Is it possible to use Axiom of Choice to prove that there would exist a sequence [itex](A_n)_{n=1,2,3,\ldots}[/itex] with the properties: [itex]A_n\subset\mathbb{R}[/itex] for all [itex]n=1,2,3,\ldots[/itex],(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

A_1\subset A_2\subset A_3\subset\cdots

[/tex]

and

[tex]

\lim_{k\to\infty} \lambda^*(A_k) < \lambda^*\Big(\bigcup_{k=1}^{\infty} A_k\Big)

[/tex]

where [itex]\lambda^*[/itex] is the Lebesgue outer measure?

If we assume that all sets [itex]A_1,A_2,A_3,\ldots[/itex] are Lebesgue measurable, then it is known that

[tex]

\lim_{k\to\infty} \lambda(A_k) = \lambda\Big(\bigcup_{k=1}^{\infty} A_k\Big)

[/tex]

If we don't assume that the sets are measurable, the direction "[itex]\leq[/itex]" can still be proven easily, but the direction "[itex]\geq[/itex]" is more difficult.

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# I Ascending subset sequence with axiom of choice

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