# $f$ absolutely continuous, $\hat f(n) \downarrow 0$, $\hat f(n)$ positive and even, then $\sum \hat f_n <\infty$

Suppose $f$ is absolutely continuous and the Fourier coefficients of $f$ satisfy $$\hat f(n)= \frac{a_n sgn(n)}{n} \ge 0$$
where the $a_n$ are postive, even, and decreasing to zero as $|n| \to \infty$, so that the Fourier coefficients of $f$ are even and positive. Show $$\sum_{n=-\infty}^{\infty} \hat f(n)<\infty$$

This was answered in this post. The post says:

A standard theorem says that the Fourier series of an absolutely continuous function converges to it uniformly so taking $x=0$ you get the result.

I suspect the theorem holds almost everywhere, so I don't see why taking $x=0$ is justified.
If possible, can you point me to a proof for this theorem?
Or explain why the choice is possible.

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Since $f$ is absolutely continuous, we can write $$f(x)=f(0)+\int_0^x g(t)\,\mathrm{d}t$$ where $g\in L^1$.
Since $f$ is even, it's Fourier coefficients are real. Since $\hat{f}(n)\downarrow0$, we have $\hat{f}(n)\ge0$
Therefore, \begin{align} \sum_{n\in\mathbb{Z}}\hat{f}(n) &=f(0)+\sum_{n\in\mathbb{Z}}\int_0^1\int_0^x g(t)\,\mathrm{d}t\,e^{-2\pi inx}\,\mathrm{d}x\\ &=f(0)+\sum_{n\in\mathbb{Z}}\int_0^1\int_t^1 g(t)\,e^{-2\pi inx}\,\mathrm{d}x\,\mathrm{d}t\\ &=f(0)+\sum_{n\in\mathbb{Z}}\int_0^1 g(t)\frac{e^{-2\pi int}-1}{2\pi in}\,\mathrm{d}t\\ &=f(0)-\sum_{n\in\mathbb{Z}}\int_0^1 g(t)\frac{\sin(2\pi nt)}{2\pi n}\,\mathrm{d}t\\ &=f(0)+\int_{1/2}^1g(t)\,\mathrm{d}t-\int_0^{1/2}g(t)\,\mathrm{d}t \end{align} where $\displaystyle\sum_{n\in\mathbb{Z}}$ is interpreted as $\displaystyle\lim_{N\to\infty}\sum_{n=-N}^N$
Therefore, $$0\le\sum_{n\in\mathbb{Z}}\hat{f}(n)\le f(0)+\|g\|_{L^1}$$