Tiling by Translation and Orthonormal Basis Let $E$ be a Lebesgue measurable subset of $\mathbb R$ such that $\bigcup_{k\in\mathbb Z}E+k=\mathbb R$ up to a set of measure zero and $E+k\cap E+j$ has measure zero for $k\neq j$.
How may I show that $\left\{e^{2\pi ikx}\right\}_{k\in\mathbb Z}$ is an orthonormal basis for $L^2\left(E\right)$?
For starters, I am stuck showing that $e^{2\pi ikx}$ has norm $1$:
$$\left\|e^{2\pi ikx}\right\|^2=\int_E\left|e^{2\pi ikx}\right|^2\,dm\left(x\right)=\int_E\left|\cos\left(2\pi kx\right)+i\sin\left(2\pi kx\right)\right|^2\,dm\left(x\right)=\int_Edm\left(x\right)=m\left(E\right).$$
Orthonormality boils down to showing that
$$\int_Ee^{2\pi ikx}\,dm\left(x\right)=0$$
for any nonzero integer $k$. I was able to show that this is the case if $E=\left[y,y+1\right]$ for $y\in\mathbb R$. However, I am not certain whether I can prove it this way without loss of generality.
 A: I can prove the identity you're struggling with, at least, and perhaps this will help with your other difficulties. Observe that \begin{align*}x\in[0,1]\cap(E-k)&\Leftrightarrow 0\leq x\leq 1,\,x+k\in E\\&\Leftrightarrow k\leq x+k\leq k+1,\,x+k\in E\\&\Leftrightarrow x+k\in[k,k+1]\cap E.\end{align*} This, combined with translation-invariance of the Lebesgue measure, means that $$m([0,1]\cap(E-k))=m([k,k+1]\cap E)\text{ for all }k\in\mathbb{Z}.$$ Then using countable additivity of the Lebesgue measure, and since the $\{E+k\}_{k\in\mathbb{Z}}$ are almost disjoint, we get \begin{align*}m([0,1])&=m\left([0,1]\cap\bigcup_{k\in\mathbb{Z}}(E-k)\right)\\&=m\left(\bigcup_{k\in\mathbb{Z}}[0,1]\cap(E-k)\right)\\(\ast)&=\sum_{k\in\mathbb{Z}}m([0,1]\cap(E-k))\\&=\sum_{k\in\mathbb{Z}}m([k,k+1]\cap E)\\&=m\left(\bigcup_{k\in\mathbb{Z}}[k,k+1]\cap E\right)\\&=m(E),\end{align*} so $m(E)=1.$
Now let's justify $(\ast)$ completely. If $\{A_{k}\}_{k\in\mathbb{Z}}$ are a collection of almost disjoint subsets of $\mathbb{R},$ i.e., for each $i\neq j,$ $m(A_{i}\cap A_{j})=0,$ then we claim that $$m\left(\bigcup_{k\in\mathbb{Z}}A_{k}\right)=\sum_{k\in\mathbb{Z}}m(A_{k}).$$ Observe that by countable subadditivity of the Lebesgue measure, the right hand side of this equality is always an upper bound for the left hand side. Therefore, we need only prove the opposite bound. The principle of inclusion-exclusion suggests that we have a lower bound given by $\sum_{k\in\mathbb{Z}}m(A_{k})-\sum_{j,k\in\mathbb{Z},j<k}m(A_{j}\cap A_{k}),$ which equals $\sum_{k\in\mathbb{Z}}m(A_{k})$ by our assumption of almost disjointness of the $A_{k},$ so let's try to prove this lower bound.
We notice that $\bigcup_{k\in\mathbb{Z}}A_{k}=\left(\bigcup_{k\in\mathbb{Z}}A_{k}\setminus\left(\bigcup_{j<k}A_{j}\right)\right)\cup\left(\bigcup_{k\in\mathbb{Z}}A_{k}\cap\left(\bigcup_{j<k}A_{j}\right)\right).$ Lets call these parts $B$ and $C$ briefly: then $m(B\cup C)\leq m(B)+m(C)=m(B),$ since $m(C)\leq \sum_{k}\sum_{j<k}m(A_{j}\cap A_{k})=0$ by repeated use of countable subadditivity and almost disjointness of the $A_{k}$, and $m(B\cup C)\geq m(B),$ so we may neglect $C$.
But $\bigcup_{k\in\mathbb{Z}}A_{k}\setminus\left(\bigcup_{j<k}A_{j}\right)$ is now a union of disjoint sets, so by countable additivity (using the equality we just proved in the first line), \begin{align*}m\left(\bigcup_{k\in\mathbb{Z}}A_{k}\right)&=m\left(\bigcup_{k\in\mathbb{Z}}A_{k}\setminus\left(\bigcup_{j<k}A_{j}\right)\right)\\&=\sum_{k\in\mathbb{Z}}m\left(A_{k}\setminus\left(\bigcup_{j<k}A_{j}\right)\right)\\&=\sum_{k\in\mathbb{Z}}m(A_{k})-m\left(A_{k}\cap\bigcup_{j<k}A_{j}\right)\\&=\sum_{k\in\mathbb{Z}}m(A_{k})-m\left(\bigcup_{j<k}A_{k}\cap A_{j}\right).\end{align*} Applying countable subadditivity to this last union, we get that $$m\left(\bigcup_{k\in\mathbb{Z}}A_{k}\right)\geq\sum_{k\in\mathbb{Z}}\left(m(A_{k})-\sum_{j<k}m(A_{k}\cap A_{j})\right),$$ as desired.
