Let $A \subset \mathbb{R^n}$ and $x \in \mathbb{R^n}$ and $A + x= :\{u: u - x \in A\}$. If $A$ is measurable, then $A + x$ is also measurable and $\mu(A) = \mu(A+x)$. This means that the Lebesgue measure is translation-invariant.
How can I start the proof? Any help is appreciated.
Thank you!
Here is a brief proof of the one dimensional case.
Theorem: If $A \subset \Bbb{R}$ is Lebesgue measurable, then for every $x \in \Bbb{R}$, $m(A)=m(A+x)$.
Proof:
If $A \subset \Bbb{R}$ is Lebesgue measurable, then define its measure to be
$$m(A):=\inf \{\sum_n l((a_n,b_n]): A \subset \bigcup_n (a_n,b_n], l((a_n,b_n])=b_n-a_n\}.$$
So we have that $A \subset \cup_n (a_n,b_n]$, then for any $x \in \Bbb{R}$,
$$A+x \subset \cup_n(a_n+x,b_n+x].$$
And since
\begin{align} l((a_n+x,b_n+x])&=(b_n+x)-(a_n+x)\\ &= b_n - a_n\\ &= l((a_n,b_n]) \end{align} Holds for every $n$, their infimums agree thus $m(A)=m(A+x)$. $\blacksquare$
Added: I am using the definition of Lebesgue measure from Bass, Real Analysis for Graduate Students. However this works if we cover $A$ by any countable union of intervals.