True or False Question on Outer measure.

Question

State TRUE or FALSE giving proper justification

1) There exist an unbounded subset $A$ of $\mathbb{R}$ s.t $m^*(A)=5$

2) There exist an open subset $A$ of $\mathbb{R}$ s.t $[1/2, 3/4] \subset A$ and $m^*(A)=1/4$

3) There exist an open subset $A$ of $\mathbb{R}$ s.t $m^*(A)<1/5$ but $A \cap (a, b) \neq \phi$ for all $a, b \in \mathbb{R}$ with $a<b$

4) If $A$ & $B$ are open subsets of $\mathbb{R}$ s.t $A \subsetneq B$, then it is necessary that $m^*(A)< m^*(B)$

My thought:

1) TRUE as we have the set $[0, 5] \cup \mathbb{Z}$

3) TRUE , if we consider the set $[0, 1/6] \cap \mathbb{Q}$

CHECK this true and please help on 2 and 4.

Answer 1

First of all, all statements are about nice Borel sets so there is no need to talk of outer measure. Your answer to 3) is wrong because your set is not open. To answer 3) arrange rational numbers in a sequence $\{r_n\}$ and take the interval around $r_n$ with length $\frac 1 {2^{n} 10}$. The union A of these intervals is open and $m(A)< \frac 1 5$. The answer to 4) is no. Let A be the union of the intervals removed in the construction of Cantor set and $B=(0,1)$. Then $m(A)=m(B)=1$ even though A is proper subset of B.

Answer 2

Any open set $S$ containing the point $1/2$ will have a subset $(-d+1/2,\;d+1/2)$ for some $d>0,$ whereupon if also $S\supset [1/2,\;3/4]$ then $$S\supset (-d+1/2,\;d+1/2)\cup [1/2,\;3/4]\supset (-d+1/2,\;3/4] $$ which implies $m^*(S)\geq m^*(\;(-d+1/2,\;3/4]\;)=3/4-(1/2-d)=1/4+d>1/4.$

Measure and outer measure are meant to be generalizations of the notion of length, so it is helpful that their definitions imply that the measure, and the outer measure, of an interval, are, in fact, its length.