Forgive, for the title didn't know how to name this questions. Please change to something better.
Let $B_1(n)$ denote a unit ball around $n\in \mathbb{Z}^{+}$. Suppose that for every $n$ there exists a set $E_n \subset B_1(n)$ such that $\mu(E_n)=c$, that is every unit interval around an integer has a subset of Lebesgue measure $c$.
Let $v$ be some probability measure. Is it correct that the set $E= \cup_{n\in \mathbb{Z}^{+}} E_n$ has probability measure $\nu(E)=c$? In other words, the probability of 'hitting' a set $E$ is $c$.
Here is my argument: I used the following theorem from wiki ( I would appreciate a good reference)
Theorem: Any σ-finite measure μ on a space X is equivalent to a probability measure on X: let $V_n$, $n ∈ N$, be a covering of X by pairwise disjoint measurable sets of finite μ-measure, and let $w_n$, $n ∈ N$, be a sequence of positive numbers (weights) such that
\begin{align} \sum_{n = 1}^\infty w_n = 1. \end{align} The measure ν defined by
\begin{align} \nu(A) = \sum_{n = 1}^\infty w_n \frac { \mu (A \cap V_n) } {\mu (V_n)} \end{align} is then a probability measure on X with precisely the same null sets as $μ$.
From the above result, letting $X=\mathbb{R^+}$ and $A=E$ and using the fact that Lebesgue measure is $\sigma$-finite.
Also, letting $V_n= B_1(n)$, for $n=0$ let $V_0= [0, 1/2)$.
We have that \begin{align} \nu(E)= \sum_{n = 1}^\infty w_n \frac { \mu (E \cap B_1(n)) } {\mu ( B_1(n))} = \sum_{n = 1}^\infty w_n \mu (E \cap B_1(n)) = \sum_{n = 1}^\infty w_n \mu (E_n)= \sum_{n = 1}^\infty w_n c=c \end{align}
Is this argument correct? If so, how can I improve it? If not correct please help. I would also appreciate a reference to the theorem above.
Thank you very much.
Bounty: For the bounty I would like to have a reference ( a book or a paper) that shows the above theorem.
Strictly speaking your application of the theorem that you cite is not correct, because your $V_n$ are not pairwise disjoint: $(0, \dfrac 1 2) \subset V_0 \cap V_1$ and $(n, n+1) \subset V_n \cap V_{n+1}$. If I were you I would choose $V_0$ as you do and then $V_n = [n - \dfrac 1 2, n + \dfrac 1 2)$.
This mistake is not really severe, though, and you may leave your $V_n$ as they are provided you use the slightly different form of the same theorem that can be found on PlanetMath and that does not require disjointness (if you see a web error at the given address, just scroll down that page).
Save for this detail, your proof is correct.