Sub-additivity of Lebesgue Outer Measure

Question

I got stuck reading a proof of the sub-additivity of the Outer Measure $\lambda^*$, namely the proof of the statement that $$ \lambda^*\left(\bigcup_{i=1}^\infty E_i\right)\le\sum_{i=1}^\infty\lambda^*(E_i). $$ If $\epsilon>0$, there is a countable covering $\{R_{ij}\}_{j\in\mathbb{N}}$ of $E_i$ such that $$ \sum_{j=1}^\infty \lambda(R_{ij})\le\lambda^*(E_i)+\frac{\epsilon}{2^i}, \quad E_i\subset\bigcup_{j=1}^\infty R_{ij}. $$ Then, after doing some manipulation we get the result $$ \lambda^*(E)\le\sum_{i=1}^\infty\lambda^*(E_i)+\varepsilon, $$ allowing us to conclude that, because $\varepsilon>0$ was arbitrary, our condition is proved. Why is this so? Just because it is arbitrary does not mean anything does it?

Answer 1

If $\forall \epsilon > 0, a \leq b + \epsilon$ then $ a \leq b $.

Indeed, if $a > b$, then $\epsilon = \frac{a-b}{2} > 0$, and the hypothesis imply that

$a \leq b + \frac{a-b}{2}$

And this is equivalent to

$\frac{a}{2} \leq \frac{b}{2}$

Contradiction !