# Measure Theory - working with unusual measures and set functions

1. Let $m$ define the Lebesgue measure. Let $\mu$ define the measure $\mu(A)=m(A\cap(0,1))$ for a Borel set $A$. Let $K=\bigcap \{A:A$ is closed, $\mu(A)=1\}$, $D=\bigcap \{G:G$ is open, $\mu(G)=1\}$. Determine which points belong to $K$ and which points belong to $D$.

2. Let $M$ be the algebra consisting of finite unions of sets of the form $\mathbb{Q} \cap (a,b]$ where $0 \leq a < b \leq 1$. Define the finitely additive set function $\mu$ on $M$ as $\mu(\mathbb{Q} \cap (a,b])=b-a$ and $\mu \big(\bigcup_{i=1}^n A_i \big)=\sum_{i=1} ^m \mu (A_i)$ with $A_i \in M$ pairwise disjoint. Is $\mu$ countably additive on $M?$

These are a few more old qualifying exam practice questions. I wanted to take a stab and see if my direction was solid.

For number 1, my answer for $K$ is the interval $[0,1]$. First, for any element $a$ outside of this interval, we can construct a closed set that does not contain it but contains $[0,1]$ (easy enough to do). Then, for $a=0$ note any closed set without zero, $C-{0}$ containing $(0,1)$ would lose a limit point and thus no longer be closed. So all closed sets containing $(0,1)$ must also contain $0$ (and $1$ likewise).

Furthermore, there are no sets containing a closed set in $[0,1]$ that is a proper subset of $[0,1]$. The reason for this is that, if so, then the compliment of that closed set in $[0,1]$ is open and of measure zero in $[0,1]$, and if it were nonempty, as an open set it would contain at least one interval of positive length in $[0,1]$, a contradiction, since the compliment in $[0,1]$ must be of measure zero and open.

My proposition is that $D$ is empty. For any $x,$ we have $(-\infty, x)\cup (x, \infty) \in D$. Thus there are no points in the common intersection. $////$

For number 2, I am very confused on how this one is supposed to work - it is almost like the information about the intersection with the rationals is irrelevant. Since the rationals are dense in $\mathbb{R}$, $\mu$ works just as the Lebesgue measure and so it is additive. That answer is raw, but I don't want to flesh it out because it seems too easy and incomplete.

For the second one, you can enumerate the rationals in $(0, 1]$, call them $\{a_i\}_{i\in\mathbb{N}}.$ For each such $a_i$ make $A_i = (a_i-7^{-2i-2}, a_i+7^{-2i-2}] \cap (0, 1] \cap \mathbb{Q}$. Now, the union is $(0, 1] \cap \mathbb{Q}$ which has a measure of 1, but the sum of the measures is strictly less than 1, and you can make it as small as you like.