Rudin's Real and Complex Analysis 9.16: Why is each $E_f$ defined only up to sets of measure $0$?

Question

I have some difficulties to understand the following sentences found from Rudin's RCA 9.16 (p. 188):

The obvious way of constructing $E$ from $M$ is to associate with each $f\in M$ the set $E_f$ consisting of all points at which $\hat{f}(t) = 0$, and to define $E$ as the intersection of these sets $E_f$. But this obvious attack runs into a serious difficulty: Each $E_f$ is defined only up to sets of measure $0$.
$\cdots$
But there are uncountably many $f\in M$, so we lose all control over $\bigcap E_f$.

Here, $M$ is a closed translation-invariant subspace of $L^2$ and $\hat{f}$ is the Fourier transform of $f$. I understand that this method has a problem because the measurability of the set $\bigcap_{f\in M}E_f$ is not guaranteed because $M$ is uncountable. However, I do not understand the sentence "Each $E_f$ is defined only up to sets of measure $0$." I guess that it means \begin{equation*} m(E_f) = m(\{t\in R^1:\hat{f}(t) = 0\}) = 0 \end{equation*} where $m$ is the Lebesgue measure devided by $\sqrt{2\pi}$. But, if $f = 0$, then $\hat{f} = 0$ and $E_f = R^1$. Can anybody explain what I am missing here?

Answer 1

If $f \in L^{2}$ then $\hat {f} \in L^{2}$. This makes $\hat {f} $ an equivalence class of functions, not a single function. So the points where $\hat {f} $ vanishes is not well defined. If you choose two different functions in the same equivalence class then the corresponding sets differ by a set of measure $0$.

Answer 2

An element of $L^2$ is not a function, but rather an equivalence class of functions. If you pick two different representatives $f_1$ and $f_2$ of some $\hat{f} \in L^2$, they agree away from a set of measure $0$, so the symmetric difference of $E_{f_1}$ and $E_{f_2}$ has measure $0$, but of course $E_{f_1}$ and $E_{f_2}$ need not be literally equal!