Is Lebsegue Measure Translation Invariant? I am trying to prove that the Lebsegue measure is translation-invariant. Namely, given a set $X\subseteq\mathbb{R}$, I'd like to show $X + y$ is measurable and $\mathit{m}(X + y) = \mathit{m}(X)$. Namely, that the measures -- not the outter measures alone -- agree. I am mostly stuck on demonstrating that the translation $X + y$ is measurable to begin with. Any ideas?
 A: Observe that the open sets on $\mathbb{R}$ are invariant under translations. That is, for any open $S \subseteq \mathbb{R}, S + y$ is open. Thus the Borel sets are invariant under translations, too. This establishes the measurability.
Now define $m^y(S) = m(S + y)$. We wish to show that $m^y = m$. By definition, Lebesgue measure is generated by the premeasure $m_0$ on the algebra of intervals. Then clearly $m_0 = m_0^y$, the translated premeasure. Then, observe that $\mathbb{R}$ is the countable union of sets with finite measures, namely $\mathbb{R} = \bigcup_{j=-\infty}^\infty [j,j+1)$, so $\mathbb{R}$ is $\sigma$-finite.
The extension of premeasure to a measure is unique when the space is $\sigma$-finite, so $m^y = m$. Finally, we need to show that Lebesgue null sets are preserved by translation. From the conclusion above, any Borel set $S$ satisfies $m(S) = m^y(S)$, and this remains true for sets with zero measure. By completeness any null set is measurable, and the proof is complete.
A: The Lebesgue measure is defined in terms of some basic sets, the open intervals, for example. The measure of a translated open set is the same
as the measure of the set, so this structural property carries through to the Lebesgue measure.
That is, if $A$ is a set and $U_k$ forms an open cover by intervals, then
$U_k+ \{x\}$ forms an open cover of $A+ \{x\}$. Since the length of
$U_k$ and the length of $U_k+\{x\}$ is the same, then just applying the
definition shows that $mA = m (A + \{x\})$.
The result is true in $\mathbb{R}^n$ of course.
This sort of approach is used a lot in measure theory.
