If $[0,1[$,the sum modulo 1,$\oplus$ of $x$ & $y$ is defined by $$x\oplus y = \begin{cases} x+y, & \text{if $x+y<1$} \\[2ex] x+y-1, & \text{if $x+y\geq1$ } \end{cases}$$ If $E\subset[0,1[$,then the translated set modulo 1 of $E$ by $y$ is defined to be the set given by $E\oplus y$ ={$z:z=x \oplus y,x\in E$}
Theorem:
Let $E\subset[0,1[$ be a measurable set and $y\in [0,1[$ be given.Then the set $E\oplus y$ is measurable & $m(E\oplus y)=m(E)$.
Proof:Define $$\begin{cases} E_1=E\cap[0,1-y[ \\[2ex] E_2=E\cap [1-y,1[ \\[2ex] . \end{cases} $$
Then,$E_1 \cap E_2 =\phi$ & $E_ \cup E_2=E$.Hence $m(E)=m(E_1)+m(E_2)$.
Now,
$E_1 \oplus y=E_1 +y$ & $E_2\oplus y=E_2 +y-1$.
So,$E_1 \oplus y$ & $E_2\oplus y$ are disjoint measurable sets with
$m(E_1 \oplus y)=m(E_1 +y)=m(E_1)$
$m(E_2\oplus y)=m(E_2 +y-1)=m(E_2)$.
Since $m$ is translation invariant.
Also,$E\oplus y=[E_1 \cup E_2] \oplus y=(E_1 \oplus y) \cup (E_2 \oplus y)$.
Hence,$E \oplus y$ is measurable. $m(E\oplus y)=m(E_1 +y)+m(E_2 +y)=m(E_1)+m(E_2)=m(E)$
My queries are:
1.What is the significance of sum modulo 1 & translate modulo 1?
2.How does the coloured stuff in the proof happen?
To see $E_1\oplus y=E_1+y$, note that $$E_1\oplus y=\{x\oplus y\;|\;x\in E_1\}.$$ But $0\leq x<1-y$ for $x\in E_1$ and so $x+y<1$. Thus $x\oplus y=x+y$ as defined and so $$E_1\oplus y=\{x\oplus y\;|\;x\in E_1\}=\{x+ y\;|\;x\in E_1\}=E_1+y.$$
Similarly, $1-y\leq x<1$ for $x\in E_2$ and so $1\leq x+y$ which gives $$E_2\oplus y=\{x\oplus y\;|\;x\in E_2\}=\{x+ y-1\;|\;x\in E_2\}=E_2+y-1.$$