Caratheodory Extension theorem proof the role of ϵ I would be very thankful if someone could answer this question. I am new here so I copied a similar question and I asked my doubt.
$$\lambda(A) = \lambda (A \cap E) + \lambda (A \cap E^c)$$
where $\lambda$ is an outer measure, $A \subset R$, $E \subset R$, and $E$ is an elementary set. That is, $E$ is a union of finitely many bounded intervals.
By countable sub-additivity, I know 
$$ \lambda(A) \leq \lambda(A \cap E) + \lambda(A \backslash E) $$
since $A = (A \cap E)\cup (A \backslash E)$. 
I need help understanding the opposite inequality.
By the definition of the Lebesgue outer measure, for each $\epsilon>0$, there an exists open set $O$, with $O \supset A$ such that $\lambda(O) \leq \lambda(A) + \epsilon$. Now observe that 
$$ \lambda(A) + \epsilon \geq \lambda(O) = \lambda(O \cap E) + \lambda (O \backslash E) \geq \lambda(A \cap E) + \lambda (A \backslash E) $$
The equality above comes from the fact that the Lebesgue outer measure is countably (and finitely) additive for Lebesgue measurable sets.
Since $\epsilon$ is arbitrary, we have $\lambda (A) \geq \lambda(A \cap E) + \lambda(A \backslash E)$.
I do not understand the role of $\epsilon$. How can $\lambda(A) \geq \lambda(A \cap E) + \lambda(A \backslash E)$ when $\epsilon$ is zero? 
 A: This is a basic argument from real analysis/advanced calculus. The chain of inequalities you derived gives you that, for all $\epsilon > 0$,
$$ \lambda(A) + \epsilon \geq \lambda(A \cap E) + \lambda(A \cap E^c) $$
Since this is true for all $\epsilon > 0$, it must be true for $\epsilon = 0$ as well.
Suppose not. Then:
$$ \lambda(A) < \lambda(A \cap E) + \lambda(A \cap E^c) $$
which means we can find $\epsilon$ small enough so that 
$$ \lambda(A) + \epsilon < \lambda(A \cap E) + \lambda(A \cap E^c) $$
which is a contradiction.
Alternatively, recall that, in $\mathbb R$, limits preserve inequalities. (That being said, I suspect you'll need to employ an argument by contradiction similar to the one I just gave to prove this fact if you don't already know it.)
A: The choice of open set $O$ depends on your $\epsilon$. You initially use $O$ to deduce some inequalities but then you get an inequality involving $\epsilon$ but no $O$. So the $\epsilon$ dependence is now missing, as a result the inequality is true for all $\epsilon>0$. Take $\epsilon\to0$ to get the answer finally.
