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Show $\exists f \in C \ni \|f\|_\infty \le 1$, but it's Fourier series diverges.

The proof is in our textbook (Katznelson, Harmonic analysis). It uses this argument.

Let $D_n(t)=\sum_{k=-n}^ne^{ikt}$ be the Dirichlet Kernel, $g=\text{sgn}(D_n(t))$, and $E=\{f \in C: \|f\|_\infty \le 1\}$.
Choose $f$ in $E$ for which $ f(t)=g(t) \text{, except around the discountinuity of g,} $ and for which the sum $S$ of the length of the intervals where $f$ and $D_n(t)$ differ is less than $\epsilon/2n$.

Why is choosing an $f$ with $S \le \epsilon/2n$ possible?

If the number of discontinuity of $g$ is countable, I understand.
But if it's not, I don't see how to justify this choice.
And I don't know the cardinality of the discontinuities of $g$.

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    @ChristopherA.Wong I agree with you fully, in particular things like $L^p\ni f_n\to f, \text{ as $n\to\infty$}$ are common (but I can not point at a source, except some less reliable black board which might be erased by now).2012-10-13

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Since $D_n(t)= \frac{\sin (n + \frac{1}{2}) t }{\sin \frac{t}{2}}$ (taken as the limit when $\sin \frac{t}{2} = 0$), you can see that $D_n$ has discontinuities when $\sin (n + \frac{1}{2}) t = 0$, ie, when $t \in \{ \frac{k \pi}{n + \frac{1}{2}} \}_{k=-\infty}^\infty$. So the discontinuities of $g$ are nicely spaced.