# I Ascending subset sequence with axiom of choice

1. Aug 9, 2017

### 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.

2. Aug 14, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Aug 15, 2017

### 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?

4. Aug 15, 2017

### Someone2841

5. Aug 17, 2017

### 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.

Last edited: Aug 17, 2017