Let $f:\mathbb R \to \mathbb C$ be $2\pi$ periodical and integrable over $[0,2\pi]$, and the fourier coefficients $(c_{k})_{k\in\mathbb Z}$ , whereby $c_{k}=0$ $\forall k \in \mathbb Z$.
Show that a Lebesgue Null Set $\nu \subset \mathbb R$ exists, such that $f(x)=0$ $\forall x \in([0,2\pi]-\nu), $ in other words $\forall x \in \nu^{c}$.
I have a tough time with the existence of null set $\nu$, let alone the importance thereof. I mean if $c_{k}=0$ $\forall k \in \mathbb Z$, then $(Ff)(x):=\sum^{\infty}_{k=-\infty} c_{k}e^{ikx}=\sum^{\infty}_{k=-\infty} 0e^{ikx}=0$, so therefore $f(x)=0$ $\forall x\in \mathbb R$. What significance does the Null Set $\nu$ have?
Hint: If $\hat f(k)=0$ for all $k,$ then
$$\int_0^{2\pi} f(t)p(t)\,dt=0$$
for all trigonometric polynomials $p.$ Because trigonometric polynomials are dense in the continuous $2\pi$-periodic functions (with respect to the sup-norm on $[0,2\pi]),$ we have
$$\int_0^{2\pi} f(t)g(t)\,dt=0$$
for every continuous $2\pi$-periodic function $g.$ This will lead to
$$\int_0^{2\pi} |f(t)|\,dt=0$$
if you invoke some well known theorems in measure theory.