1. ## Bounded, measurable?

1. Suppose $A$ is a bounded set and that $m^{*}(
A\cap I)\leq am^{*}(I)$
for every interval I and $0 < a < 1$. Prove that $m^{*}(A)=0$. What if A is unbounded?

2. $A$ is a subset of $\mathbb{R}$ with the following property: for every $\epsilon > 0$ there exist measurable sets B and C such that :

$B \subset A \subset C$ and $m(c \cap B^c) < \epsilon$. Prove that A is measurable.

thanks.

2. Originally Posted by davidmccormick
1. Suppose $A$ is a bounded set and that $m^{*}(
A\cap I)\leq am^{*}(I)$
for every interval I and $0 < a < 1$. Prove that $m^{*}(A)=0$. What if A is unbounded?

2. $A$ is a subset of $\mathbb{R}$ with the following property: for every $\epsilon > 0$ there exist measurable sets B and C such that :

$B \subset A \subset C$ and $m(c \cap B^c) < \epsilon$. Prove that A is measurable.

thanks.
1.

Let { $I_k$} be a collection of open intervals such that $A \subset \bigcup I_k$

Now, note that $A = \bigcup_k (A \cap I_k)$, so by subadditivity and the assumption we have:

$m^*(A) \leq \sum_k m^*((A \cap I_k) \leq a*\sum_k m^*(I_k)$. I.e.

$\sum_k m^*(I_k) \geq \frac{m^*(A)}{a}$ for every collection of open intervals that cover A. Thus, we have that:

inf{ $\sum_k m^*(I_k)$; $A \subset \bigcup I_k$} $\geq \frac{m^*(A)}{a}$.

But, the term on the left is m*(A)! Which gives us:

$m^*(A) \geq \frac{m^*(A)}{a}$ which is only possible in m*(A) = 0 (try it out)

Notice that I never mentioned once that A is bounded

EDIT: Monotonicity replaced with countable sub-additivity