Cantor–Lebesgue theorem

Remarks

Proof idea: write $a_n \cos nx + b_n \sin nx = \rho_n \cos(nx + \varphi_n)$ with $\rho_n = \sqrt{a_n^2 + b_n^2}$. If $\rho_n \not\to 0$, then $\rho_{n_j} \ge \varepsilon > 0$ along a subsequence, so $\cos^2(n_j x + \varphi_{n_j}) \to 0$ on $E$. Replace $E$ by a bounded subset of positive measure. By dominated convergence, $\int_E \cos^2(n_j x + \varphi_{n_j}) \, dx \to 0$. But this integral equals $\frac{1}{2} \abs{E} + \frac{1}{2} \int_E \cos(2 n_j x + 2 \varphi_{n_j}) \, dx$, and the last integral tends to $0$ by the Riemann–Lebesgue lemma, a contradiction.

In complex form: if $c_n e^{inx} + c_{-n} e^{-inx} \to 0$ for every $x$ in a set of positive measure, then $c_n \to 0$ as $\abs{n} \to \infty$. So a trigonometric series that converges on a set of positive measure has coefficients tending to $0$.

Example: if $\sum_n (a_n \cos nx + b_n \sin nx)$ converges for every $x$ in $[0, 1]$, its terms tend to $0$ there, so $a_n, b_n \to 0$.

Positive measure is needed: at the single point $x = 0$ the terms $n \sin nx$ are all $0$, but $b_n = n \not\to 0$.

Going deeper

For a student

The theorem is the first step in the proof of Cantor's uniqueness theorem. It reduces questions about convergent trigonometric series to series whose coefficients tend to $0$, where Riemann's methods apply. Its proof is a typical use of the Riemann–Lebesgue lemma together with a convergence theorem for integrals.