# Are following set functions outer measures?

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?

• Another outer measure you may see is the Counting Measure: If $A$ is finite then $v(A)$ is the number of members of $A.$ If $A$ is infinite then $v(A)=\infty.$ – DanielWainfleet Nov 13 '20 at 23:49

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, $$\Sigma_i \mu(A_i) \geq \mu(A_k) = 1 = \mu(\cup_{i} A_i)$$