Prove that $\mathcal M : = \left \{E \in \mathcal B_{\Bbb R^2}\ |\ \lambda_{\Bbb R^2} (E+x) = \lambda_{\Bbb R^2} (E) \right \}$ is a monotone class. 
Let $(\Bbb R^2, \mathcal L_{\Bbb R^2}, \lambda_{\Bbb R^2})$ be the Lebesgue measure space on $\Bbb R^2.$

Prove that $$\mathcal M : = \left \{E \in \mathcal B_{\Bbb R^2}\ |\ \lambda_{\Bbb R^2} (E+x) = \lambda_{\Bbb R^2} (E) \right \}$$ is a monotone class of subsets of $\Bbb R^2,$ for any $x \in \Bbb R^2.$
I have proven that $\mathcal M$ is closed under countable increasing unions. But I find it difficult to prove that it is closed under countable decreasing intersections as well. Let $\{F_n \}_{n=1}^{\infty}$ be a decreasing sequence of subsets in $\mathcal M$ i.e. $F_n \in \mathcal M$ and $F_n \supseteq F_{n+1},$ for all $n \geq 1.$ Let $F = \bigcap\limits_{n=1}^{\infty} F_n.$
We need to show that $F \in \mathcal M$ or in other words $\lambda_{\Bbb R^2} (F+x) = \lambda_{\Bbb R^2} (F).$ Let $E_n = F_n^c,$ $n \geq 1.$ Since $F_n$'s are decreasing, $E_n$'s are increasing. Let $E = \bigcup\limits_{n=1}^{\infty} E_n.$ Then by De Morgan's laws it is easy to see that $F = E^c.$
Since $E_n$'s are increasing to $E$ and since $\mathcal M$ is closed under countable incresaing unions (I assume it as I have proved it already) so it follows that $\lambda_{\Bbb R^2} (E + x) = \lambda_{\Bbb R^2} (E).$ Therefore \begin{align*} \lambda_{\Bbb R^2} (F^c+x) & = \lambda_{\Bbb R^2} (E + x) \\ & = \lambda_{\Bbb R^2} (E) \\ & = \lambda_{\Bbb R^2} (F^c) \end{align*}
This shows that $F^c \in \mathcal M.$ So if we can show that $\mathcal M$ is closed under complementation then we are through. But I find difficulty to prove this part. I know that for any $E \subseteq \Bbb R^2$ and for any $x \in \Bbb R^2$ we have $(E+x)^c = E^c +x.$ If $E \in \mathcal M$ then by using the previous equality I get $$\lambda_{\Bbb R^2} (E^c + x) = \lambda_{\Bbb R^2} ((E+x)^c).$$ But now I got stuck. Because if $\lambda_{\Bbb R^2} (E) = +\infty$ then we don't have a valid expression for $\lambda_{\Bbb R^2} ((E+x)^c).$ Can anybody please help me in this regard?
Thanks in advance.
 A: We have well known generators for $B_{\Bbb R^2}$ which are stable under intersections, e.g. all rectangles.
If we consider a new measure $\tilde{\lambda}$ defined by $$\tilde{\lambda}(E) := \lambda_{\Bbb R^2} (E+x)$$ it's easy to show that $\tilde{\lambda}$ and $\lambda_{\Bbb R^2}$ are $\sigma$-finite and equal on those generators so it follows that they are equal on $B_{\Bbb R^2}$ and we get $\mathcal{M} = B_{\Bbb R^2}$
A: Here's a different way of getting the invariance of the Lebesgue measure:

*

*First, argue that for every rectangle $R$, $\lambda(R+x)=\lambda(R)$.


*Call $R\subset \mathbb{R}^2$ a dyadic rectangle if it is of the form $R=I\times J$, where $I$ and $J$ are half-open intervals (of the form $[,)$) whose endpoints are dyadic rationals (i.e. of the form $k/2^n$ for some $k\in\mathbb{Z}$ and $n\in\mathbb{N}$). Show that every open set $U$ is a countable union of pairwise disjoint dyadic rectangles.


*Put (1) and (2) together to show that $\lambda(U+x)=\lambda(U)$ for every open set $U\subset\mathbb{R}^2$. This uses that, if $R$ and $S$ are disjoint, then so are $R+x$ and $S+x$.


*Use the outer regularity of the Lebesgue measure (plus the fact that $U+x$ is open whenever $U$ is) to show that $\lambda(A+x)=\lambda(A)$ for any measurable set $A$.
By the way, this holds for any dimension, not just $2$. You just need to replace rectangles by their higher dimensional analogues.
