Let $I_{n}=(n,n+1]$ defined for $n\in\Bbb{Z}.$ I need to prove a subset $E\subset\Bbb{R}$ is Lebesgue-measurable $\iff E\cap I_{n}$ is Lebesgue measurable.
I denote $l^{*}$ as the Lebesgue outer measure. By definition, a set $E$ is $l^{*}-$measurable if $$l^{*}(A)=l^{*}(A\cap E)+l^{*}(A\,\backslash\, E)\;\forall \, A\subset\Bbb{R}.$$
At the beginning of the question, it asks me to prove that if $E$ is contained in a finite union of $I_{n}$, so $l^{*}(E)<\infty,$ which was pretty simple.
Next, asks me to build a Lebesgue-Measurable set $E$ such that $l^{*}(E)<\infty$ and $l^{*}(E\cap I_{n})>0\forall n\in\Bbb{Z}.$ I did it like follows:
For each $n\in\mathbb{Z}$, define $a_{n}=n+\frac{1}{2}$ (the mid term in $I_{n})$. Now, define $$r_{n}= \begin{cases} \frac{1}{2^n} \quad\textrm{if}\quad n>0, \\ -\frac{1}{2^n}\quad\textrm{if}\quad n<0 ,\\ \frac{1}{2}\quad\textrm{if}\quad n=0. \end{cases}$$ The $r_{n}$ will be the radius of subintervals of $I_{n}$ centered in $a_{n}.$ Finally, define $E_{n}=(a_{n}-\frac{r_{n}}{2},a_{n}+\frac{r_{n}}{2})$ and $E=\bigcup_{n\in\mathbb{Z}}E_{n}.$ By construction, $E_{n}\subset I_{n},$ so
$$l^{*}(E\cap I_{n})=l^{*}(E_{n})=r_{n}>0\; \forall\, n\in\Bbb{Z}.$$
Together, we have
$$l^{*}(E)=l^{*}\left(\bigcup_{n\in\mathbb{Z}}E_{n}\right)\leq\sum_{n\in\mathbb{Z}}l^{*}(E_{n})=\sum_{n=1}^{\infty}l^{*}(E_{n})+\sum_{n=-\infty}^{-1}l^{*}(E_{n})+l^{*}(E_{0})=2\sum_{n=1}^{\infty}r_{n}+\frac{1}{2}=2\sum_{n=1}^{\infty}\frac{1}{2^{n}}+\frac{1}{2}=4+\frac{1}{2}<\infty\implies l^{*}(E)<\infty. $$
I have two questions:
1) Is my construct of $E$ consistent? If not, what need I to change?
2) Can I use some of these facts to prove the sufficient and necessarilly conditions for $l^{*}-$measurable ? How to approach this question? I really don't know.