Trigonometric series which is dominated pointwise convergent to Lebesgue integrable but not improperly Riemann integrable function

60 Views Asked by At

I'm looking for an example of a trigonometric series $$ \sum_{n=0}^{\infty} A_n(x) = \sum_{n=0}^{\infty} (a_n \sin(nx) + b_n \cos(nx)) $$ such that:

  1. The series converges pointwise on $[0,2\pi]$
  2. The sum of the series is Lebesgue integrable on $[0,2\pi]$.
  3. The sum of the series is not improperly Riemann integrable on $[0,2\pi]$.
  4. The partial sums satisfy $|s_N| \leq g$ for some $g \in L^1([0,2\pi])$

Motivation: I'd like an example of a trigonometric series that cannot be integrated as either a Riemann or improper Riemann integral, but can be integrated term-by-term as a Lebesgue integral. This would illustrate the power of the Lebesgue integral with a (hopefully) natural example. (I know a standard example is a sequence converging pointwise to the indicator function of the rationals, but I don't consider this very natural.)

Even better if the partial sums satisfy $$ \require{enclose} \enclose{horizontalstrike}{\sum_{n=0}^{N} |a_n \sin(nx) + b_n \cos(nx)| \leq g(x),} $$ so that the dominated convergence theorem lets us use term-by-term integration to show that $a_n,b_n$ are the Fourier coefficients of the sum.

Edit: Conditions 1 and 4 already allow us to use the dominated convergence theorem to compute the coefficients $a_k$ and $b_k$ by term-by-term integration of $\sin(kx)\sum_{n=0}^{\infty} A_n(x)$ and $\cos(kx)\sum_{n=0}^{\infty} A_n(x)$.