$m(E_i)=\infty \space \forall i$ and $m(E)=const$

Question

$E ↘ \bigcap_n^\infty E_k$. Is it possible to construct the sequence $E_k$, $E_k \supset E_{k+1}$, such that $m(E_i)=\infty\quad\forall i$ and

1. $m(E)=\infty$ ?
2. $m(E)=0$ ?
3. $m(E)=\text{const} \neq0$ ?

I'm given this problem and I try to find such sequences.

1. It might is this sequence: $E_n=\big [- \frac{1}{n},\infty \big )$

2. It is easy to see that $E_n=(n,\infty)$ is a good sequence.

But what about the 3 one? Any help would be wonderful.

Answer 1

For 3), try $E_{n}=(-\infty,-1-n]\cup[-1,1]\cup[1+n,\infty)$.

Answer 2

If you have 2. (with say $E_i \subseteq [0,\infty)$ for all $i$), you get 3. by setting $E_i' = [-2,-1]\cup E_i$ for all $i$. Then the constant is $1$.

For another constant, say $\alpha$, replace $[-2,-1]$ by $[-1-\alpha,-1]$.