The function is $\nu:\mathcal{P}(X) \rightarrow \mathbb{R}$
I want to show if a),b),c) are outer measures.
a) X is anything, $\nu(A)=\begin{cases}
0 & \text{für } A = \emptyset \\
1 & \text{für } A \neq \emptyset \\
\end{cases}$
I'd say this is an outer measure. 1.) $\nu(\emptyset)=0$ is trivial, 2) $A\subseteq B \Rightarrow \nu(A) \leq \nu(B)$ is also pretty obvious and 3) $\nu(\bigcup_{i=1}^{\infty} A_i) \leq \sum_{i=1}^{\infty} \nu(A_i)$ also seems correct since the left side can either be 1 or 0. If it's zero, the right side is also zero. If it's 1, the right side is bigger or equal 1.
b) X is infinite, $\nu(A)=\begin{cases}
0 & \text{für } |A| < \infty \\
1 & \text{für } |A| = \infty \\
\end{cases}$
I'd say this is not an outer measure since 3) $\nu(\bigcup_{i=1}^{\infty} A_i) \leq \sum_{i=1}^{\infty} \nu(A_i)$ isn't always true.
Assume you split the infinite set in single-element subsets. Meaning that $X = \bigcup_{i\in\mathbb{N}}^{\infty} \{i\}$ with $A_i = \{i\}$
Then…
$1=\nu(\bigcup_{i=1}^{\infty} A_i) \leq \sum_{i=1}^{\infty} \nu(A_i) = 0$ is obviously wrong. Therefore I'd say that b) isn't an outer measure
c) X is uncountable, $\nu(A)=\begin{cases}
0 & \text{für } |A| \leq \aleph_0\\
1 & \text{für } |A| > \aleph_0\\
\end{cases}$
I'd say 1) and 2) are pretty trivial again. 3.) is clear if every set $A_i \subseteq \mathcal{P}(X)$ is finite or countable. If there exists an $A_i$ that is uncountable and without loss of generality $A_i$ is disjoint to a set $A_j$ (of course, $|A_j|=\lambda \leq \kappa = |A_i|$), then: $|A_i \cup A_j| = |A_i|+|A_j| = \max(\lambda, \kappa) = \kappa$
Therefore $1 = \nu(\bigcup_{i=1}^{\infty} A_i) \leq \sum_{i=1}^{\infty} \nu(A_i) = \#\{A_i: A_i \text{ is uncountable}\} \geq 1$
From that I conclude that c) is an outer measure
Is that correct?
Best Answer
This looks fine. A remark about (c). I believe you have the right idea but the way you wrote it seems slightly suspect. A better way to do it would be to note the following fact: a countable union of countable sets is countable. This just uses the fact that if there $A_{i}$ is countable for every $i \geq 1$, then there is an obvious surjection from $\mathbb{N} \times \mathbb{N}$ onto $\cup{i}A_i$, and then $|\mathbb{N} \times \mathbb{N} | = |\mathbb{N}|$.
Given this one can do the last line of your solution (c), i.e, $\mu(\cup_{i} A_i) = 1$ iff it is uncountable iff it has at least one uncountable $A_i$, call it $A_k$ in which case, $\Sigma_i \mu(A_i) \geq \mu(A_k) = 1 = \mu(\cup_{i} A_i)$