Bernstein sets are sets of uniqueness

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Let $B\subset\mathbb{R}$ a Bernstein set, that is, $B$ and $\mathbb{R}\setminus B$ meet every uncountable closed set of $\mathbb{R}$. I want to show that $B$ is a set of uniqueness, i.e. if $S\sim \sum c_n e^{inx}$ is a trigonometric series such that

$$ \sum_{n=-\infty}^\infty c_ne^{inx} =0 \qquad \forall x\in \mathbb{R}\setminus B$$

then it must be the case that $c_n=0 \ \forall n\in\mathbb{Z}$.

Any hints on how to get started? I've already shown that ($\lambda$ denotes Lebesgue measure):

$$ \text{$A$ is a measurable set of uniqueness}\implies \lambda(A)=0$$

and I'm using Bernstein sets to show that the measurability condition cannot me dropped.

Thanks in advance!

EDIT:

According to R. Cooke (see page 306 here), this was proved by Bernstein in 1908, who showed that a lemma of Schwarz (page 9 of this paper) can be extended by allowing the hypothesis to fail in a set which contains no (non empty) perfect subset (i.e. a Bernstein Set). According to Cooke, Bernstein showed this using a result of Hölder which can be found on page 183 of

HÖLDER, O.

1884 "Zur Theorie der trigonometrischen Reihen," Mathematische Annalen, 24, 181-216.

However, I

a) Haven't been able to find Hölder's text.

b) Don't read German.