Proof of subadditivity of the Lebesgue measure In the proof of the following proposition:  Let $\{E_n\}$ be a countable collection of sets of real numbers. Then $$ m^\ast\left(\bigcup E_n\right)\leq \sum m^\ast\left(E_n\right)~,$$ 
we  suppose that $m^\ast(E_n)$ is finite for all $n$.
Then for each $E_n$, there is a countable collection $\{I_{k}^{n}:k\geq 1\}$ such that $E_n\subset \bigcup_{k}I_{k}^{n}$ $~~$ and $$\sum l\left(I^{n}_{k}\right) \leq m^\ast(E_n)+\frac{\epsilon}{2^n},~~~\epsilon >0.$$ 
Here is my question. I don't understand why the last inequality true. Explanations will be very much appreciated.
Thanks.
 A: $m^*(E_n)=\inf{\left\{\displaystyle\sum_{k=0}^\infty \ell(I_k)\mid E_n\subseteq\bigcup_{i=0}^\infty I_k \right\}}$, where $I_k$ is an open bounded interval. Since what is on the left hand side of your inequality is one of these possible coverings, the infimum, $m^*(E_n)$ is less than or equal to that. This means you can find an $\epsilon>0$ such that $m^*(E_n) + \epsilon > \displaystyle\sum_{k=0}^\infty \ell(I_k)$, where $\{I_k\}_{k=0}^\infty$ is one possible colleciton of bounded open sets containing $E_n$. Since we have a countable collection of these sets, $E_n$, we can associate to each set a natural number $n$ so that your inequality holds for the choice of $\dfrac{\epsilon}{2^n}$ for the $n$th set.
A: From the foundation of mathematical analysis, I.e from the definition of infimum. Which is defined as follow
Definition: let $X$ be a non empty set $\alpha=\inf(X)$ if for every $\epsilon>0$ there exists a $d_{\epsilon}\in\,X$ such that $\alpha\le\,d_{\epsilon}<\alpha+\epsilon$.
Now, using the definition of exterior measure, which is defined to be 
$m^*E=\inf\left\{\sum \,l(I_n):E\subset\bigcup\,I_n\right\}$ hence since $m^*E$ is the inf of all d sum of the intervals we can define it like inf I.e. given any $\epsilon>0$ we can find and interval $I_n^{\epsilon}$ such that it covers E, then 
$m^*E\le\sum l(I_n^{\epsilon})<m^*E+\epsilon$ but since its for any $\epsilon>0$ we can choose $\epsilon=\frac{\epsilon}{2^n}$ hence it becomes
$m^*E\le\sum l(I_n^{\epsilon})<m^*E+\frac{\epsilon}{2^n}$. They just used the last inequality. 
