# Prove that $\mu_{\star}(A) = \frac{\sup A - \inf A}{2}$ is outer measure.

Let $X = \mathbb{N}$ and $\mu_{\star}: \mathcal{P}(\mathbb{N}) \rightarrow [0,\infty]$ such that $$\mu_{\star}(A) = \frac{\sup A - \inf A}{2}$$where $\sup \emptyset = \inf \emptyset = 0$. Prove that $\mu_{\star}$ is outer measure.

The first condition is obviously, because we have $\mu_{\star}(\emptyset) = 0$. So let $$A \subseteq \bigcup_{n=1}^{\infty} A_n$$we would like show that $$\mu_{\star}(A) \le \sum_{n=1}^{\infty} \mu_{\star}(A_n)$$

Hence from definition of $\mu_{\star}(A)$ we have to show:

$$\frac{\sup A - \inf A}{2} \le \sum_{n=1}^{\infty} \frac{\sup A_n - \inf A_n}{2}$$ So $$\sup A - \inf A \le \sum_{n=1}^{\infty} \sup A_n - \sum_{n=1}^{\infty} \inf A_n \quad (\dagger)$$

But as you can see in this topic: Inequality with infimum and supremum for $A \subseteq \bigcup_{n=1}^{\infty}A_n$

inequality $( \dagger)$ is not true. So this exercise is wrong? I am really confused, because it is next mistake today...

I will grateful for your help.

• The series $\sum_{n\ge1}(\sup A_n-\inf A_n)$ may converge also when the series $\sum_{n\ge1}\sup A_n$ and $\sum_{n\ge1}\inf A_n$ don't. Commented Dec 31, 2013 at 17:31
• Expanding on @egreg's comment, here's an example: Let $A_{n} = \left\{1\right\}$ for $n\geq 1$. Then, $\sup A_{n} = 1 = \inf A_{n}$, and so $\sum_{n\ge1}(\sup A_{n}-\inf A_{n}) = \sum_{n\ge1}0 = 0$ whereas the sums $\sum_{n\ge1}(\sup A_{n})$ and $\sum_{n\ge1}\inf A_{n}$ diverge. Commented Dec 31, 2013 at 17:34
• Ok, so I have to show that $\sup A - \inf A \le \sum_{n=1}^{\infty} ( \sup A_n - \inf A_n)$. Do you know how can I get it? Commented Dec 31, 2013 at 17:37
• But for $A = \{{0,1,10,11\}}$, $A_1 = \{{0,1\}}$, $A_2 =\{{10,11\}}$, $A_3 = A_4 = \dots = \{{0\}}$ this inequality also is not satisfied... I am really confused... Commented Dec 31, 2013 at 18:38
• @Thomas Where did this problem come from? Commented Jan 7, 2014 at 21:02

For $$A = \left\{ 0,1,10,11 \right\} \\ A_1 = \left\{ 0,1 \right\} \\ A_2 = \left\{ 10,11 \right\} \\ A_3 = A_4 = ... = \left\{ 0 \right\}$$
we have: $$\mu_{\star}(A) = \frac{11}{2}$$ but $$\sum_{n=1}^{\infty} \mu_{\star}(A_n) = 1$$
So do we have mistake in task?$\mu_{\star}$ is not outer measure?