Query regarding a proof of translation invariance of lebesgue measue Here $\mathcal{B}(\mathbb{R})$ will denote the borel sigma field in $\mathbb{R}$ and $\lambda$ denotes the legesbue measure.
I have already proved that $A+x\in\mathcal{B}(\mathbb{R})\ \forall A\in\mathcal{B}(\mathbb{R}),\ x\in\mathbb{R}$ by showing that the collection of "good sets" here $\{A:\ A+x\in\mathcal{B}(\mathbb{R})\}$ is a $\sigma$-field.
Now again I use good set trick and construct $S=\{A\in\mathcal{B}(\mathbb{R}):\ \lambda(A+x)=\lambda(A)\}$. It's easy to see that $S$ contains all the intervals in $\mathbb{R}$. I want to show that $S$ is a $\sigma$-field. 
But I'm stuck with only one step i.e. if $A\in S$ then $A^c\in S$.
So we have $\lambda(A+x)=\lambda(A)$, we have to show that $\lambda(A^c+x)=\lambda((A+x)^c)=\lambda(A^c)$.
Here $\lambda$ is not finite measure, so I'm not getting any wayout to prove the above claim.
Can anyone help me in this regard? Thanks for assistance in advance.
 A: To prove the translation invariance of Lebesgue measure, the last step is a little more subtle, exactly to take care of sets of infinite measure. The  $\sigma$-field $S$ must be defined in a slightly different way. Let us see it.
Define
$$S=\{A\in\mathcal{B}(\mathbb{R}): \textrm{ for all } n \in \Bbb N,  \lambda((A\cap [-n,n])+x)=\lambda(A\cap [-n,n])\}$$
It easy to see that $S$ contains all the intervals in $\mathbb{R}$ and that $S$ is a $\sigma$-field. So $\mathcal{B}(\mathbb{R}) \subseteq S$ (in fact, $\mathcal{B}(\mathbb{R}) = S$).
So we have that, for all $A \in \mathcal{B}(\mathbb{R})$ and all $n \in \Bbb N$, $\lambda((A\cap [-n,n])+x)=\lambda(A\cap [-n,n])\}$.
Now, note that, for all $A \in \mathcal{B}(\mathbb{R})$, $A\cap [-n,n] \nearrow A$ and $(A\cap [-n,n])+x \nearrow A+x$. So we have
$$\lambda(A+x)= \lim_{n\to \infty}\lambda((A\cap [-n,n])+x)=\lim_{n\to \infty}\lambda(A\cap [-n,n])= \lambda(A) $$
Remark: Let us prove in detail that $S$ is closed under complement.
Suppose $A \in S$. It means that $A \in \mathcal{B}$ and, for each $n \in \Bbb N$, $\lambda((A\cap [-n,n])+x)=\lambda(A\cap [-n,n])\}$.
So we have that $A^c\in \mathcal{B}$ and for each $n \in \Bbb N$,
\begin{align*}
\lambda( (A^c\cap [-n,n]) +x) & = \lambda((A^c+x)\cap [-n+x,n+x]) \:\:\:\:\text{ by 1}\\
& = \lambda((A+x)^c\cap [-n+x,n+x]) \:\:\:\:\text{ by 2}\\
&= \lambda([-n+x,n+x]) - \lambda((A+x)\cap [-n+x,n+x]) \\
&= 2n - \lambda((A+x)\cap [-n+x,n+x]) \\
& = 2n - \lambda((A\cap [-n,n])+x)  \:\:\:\:\text{ by 3}\\
&= 2n -  \lambda(A\cap [-n,n]) \\
&= \lambda([-n,n]) -  \lambda(A\cap [-n,n]) \\
& = \lambda(A^c\cap [-n,n])
\end{align*}
where we used:

*

*$(A^c\cap [-n,n]) +x = (A^c+x)\cap [-n+x,n+x]$

*$A^c+x =(A+x)^c$

*$(A+x)\cap [-n+x,n+x]= (A\cap [-n,n])+x$
