Translation of a union of sets equals the union of a translation of sets. I'm wondering whether or not this is true for any arbitrary set of real numbers. It seems pretty straightforward to me. 
Sorry if this is overly pedantic, I've just been scarred by enough measure theory this semester that I've learned everything which seems obvious is true, except for the things which aren't; for those things, there is some pathological counter example that you forgot to consider.
If $\{S_{\alpha}\}_{\alpha \in I}$ is some arbitrary collection from the powerset of $\mathbb{R}$, then we should have, for any $x\in \mathbb{R},$ $$x+\bigcup_{\alpha \in I}S_{\alpha} = \bigcup_{\alpha \in I}x+S_{\alpha};$$ 
A typical element of the LHS is of the form $x+s$, where $s\in S_{\alpha}$ for some $\alpha\in I$. It then follows that $x+s \in x+S_{\alpha} \subseteq \bigcup_{\alpha \in I}x+S_{\alpha}$.
On the other hand, if $t\in\bigcup_{\alpha \in I}x+S_{\alpha}$, then $t\in x+S_{\alpha}$ for some $\alpha \in I$, and so $t=x+s$ (where $s\in S_{\alpha}$). Since $s\in S_{\alpha}$, $s\in \bigcup_{\alpha \in I}S_{\alpha}$, and so $t=x+s \in x+\bigcup_{\alpha \in I}S_{\alpha}.$
Is there something I'm missing here, or am I just being gun-shy at this point?
Thanks in advance.

For any function $f: \mathbb R \to \mathbb R$ and any collection $(S_{\alpha})_{\alpha \in I}$ we have $f(\bigcup S_{\alpha})_{\alpha \in I})=\bigcup_{\alpha \in I} f(S_{\alpha})$. In particular for the function $f(y)=x+y$. 
