# Counterexample for a premeasure on a semiring over Q

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In summary, the problem is to show that \mu_{\alpha} is not countably additive by using the incompleteness and measure zero properties of the rationals. This can be done by constructing a sequence of intervals with measures approaching zero, but whose union is not in the family P. The key is to use the fact that the rationals can be covered by countably many intervals with arbitrarily small total length.
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## Homework Statement

Let $$\alpha(r)=r$$ and let $$P$$ be the family of intervals $$[a,b)$$ in $$\mathbb{Q}$$. Define $$\mu_{\alpha}([a,b))=\alpha(b)-\alpha(a)$$. Show by example that $$\mu_{\alpha}$$ is not countably additive.

## Homework Equations

$$\mu$$ is countably additive if for any sequence of mutually disjoint subsets $$E_1, E_2, ...$$ such that $$\bigcup_{i=1}^{\infty}E_i\in P$$, we have $$\mu\left(\bigcup_{i=1}^{\infty}E_i\right)=\sum_{i=1}^{\infty}\mu(E_i)$$.

## The Attempt at a Solution

This problem is driving me insane. I know I have to somehow use the incompleteness of the rationals, so I tried to use sequences which I know do not converge in $$\mathbb{Q}$$. For example, I tried defining $$x_{n+1}=x_n + 1/n^2$$ and $$E_n=[x_n,x_{n+1})$$, where $$x_1=0$$. Then $$\mu(E_n)=1/n^2$$, so $$\sum_{i=1}^{\infty}\mu(E_i)=\pi^2/6$$, which is not in $$\mathbb{Q}$$. But this does not work because $$\bigcup_{i=1}^{\infty}E_i=[0,\pi^2/6)\notin P$$.

Basically I keep running into two problems. Either this happens, or I find a sequence whose union is in $$P$$ but $$\mu$$ is countably additive. Any hints or suggestions would be greatly appreciated.

The hint is that you need to use not only the incompleteness of the rationals, but also the fact that they have measure zero. That is, for every epsilon>0, it is possible to cover Q by countably many intervals s.t the sum of their lengths is less than epsilon. That's because Q is countable. Thus, these intervals will have union [0,1)$$\cap$$Q, but the sum of their measures will be as small as you like.

## 1. What is a counterexample for a premeasure on a semiring over Q?

A counterexample for a premeasure on a semiring over Q is a specific example that disproves the validity of a certain statement or theorem regarding premeasures on semirings over the rational numbers (Q). This counterexample can be used to show that the statement or theorem is not universally true.

## 2. How is a premeasure defined on a semiring over Q?

A premeasure is a function that assigns a non-negative value (usually interpreted as a measure or size) to a specific set. In the case of a semiring over Q, this function must satisfy certain properties such as being countably additive and having the empty set as its smallest value.

## 3. Why is a counterexample important in mathematics?

Counterexamples are important in mathematics because they help to refine and strengthen our understanding of mathematical concepts. By providing a specific example that disproves a statement or theorem, counterexamples help to identify the limitations and boundaries of mathematical theories and can lead to the development of new and improved theories.

## 4. How can a counterexample be constructed for a premeasure on a semiring over Q?

A counterexample for a premeasure on a semiring over Q can be constructed by carefully choosing a specific set and assigning values to its subsets in a way that violates the properties of a premeasure. This may involve using irrational numbers or non-standard operations to manipulate the values of the subsets.

## 5. Can a counterexample be used to disprove a premeasure on a semiring over Q?

Yes, a counterexample can be used to disprove a premeasure on a semiring over Q if it clearly violates one or more of the properties required for a premeasure. However, it is important to note that a single counterexample does not necessarily prove that a premeasure is invalid; further analysis and counterexamples may be needed to fully understand the limitations of the premeasure.

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