Ascending subset sequence with axiom of choice

[jostpuur]

Is it possible to use Axiom of Choice to prove that there would exist a sequence (A_n)_{n=1,2,3,\ldots} with the properties: A_n\subset\mathbb{R} for all n=1,2,3,\ldots,

<br /> A_1\subset A_2\subset A_3\subset\cdots<br />

and

<br /> \lim_{k\to\infty} \lambda^*(A_k) &lt; \lambda^*\Big(\bigcup_{k=1}^{\infty} A_k\Big)<br />

where \lambda^* is the Lebesgue outer measure?

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

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

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

 

[Someone2841]

Sorry if this comes from a place of ignorance, but I notice something and I wonder if you'd comment on it.

Since every $A_n \subset A_{n+1}$, it seems that it follows by induction that $A_n = \bigcup_{k=0}^n A_k$. If this is the case, it seems that $\lim_{k\to \infty}A_k=\bigcup_{k=0}^\infty A_k$ and so naturally $\lim_{k\to \infty}\lambda(A_k)=\lambda(\bigcup_{k=0}^\infty A_k)$. If this is true, how would the AoC avoid this?

 

[Someone2841]

The flaw on my reasoning was assuming that $\lambda *$ was continuous from below even with non-measurable sets. This ends up being true: https://math.stackexchange.com/questions/116847/continuity-from-below-for-lebesgue-outer-measure. So no, no such sequence of sets may be constructed.

 

[jostpuur]

The proof on math stack exchange contains a little mistake, but I got a feeling that the proof works anyway, and the little mistake can be fixed. The mistake is that the guy forgets that the sets G_k depend on epsilon, so they are like sets G_k(\epsilon). He first proves that

<br /> m\Big(\bigcup_{k=1}^n G_k\Big) &lt; m^*(E_n) + \Big(1 - \frac{1}{2^n}\Big)\epsilon<br />

and then states that because the epsilon is arbitrary, also

<br /> m\Big(\bigcup_{k=1}^n G_k\Big) \leq m^*(E_n)<br />

would be true, but that does not make sense.

I did not find any mistake before the point where the (1-\frac{1}{2^n})\epsilon spread is present, so I believe that it is true. That inequality implies

<br /> m\Big(\bigcup_{k=1}^{\infty} G_k\Big) \leq \lim_{n\to\infty}m^*(E_n) + \epsilon<br />

and this is sufficient to complete the proof.