Translate of $F_{\sigma}$, $G_{\delta}$ and a set of measure $0$ I want to prove the following:
Show that (i) the translate of an $F_{\sigma}$ set is also $F_{\sigma}$,(ii) the translate of an $G_{\delta}$ set is also $G_{\delta}$, and (iii) the translate of a set of measure zero also has measure zero.
First, if $E$ has measure zero, then since the measure if translation invariant, we have:
$$m(E) = 0 \wedge m(E) = m(E + y) \Rightarrow 0 = m(E+y).$$
Now, let $G \subset \mathbb{R} $ to be a $G_{\delta}$ set, then there exist a countable family $\{\mathcal{O}_k: k \in \mathbb{N}\}$ of open sets such that $G = \bigcap_{k \in \mathbb{N}} \mathcal{O}_k$. Now, we have that if $\mathcal{A}$ is a set of subsets of $\mathbb{R}$, then for any element $y \in \mathbb{R}$, we have:
$$\left(\bigcap_{A\in\mathcal A}A\right)+y=\bigcap_{A\in\mathcal A}(A+y). \quad \quad (1)$$
Proof:
Let $x \in \left(\bigcap_{A\in\mathcal A}A\right)+y$. Then, we have that $x = a + y$, where $\forall A \in \mathcal{A}(a \in A)$, but then this implies that 
$$\forall A \in \mathcal{A}(a \in A \wedge x = a+y) \Rightarrow x \in \bigcap_{A\in\mathcal A}(A+y).$$
Now, let $x \in \bigcap_{A\in\mathcal A}(A+y)$, then $\forall A \in \mathcal{A}(x = a + y)$, but this means that $a \in \bigcap_{A \in \mathcal{A}}A$, thus $$x \in \left(\bigcap_{A\in\mathcal A}A\right)+y$$
On the other hand, we have that the translate of an open set remains open, in other words,
$$ \text{If }I \subset \mathbb{R} \text{ is open } \Rightarrow I+y \text{ is open } \quad \quad (2)$$
Proof:
Let $x \in I+y \Rightarrow x-y \in I,$ since $I$ is open there exist $c',d' \in I$ such that $x-y \in (c',d')$,but then $x \in (c,d)$, where $c = c'+y$ and $d = d'+y$. Therefore, $I + y$ is open.
Now,using (1) and (2), we have that:
$$G + y = (\bigcap_{k \in \mathbb{N}} \mathcal{O}_k) + y = \bigcap_{k \in \mathbb{N}} (\mathcal{O}_k+y),$$
which is a $G_{\delta}$ set, since is a countable intersection of open sets. We thus conclude that $G_{\delta}$ is translation invariant. Now, to prove that $F_{\sigma}$ is translation invariant, it's enough to see that the complement of a $G_{\delta}$ set is a $F_{\sigma}$ set, that is
$$ G^c = (\bigcap_{k \in \mathbb{N}} (\mathcal{O}_k))^c  = \bigcup_{k \in \mathbb{N}} \mathcal{O}_k^c,$$
where $\mathcal{O}_k^c$ is closed, since $\mathcal{O}_k$ is open.It remains to show that if $A \subset \mathbb{R}$ and $y \in \mathbb{R}$, then:
$$(A+y)^c = A^c + y.$$
Proof:
$$\begin{align*}
x \in A^c +y &\Leftrightarrow x-y \in A^c \\ 
 &\Leftrightarrow x-y \notin A \\ 
 &\Leftrightarrow x \notin A + y \\ 
 &\Leftrightarrow x \in (A+y)^c
\end{align*}$$
We thus conclude that a $F_{\sigma}$ set is translation invariant, since:
$$G^c+y = (\bigcap_{k \in \mathbb{N}} (\mathcal{O}_k+y))^c = \bigcup_{k \in \mathbb{N}} (\mathcal{O}_k+y)^c = \bigcup_{k \in \mathbb{N}} \mathcal{O}_k^c+y$$
Alternative way:
Let $A \subset \mathbb{R}$ to be a $F_{\sigma}$ set, then there exist a family $\{F_k: k \in \mathbb{N}\}$ of closed sets such that:
$$A = \bigcup_{k \in \mathbb{N}}F_k.$$
We want to prove that for $y \in \mathbb{R}$, we have:
$$A + y \text{ is closed }$$
Clearly, we have $A +y = (\bigcup_{k \in \mathbb{N}}F_k) +y$. Now, notice the following:
$$\begin{align*}
(\bigcup_{k \in \mathbb{N}}F_k) +y &= \{a+y | a\in \bigcup_{k \in \mathbb{N}}F_k\} \\
 &= \exists k \in \mathbb{N}\{a+y|a \in F_k\} \\ 
 &=\bigcup_{k \in \mathbb{N}}\{a+y|a \in F_k\} \\
 &= \bigcup_{k \in \mathbb{N}}(F_k +y)
\end{align*}$$
It is enough to prove that $F_k+y$ is closed.Indeed, if $x \notin F_k + y \Rightarrow x-y \notin F_k$. Now, since $F_k$ is closed, there exist $r > 0$ such that
$$[((x-y)-r,(x+y)+r)]\cap F_k = \emptyset \Rightarrow (x-r,x+y)\cap (F_k + y) = \emptyset,$$
thus $F_k + y$ is closed.
 A: You seem to have gone off-track. You are quite correct that since $G$ is $G_\delta,$ then there exists a family of open sets $\{\mathscr{O}_k: k \in \mathbb{N}\},$ and I assume that you are wanting to say that the intersection of all of these open sets is exactly $G.$ Unfortunately, that's not what you've said. Rather, you should say that $G=\bigcap_{k\in\Bbb N}\mathscr O_k.$ Then we easily have $$G+y=\left(\bigcap_{k\in\Bbb N}\mathscr O_k\right)+y,$$ but we should justify that this is equal to $\bigcap_{k\in\Bbb N}(\mathscr O_k+y)$.
Fortunately, it isn't too difficult to prove that if $\mathcal A$ is a set of subsets of $\Bbb R,$ then for any $y\in\Bbb R,$ we have $$\left(\bigcap_{A\in\mathcal A}A\right)+y=\bigcap_{A\in\mathcal A}(A+y).$$
Now, your claim that $G+y$ is open is unfounded. After all, $\{0\}$ is a $G_\delta$ set, and no translate of $G$ is open! Rather, you need to justify that $\mathscr{O}_k+y$ is open for all $k\in\Bbb N.$ It is enough to prove that the translate of an open interval is open. (Why?) Thus, as a countable intersection of open sets, $G+y$ is a $G_\delta$ set, as desired.
All that is left is to prove that if $A\subseteq\Bbb R$ and $y\in\Bbb R,$ then $(A+y)^c=A^c+y,$ whence we can apply DeMorgan's Laws together with the previous results to get the rest.
Edit: It looks better! Again, I must note that $G+y$ is not (necessarily) an open set, but a $G_\delta$ set.
Regarding your proof that $I+y$ is open: you should show (unless you're relying on previous results) that for any $x\in I+y,$ there exist $c,d\in I+y$ such that $x\in(c,d).$ Use openness of $I$ to find $c',d'\in I$ such that $x-y\in(c',d'),$ and proceed.
Regarding your proof that $A^c+y=(A+y)^c,$ we should proceed in a related fashion. Since $x\in A^c+y$ iff $x-y\in A^c$ iff $x-y\notin A$ iff $x\notin A+y$ iff $x\in(A+y)^c$ (which is fairly straightforward to justify), then we're done. Also, "$a+y^c$" doesn't really make sense--what would be the complement of a number?
It looks good, otherwise (from what these tired eyes can see).
