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I have a homework problem that I've been stuck on for some time. Define $\mathcal{A}$ to be an algebra on $\mathbb{Q}$ generated by the set $S = \{ (a, b] \cap \mathbb{Q} : a, b \in \mathbb{Q} \}$, and define the function $\mu_0 : \mathcal{A} \to [0, \infty]$ by $$ \mu_0 \left( \bigcup_{i=1}^{\infty} (a_i, b_i] \cap \mathbb{Q} \right) = \sum_{i=1}^{\infty} (b_i - a_i) $$ for a disjoint collection of sets $\{ (a_i, b_i] \cap \mathbb{Q} \} \subseteq S$. The question asks to show that $\mu_0$ is not a premeasure on $\mathcal{A}$.

First I see that $$ \mu_0(\varnothing) = \mu_0 \left( \bigcup_{i=1}^{\infty} (i, i] \cap \mathbb{Q} \right) = \sum_{i=1}^{\infty} (i - i) = 0. $$ Also, $$ \mu_0((a, b] \cap \mathbb{Q}) = b - a \quad (1) $$ by taking the union with infinitely many empty sets. So now for a disjoint collection of sets $\{ (a_i, b_i] \cap \mathbb{Q} \} \subseteq S$ whose union is in $\mathcal{A}$, why doesn't $(1)$ imply we have $\sigma$-additivity: $$ \mu_0 \left( \bigcup_{i=1}^{\infty} (a_i, b_i] \cap \mathbb{Q} \right) = \sum_{i=1}^{\infty} (b_i - a_i) = \sum_{i=1}^{\infty} \mu_0((a_i, b_i] \cap \mathbb{Q})? $$

tj1991
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    Whats the definition of a premeasure ? – Rene Schipperus Sep 27 '14 at 19:53
  • The definition I am working with says $\mu_0$ is a premeasure on $\mathcal{A}$ if $\mu_0(\varnothing) = 0$, and if ${A_i} \subseteq \mathcal{A}$ is a sequence of disjoint sets whose union is in $\mathcal{A}$, then $\mu_0 \left(\bigcup_i A_i \right) = \sum_i \mu_0(A_i)$. – tj1991 Sep 27 '14 at 19:58

1 Answers1

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Let $(\varepsilon_n)_{n\in\mathbb{N}}$ be a sequence of rational numbers with $\sum_{n=1}^\infty\epsilon_n<1$, and let $(q_n)_{n\in\mathbb{N}}$ be an enumeration of $(0,1]\cap\mathbb{Q}$.

Define intervals $I_n$ for $n=1,2,\ldots$ as follows: If $q_n\in\bigcup_{k=1}^{n-1}I_n$, lelt $I_n=\emptyset$. Otherwise, let $I_n=(p_n,q_n]$ where $p_n$ is the largest of the numbers $0$, $q_n-\varepsilon_n$, $q_1$, … $q_{n-1}$ less than $q_n$.

Now $(0,1]\cap\mathbb{Q}$ is the disjoint union of the rational intervals $I_n\cap\mathbb{Q}$, and the sum of the lengths of these intervals is less than $1$.

Harald Hanche-Olsen
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  • So, basically this shows that $\mu_0$ is not even well-defined, doesn't it? – PhoemueX Sep 27 '14 at 21:17
  • @PhoemueX Yes, good point. The normal way to do the example would be to consider the algebra generated by the half open intervals, as the text indeed implies – and not the σ-algebra. And then this example shows that $\mu_0$, though being additive on this algrebra, is not countably additive. – Harald Hanche-Olsen Sep 28 '14 at 09:06
  • What is $k$ in the Harald's answer ? – user174318 Sep 29 '14 at 00:51
  • I don't understand why $(0,1]$ is the disjoint union of $I_n$'s and why the sum of the lengths of these inetrvals less than $1$. Can anyone explain it more clearly? – user174318 Sep 29 '14 at 01:03
  • @user174318 Actually, there was a typo: I had meant to write $(0,1]\cap\mathbb{Q}$, not $(0,1]$. That is fixed now; thanks for asking the question that helped me notice the typo. Any number in this set is $q_n$ for some $n$. If $q_n$ is already in the union of the first $n-1$ intervals, I don't add a new interval (technically, put $I_n=\emptyset$). Otherwise, I do add a new interval with $q_n$ in it, so that $q_n$will belong to the union. So now any rational number in $(0,1]$ is in the union of the I_n. And I pick the new interval small enough to be disjoint from the previous one … – Harald Hanche-Olsen Sep 29 '14 at 05:26
  • @HaraldHanche-Olsen, if instead of $(0,1] \cap \mathbb{Q}$, I let the interval be $(0,1]$ in $\mathbb{R}$. Suppose that I apply $\mu_0$ to this set, and use your construction (mentioned above), I should still get the same result right? i.e. the infinite intervals ${I_n}$ will cover $(0,1]$, but the sum of their lengths is less than $1$, i.e. $\mu_0$ is not countably additive as well in this set... – Link L Sep 12 '19 at 04:26
  • @HaraldHanche-Olsen [PEDANTRY] A bit late to the party but I think there's another small typo (that's not really consequential at all). I think you meant $q_n \in \cup_{k=1}^{n-1}I_k$, not $q_n \in \cup_{k=1}^{n-1}I_n$. – Ray Mar 23 '23 at 14:53
  • I have trouble rigorously extending the case to $\mathbb{R}$. I mean, I can't prove that your construction actually covers $(0, 1]$ in $\mathbb{R}$. I mean I totally understand that it covers all of $(0, 1] \cap \mathbb{Q}$, because every $q \in (0, 1] \cap \mathbb{Q}$ either "has its own interval" $(p, q]$ for some $p < q$ or it was "already covered by another interval", but how do I show that some $r \in \mathbb{R}$, $r \notin \mathbb{Q}$ wasn't "missed out" by this construction? – Ray Mar 23 '23 at 15:08
  • But on second thought, I guess you could argue that the set of reals in $(0, 1]$ that aren't covered by your construction has measure 0, as follows. Let $S$ be the set covered by your construction, then $\mathbb{Q}\cap(0, 1] \subseteq S \subseteq (0, 1]$. Then any $r \in (0, 1]\backslash S$ is a singleton, that is, $\nexists a < r < b$ s.t. $(a, b) \subseteq (0, 1] \backslash S$. Note that $\mu({r}) = 0$. Then $\mu (0, 1] - \mu(S) = \mu((0, 1]\backslash S) = \mu(\cup_{r\in(0, 1]\backslash S}{r} = \sum_{r\in(0, 1]\backslash S}\mu({r}) = \sum_{r\in(0, 1]\backslash S} 0 = 0$. – Ray Mar 24 '23 at 13:01