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Without using $$m^*(\bigcup_{i=1}^\infty A_i)=\lim_{n\to\infty}m^*{(A_1\cup...\cup A_n)}$$ how to prove it?

$m^*$ is the Lebesgue outer measure.

Linked: Is $\mu(\cup A_n)=\sum\mu(A_n)$ an overstatement?.

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It would follow from the definition of measurable set that $$m^*(A \cup B)= m^*(A)+m^*(B)$$ for every disjoint pair of subsets $A, B \subset \mathbb{R}$ implies that every subset of $\mathbb{R}$ is measurable. However, this is a contradiction, since there is at least one measurable set, eg the Vitali set on $[0,1].$

For the second part, use induction. Certainly the case is true if $n=1$. Now assume it is true for the $(n-1)th$ case, then noting that $(\bigcup_{k=1}^nA_k)\cap A_n=A_n$ and $(\bigcup_{k=1}^nA_k)\cap {A_n}^{C}=\cup_{k=1}^{n-1}A_k$ are disjoint, $$m^*(\cup_{k=1}^{n} A_k)=m^*(A_n)+m^*\left(\cup_{k=1}^{n-1}(A_k)\right)=m^*(A_n)+\sum_{k=1}^{n-1}m^*(A_k)=\sum_{k=1}^{n}m^*(A_k).$$

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  • $\begingroup$ Reference: see Royden, Real Analysis 4th Edition, pp.36-37 and pp.48-49. $\endgroup$ – Darrin Mar 6 '14 at 4:59

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