I am supposed to consider $L^2(-1,1)$ and the subspace $V= \{ u \in H : u \operatorname{is} \mathcal{M}_{sym}-measurable \}$ where $\mathcal{M}_{sym}$ is the $\sigma-$algebra generated by: $\{E \operatorname{Lebesgue measurable and } E=-E \}$. No I am supposed to say if this is closed and give an explicit projector. But I struggle to unserstand what exactly $V$ is. My solution manual says it's just the even functions, but why couldn't we consider f.e.:
$$ u: (-1,1) \rightarrow \mathbb{C}, \quad x \mapsto \begin{cases} 1, \quad x \in \{-0.5,-0.25,0.5\} \\ -1, \quad x \in \{0.25\} \\ 0, \quad \operatorname{elsewhere} \end{cases} $$
Then $u$ is clearly in $L^2$ and:
$$ u^{-1}(E)=\begin{cases} \{-0.5,-0.25,0.25,0.5 \}, \quad 0 \notin E \supseteq \{-1,1 \} \\ (-1,1) \setminus \{-0.5,-0.25,0.25,0.5 \}, \quad \{-1,1 \} \nsubseteq E \ni \{0\} \\ (-1,1), \quad \{-1,1 \} \subseteq E \ni \{0\} \\ \{\}, \quad \operatorname{elsewise} \end{cases} $$
Which is always in $\mathcal{M}_{sym}$. What am I doing wrong?
I get $u^{-1}\big(\{-1\}\big) = \{1/4\} \notin \mathcal M_{\text{sym}}$. Note "$u$ is $\mathcal M_{\text{sym}}$-measurable" means $u^{-1}(E) \in \mathcal M_{\text{sym}}$ for all Borel sets $E$.