A translation invariant sigma algebra Let $X$ be a complex vector space. Assume that $M_0$ is a translation invariant subset in $2^X$ in the sense that: For any $E$ in $M_0$ and $x_0$ in $X$,   $E+x_0$ is also in $M_0$. We denote $M$ by the sigma algebra generated by $M_0$. 
Question: Let $A$ be in $M$ containing zero. Does there exist any sequence $\{E_n\}$ in $M_0$ such that zero is contained in all $E_n$'s and $A$ is also in the sigma algebra generated by $\{E_n\}$?   
 A: DEF'NS :For any family $F$ of subsets of $X$: (1) Let $b(F)$ be the Boolean algebra of subsets of $X$ generated by $F$. (2)Let $b^*(F)$ be the sigma-algebra generated by $F.$  (3) Let g(F) be the set of intersections of countable subsets of $F$.... Now (4) Let $B_0=b(M_0)$. (5) For ordinal $a$ let $B_{a+1}=b(g(B_a))$.  (6) For non-zero limit ordinal $a$  let $B_a=\cup_{c<a}B_c.$...... We have $M=b^*(M_0)=B_{\omega_1}.$ For any $A\in M$ let $h(A)=\min \{a: A\in B_a\}.$ Then $h(A)$ is a countable successor ordinal or $h(A)=0.$...... Now prove the following for any $A\in M$ : (i) If $h(A)=0$ then $A\in b(F_A)\subset b^*(F_A)$ for some finite $F_A\subset M_0 $. (ii) If $h(A)=a=c+1$ then $A\in b(g(F))$ for some countable $F\subset B_c.$ (iii) If $h(A)=a=c+1$ use (i) and (ii) to show that if each $x\in B_c$ belongs to $b^* (F_x)$ for some countable $F_x\subset M_0$ then $A\in b^*(F_A)$ for some countable $F_A\subset M_0.$ Conclude by transfinite induction  that every $A\in M$ belongs to $b^*(F_A)$ for some countable $F_A\subset M_0. $ And we have $b^*(F_A)=b^*(\{\psi (f) :f\in F_A\}),$ where $\psi (f)=f$ if $0\in f,$ and $\psi (f)=X\backslash f$ if $0\not \in f.$
