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 $n=1,2,3,\ldots$,

$$A_1\subset A_2\subset A_3\subset\cdots$$

and

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

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

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

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

$$m\Big(\bigcup_{k=1}^n G_k\Big) < m^*(E_n) + \Big(1 - \frac{1}{2^n}\Big)\epsilon$$

and then states that because the epsilon is arbitrary, also

$$m\Big(\bigcup_{k=1}^n G_k\Big) \leq m^*(E_n)$$

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

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

and this is sufficient to complete the proof.