Let $E$ be a measurable subset of $\mathbb{R}^{d}$, and suppose $S$ is the subspace of $L^{2}(\mathbb{R}^{d})$ of functions that vanish for a.e. $x \notin E$. Show that the orthogonal projection $P$ on $S$ is given by $P(f)=\chi_{E}f$ where $\chi_{E}$ is the characteristic function of $E$.
So, I have to prove this property, but I am confused especially about the "vanishing" almost everywhere part. Is this referring to convergence to zero? I suppose if it means this, then that takes care of the part of the characteristic function that equals zero...is this the right track of thinking?
Every element of $L^2$ is an equivalence class $[f]$ of Lebesgue measurable functions that are square integrable on $\mathbb{R}^d$ and are equal a.e. to each other. It makes sense to define $\chi_{E}[f]=[\chi_{E}f]$, and this is well-defined, meaning that the resulting equivalence class does not depend on the choice of $f$ in $[f]$. All functions in the equivalence class $[\chi_{E}f]$ vanish a.e. in $\mathbb{R}^d\setminus E$. And you may think of $\chi_{E} L^2$ as consisting of all absolutely square-integrable functions on $\mathbb{R}^d$ that vanish a.e. outside of $E$.