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.

  • $\begingroup$ 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. $\endgroup$
    – egreg
    Commented Dec 31, 2013 at 17:31
  • $\begingroup$ 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. $\endgroup$
    – Brian
    Commented Dec 31, 2013 at 17:34
  • $\begingroup$ 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? $\endgroup$
    – Thomas
    Commented Dec 31, 2013 at 17:37
  • $\begingroup$ 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... $\endgroup$
    – Thomas
    Commented Dec 31, 2013 at 18:38
  • $\begingroup$ @Thomas Where did this problem come from? $\endgroup$ Commented Jan 7, 2014 at 21:02

2 Answers 2


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?


Your answer is completely correct. This is not an outer measure. It just takes the convex hull of the set (which is an interval) and finds half of its Euclidean length. So unioning two small sets (i.e. small 'measure') that are far apart gives a set with large 'measure', exactly as in your example. So this can't be an outer measure.

  • 1
    $\begingroup$ (-1) this should be a comment $\endgroup$
    – Norbert
    Commented Jan 7, 2014 at 20:28
  • $\begingroup$ @Norbert Answering questions in comments has been deprecated. If you are concerned about the bounty, I can delete my answer until after it expires. Alternatively, you can upvote the OP's answer like I did or write a more satisfactory answer. $\endgroup$ Commented Jan 7, 2014 at 21:06
  • $\begingroup$ I can see why the last line (that is now edited out) could have tricked a reader into thinking it was actually a comment, but I think the first paragraph addresses the reader's question. (It confirms the OP's work.) $\endgroup$
    – rschwieb
    Commented Jan 7, 2014 at 21:36
  • 1
    $\begingroup$ Lots of people visited my topic, but only @Brian Rushton gave me an answer. So I give him my points. Thanks! $\endgroup$
    – Thomas
    Commented Jan 10, 2014 at 9:26

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