If \(f_n\) is a sequence of continuous functions \(f_n: \mathbb{R} \to \mathbb{R}\) such that \(f_n(x) \to f(x)\) for each \(x \in \mathbb{R}\), show that \(f\) is continuous at uncountably many points of \(\mathbb{R}\).
Solution
For \(\varepsilon > 0\) and \(N \ge 1\) put \[A_N(\varepsilon) = \{x \in \RR : \abs{f_m(x) - f_k(x)} \le \varepsilon \text{ for all } m, k \ge N\}, \qquad U(\varepsilon) = \bigcup_{N \ge 1} \operatorname{Int} A_N(\varepsilon).\] The space \(\RR\) is a complete metric space, hence a Baire space by the Baire category theorem.
1. Each \(A_N(\varepsilon)\) is closed, and \(\bigcup_N A_N(\varepsilon) = \RR\).
\(A_N(\varepsilon)\) is an intersection of the closed sets \(\{x : \abs{f_m(x) - f_k(x)} \le \varepsilon\}\), which are closed because \(f_m - f_k\) is continuous. For each \(x\), the sequence \((f_n(x))\) converges, hence is Cauchy, so \(x \in A_N(\varepsilon)\) for some \(N\).
2. \(U(\varepsilon)\) is open and dense in \(\RR\).
\(U(\varepsilon)\) is a union of open sets. Let \(V\) be a nonempty open interval. The closed subsets \(A_N(\varepsilon) \cap \overline{V}\) of the complete metric space \(\overline{V}\) cover \(\overline{V}\) by step 1, so by the Baire category theorem one of them has nonempty interior in \(\overline{V}\); that interior meets \(V\), so it contains a nonempty open interval \(J \subseteq V \cap A_N(\varepsilon)\). Then \(J \subseteq \operatorname{Int} A_N(\varepsilon)\), so \(V\) meets \(U(\varepsilon)\).
3. \(f\) is continuous at every point of \(C = \bigcap_{k \ge 1} U(1/k)\).
Let \(x \in C\) and \(\varepsilon > 0\); choose \(k\) with \(1/k < \varepsilon/3\). Then \(x \in W = \operatorname{Int} A_N(1/k)\) for some \(N\). For \(y \in W\) and \(m \ge N\), \(\abs{f_m(y) - f_N(y)} \le 1/k\); letting \(m \to \infty\) gives \(\abs{f(y) - f_N(y)} \le 1/k\). By continuity of \(f_N\), there is a neighborhood \(V \subseteq W\) of \(x\) with \(\abs{f_N(y) - f_N(x)} < 1/k\) for \(y \in V\). For \(y \in V\), \[\abs{f(y) - f(x)} \le \abs{f(y) - f_N(y)} + \abs{f_N(y) - f_N(x)} + \abs{f_N(x) - f(x)} < 3/k < \varepsilon.\]
4. \(C\) is uncountable.
Suppose \(C = \{c_1, c_2, \ldots\}\) is countable. The sets \(U(1/k)\) and \(\RR \setminus \{c_j\}\) form a countable family of dense open subsets of \(\RR\) by step 2, and their intersection is \(C \setminus C = \varnothing\). This contradicts the Baire property of \(\RR\). A finite \(C\) is excluded in the same way.
5. Q.E.D.
By steps 3 and 4, \(f\) is continuous at every point of the uncountable set \(C\).