Lebesgue Measurable Sets Are Translation Invariant I'd like a few proofs that translation of Lebesgue measurable sets are Lebesgue Measurable. As an example, and to add to the collection, I've included my own proof of the statement as an answer.   
Let $\mathfrak{M}(\mu)$ be the space of $\mu$-measurable sets as defined in Baby Rudin as the countable union of things in $\mathfrak{M}_F(\mu)$, finitely $\mu$-measurable sets.  
Any and all methods of proof appreciated. I suppose there's a more elegant way to prove this statement.
 A: Using the definition from Wikipedia I'd go for prooving that the outer measure is invariant under translations.
Consider the set in the definition $S(E) = \{\sum l(I_k): E \subseteq \bigcup I_k\}$, now we shall prove that $S(E)$ itself is translation invariant, that is $S(E) = S(E+\tau)$, in fact it suffices to show that $S(E)\subseteq S(E+\tau)$ because then $S(E+\tau)\subseteq S(E+\tau + (-\tau)) = S(E)$.
Assume that $x\in S(E)$, then there's a sequence $I_k$ such that $x = \sum l(I_k)$ and $E \subseteq \bigcup I_x$. Now there is a sequence $I_k+\tau$ such that $E+\tau \subseteq \bigcup (I_k+\tau)$ and $l(I_k+\tau) = l(I_k)$ implies that $\sum (I_k+\tau) = \sum(I_k) = x$ that is $x\in S(I_k+\tau)$
Now from this it follows that outer measure is invariant:
$$\mu^*(E+\tau) = \inf S(E+\tau) = \inf S(E) = \mu^*(E)$$
If Lebesgue measurability is invariant it follows that the Lebesgue measure is invariant. So with the same reasoning as for the invarianse of $S(E)$ we only have to prove the one way implication.
Assume that $E$ is Lebesgue measurable, and let $U$ be an arbitrary subset of the space, then we have:
$$\mu^*((U+\tau)\cap(E+\tau)) + \mu^*((U+\tau)\cap\overline{(E+\tau)}) = 
\mu^*((U\cap E)+\tau) + \mu^*((U\cap\overline E)+\tau)) = 
\mu^*((U\cap E)) + \mu^*((U\cap\overline E))) = \mu^*(U) = \mu^*(U+\tau)
$$
That is $E+\tau$ is Lebesgue measurable and $\mu(E+\tau) = \mu^*(E+\tau) = \mu*(E) = \mu(E)$
A: Lemma (1): $\mathfrak{M}_F$, finitely $\mu$-measurable sets, is translation invariant. 
Lemma(2): Let $A \subset \mathbb{R}$. Then, $\mu^{*}(A) = \mu^{*}(A+y)$ for any fixed $y \in \mathbb{R}$.
Proof of Lemma (2)  Let $y \in \mathbb{R}$. Let $\epsilon > 0$ and let $\{E_n\}_{n =1}^{\infty}$ be elementary sets (finite unions of intervals in $\mathbb{R}$) such that $$A \subset \bigcup_{n=1}^{\infty} E_n$$
and such that, 
\begin{eqnarray*}
\mu^{*}(A) + \epsilon &>& \sum_{n=1}^{\infty} \mu(E_n) \\
&\geq& \sum_{n=1}^{k} \mu(E_n) \\
&=& \sum_{n=1}^{k} \mu(E_n+y) \\
\end{eqnarray*}
this is due to the simple fact that the measure of intervals is easily seen to be translation invariant.
Letting $k \to \infty$, and since $\epsilon$ was arbitrary, it follows that $\mu^{*}(A+y) \leq \mu^{*}(A)$. Now, let $A'=A+y'$ with $y'=-y$, we get $\mu^{*}(A+y) \leq \mu^{*}(A)$.
Proof of Lemma (1): $A \in \mathfrak{M}_F$ if there exists a sequence of elementary sets such that $\lim \limits_{n \to \infty} d(A_n,A) = 0$ in the sense of $d(A_n,A) = \mu^{*}(S(A_n,A))$, the symmetric difference. Now $$d(A_n+y,A+y) = \mu^{*}(S(A_n+y,A+y)) = \mu^{*}(S(A_n,A)+y) = \mu^{*}(S(A_n,A))$$
since we just proved $\mu^{*}$ is translation invariant. Thus, taking $n \to \infty$, we see that $A_n +y \to A+y$ in the sense of $d$. Since $A_n+y$ is an elementary set, $A+y \in \mathfrak{M}_F$.  
Proof of Statement: If $A \in \mathfrak{M}$, then $$A = \coprod_{i \in \mathbb{N}} A_i$$ with $A_i \in \mathfrak{M}_F.$ Thus, $$A+y = \left(\coprod_{i \in \mathbb{N}} A_i \right) +y = \coprod_{i \in \mathbb{N}} (A_i +y).$$ Since each $A_i +y \in \mathfrak{M}_F$ by the lemma, $A \in \mathfrak{M}$.
