The translate of a set and measure Show (i) the translate of an $F_{\sigma}$ set is also $F_{\sigma}$, (ii) the translate of a $G_{\delta}$ set of a $G_{\delta}$, and (iii) the translate of a set of measuure zero also has measure zero.
I'm a little confused, because...
$F_{\sigma}$ is of the form $\cup^{\infty}_{n=1} [a_n, a_{n+1}]$, right? So if we translate this by y, we get $\cup^{\infty}_{n=1} [a_n+y, a_{n+1}+y]$, right? Isn't this already in $F_{\sigma}$ $ form? I'm not really sure what the question wants me to do...
Thanks in advance
 A: It appears that you have the right idea without knowing it.  Basically, you are asked to show the following: Suppose that $A \subseteq \mathbb{R}$ is a set with some property, show that all translates $A + y$ of this set have the same property.  The details become different as we go from property to property, but the idea behind the proofs are the same: we will take witnesses to the defining characteristic of the particular property in question and then translate them.  We then only have to show that these translates then witness that $A + y$ has that same property.
Consider the property "is F$_\sigma$."  Now, if $A \subseteq \mathbb{R}$ is F$_\sigma$ this means that there is a countable family $\{ E_n : n \in \mathbb{N} \}$ of closed sets such that $A = \bigcup_n E_n$.  To show that $A + y$ is also F$_\sigma$ we must find another countable family $\{ F_n : n \in \mathbb{N} \}$ of closed subsets of $\mathbb{R}$ such that $A + y = \bigcup_n F_n$.  Well, one thing we know is that $$A + y = ( {\textstyle \bigcup_n} E_n ) + y = {\textstyle \bigcup_n} ( E_n + y ).$$  Could these sets all be closed?  Yes:  If $x \notin E_n + y$, this means that $x - y \notin E_n$ and since $E_n$ is closed there is a $\epsilon > 0$ such that $( ( x - y ) - \epsilon , ( x - y ) + \epsilon ) \cap E_n = \emptyset$, and we can then show that $( x - \epsilon , x + \epsilon ) \cap ( E_n + y ) = \emptyset$; this shows that $E_n + y$ is closed.
The details for "is G$_\delta$" are almost entirely the same.  (Translates of open sets will be open sets.)
For "has measure zero" note that for $A$ to have measure zero then for each $\epsilon > 0$ there is a family $\{ I_n : n \in \mathbb{N} \}$ of open intervals in $\mathbb{R}$ such that $A \subseteq \bigcup_n I_n$, and $\sum_n \mathrm{length} ( I_n ) < \epsilon$.  (Translates of open intervals will be open intervals of the same length.)
