of them is again dense somewhere in $A_3$, say the latter.

Now take any point $y \in A_3 \cap Q$ and a very close (within distance $\min(1/k, g(y))$) to $y$ point $x$ with $g(x) > 1/k$ but $f(x) \leqslant (p+q)/2$. This pair $x, y$ contradicts the property of $f$ from the problem statement.

This solution uses the Lebesgue characterization theorem: If $f : \mathbb{R} \to \mathbb{R}$ is a function and, for all real $c$, the sublevel and superlevel sets $\{ x \mid f(x) \geqslant c \}$, $\{ x \mid f(x) \leqslant c \}$ are countable intersections of open sets then $f$ is a pointwise limit of continuous functions.

Now the solution follows from the formula with a countable intersection of the unions of intervals: $$\tag{$*$} \{ x \mid f(x) \geqslant c \} = \bigcap_{n,k=1}^{\infty} \bigcup_{\substack{y \in \mathbb{R} \\ f(y) \geqslant c}} \left( y - \min\Bigl\{ \frac1k, g\Bigl( y, \frac1n \Bigr) \Bigr\}, y + \min\Bigl\{ \frac1k, g\Bigl( y, \frac1n \Bigr) \Bigr\} \right)$$ and the similar formula for $\{ x : f(x) \leqslant c \}$. It remains to prove $(*)$.

The left hand side is obviously contained in the right hand side, just put $y = x$.

To prove the opposite inclusion assume the contrary, that $f(x) < c$, but $x$ is contained in the right hand side. Choose a positive integer $n$ such that $f(x) < c - 1/n$ and $k$ such that $g(x, 1/n) > 1/k$. Then, since $x$ belongs to the right hand side, we see that there exists $y$ such that $f(y) \geqslant c$ and $$|x - y| < \min\Bigl\{ g\Bigl( y, \frac1n \Bigr), \frac1k \Bigr\} \leqslant \min\Bigl\{ g\Bigl( y, \frac1n \Bigr), g\Bigl( x, \frac1n \Bigr) \Bigr\},$$ which yields $f(x) \geqslant f(y) - 1/n \geqslant c - 1/n$, a contradiction.