In the Littlewood’s 1st Principle, why do we need the finiteness assumption m∗ (E) < ∞?

Question

Theorem (Littlewood's 1st Principle)

Every measurable set of finite measure is nearly a finite union of disjoint open intervals, in the sense

(i) If E is measurable and m∗(E) < ∞, then for any ε > 0 there is a finite union U of open intervals such that m∗(U△E) < ε.(Clearly, the intervals can be chosen to be disjoint.)

(ii) If for any ε > 0 there is a finite union U of open intervals such that m∗ (U△E) < ε, then E is measurable.(The finiteness assumption m∗ (E) < ∞ is not essential.)

I don't understand why this theorem need the assumption m∗ (E) < ∞.

Will there be a counter example if the assumption is changed to m∗ (E) = ∞ ?

Answer 1

Consider $E = [1,2] \cup [3,4] \cup [5,6] \cup \cdots$