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.

Look up nowhere dense sets if you're interested. Sorry for the hugely late answer too. I realise that this is no help!


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