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I want to construct a generic finite interval under $\mathbb{R_+}$ - it should be bounded, closed and it should include $0$ as the lower bound. This would allow me to choose an element from this compact set (i.e. $ x \in [0, b] \subset R_+$ where b is a finite number) . I cannot do this with $\mathbb{R_+}$ because it is not bounded. How can I represent this interval?

I was looking at a text on Lebesgue outer measure and I found the following definition:

We define the $\text{Lebesgue outer measure}$ of a set $A$ in $\mathbb{R}$ by

$\lambda^*(A)$ = inf $\lbrace \sum_{j=1}^\infty |I_j|: A \subset \bigcup_{j=1}^\infty I_j$ where $I_j$ are bounded intervals $\rbrace$

They note that for a finite sum, $\lambda^*(A)$ is a (finite) nonnegative real number.

I guess I can use this. Any comments?

mathos
  • 3
  • A finite interval that contains all nonnegative real numbers? – Eff Jun 03 '15 at 19:38
  • @Eff the set of nonnegative real numbers is not bounded. I need to represent a bounded and closed (compact) interval under nonnegative real numbers. It should exclude infinity. – mathos Jun 03 '15 at 19:41
  • What do you mean construct? What is wrong with $[1,2]$? It is compact and a subset of $\Bbb R_+$. – Ross Millikan Jun 03 '15 at 19:49
  • @RossMillikan I am looking for a "generic" representation of the largest interval, $[0, b]$ where b can be any finite number . – mathos Jun 03 '15 at 19:52
  • The "largest interval" doesn't exist. Suppose for contradiction that it did exist, that you have some $b$ finite such that $m([0,b])$ is maximum (where $m$ represents the standard lebesgue measure). What can you say about $b+1$? Since $b$ is finite, $b+1$ is also finite. However, $m([0,b])=b<b+1=m([0,b+1])$ so $m([0,b])$ is not maximum. Thus, by contradiction, there is no "largest interval". – JMoravitz Jun 03 '15 at 19:55
  • @JMoravitz True. I think I made a mistake by saying the largest interval. I want to choose an element from a finite interval under $\mathbb{R_+}$ but I do not know what the upper bound is. How should I write this? Should I just say $x \in [0, b] \subset \mathbb{R_+}$ where b is any finite number ? – mathos Jun 03 '15 at 20:01
  • Suppose we have a finite (closed) interval $I$ where $I\subset \mathbb{R}+$. Since $I$ is an interval and is finite, we know there must exist some $a,b\in\mathbb{R}+$ with $a<b$ such that $I=[a,b]$. Picking some $x\in I$ then is picking some $x\in[a,b]$. However, without knowledge of what $I$ is, we cannot guess as to what $a$ and $b$ suffice. Suppose that we were to guess some $a,b$, with $a<b$ without knowledge of what $a$ or $b$ we needed to use. Then if $I=[b+1,2b+1]$ our guess not only will be wrong, but won't even intersect the true interval. – JMoravitz Jun 03 '15 at 20:05
  • The value of $a$ and $b$ is determined by what $I$ is, and we may refer to those values once we have declared the existence of $I$, though we have no further information as to what the exact value of $a$ is or the exact value of $b$ is, as their only definition comes from $I$ itself. – JMoravitz Jun 03 '15 at 20:07
  • @JMoravitz In the model that I am trying to construct I know that the element $x$ can take the value $0$ but I just don't know the upper bound and I need a generic representation for that. So $a= 0$ is given. – mathos Jun 03 '15 at 20:09
  • "$x\in[0,b]$ where $b$ is the upper bound", but as mentioned, if you are going to do any calculations involving $b$, you must know exactly what $b$ is, or must leave it as simply $b$ in the calculations, because you cannot guess a value as it will always have a chance of being wrong. – JMoravitz Jun 03 '15 at 20:12
  • @JMoravitz $x$ is used as a symbol throughout my model and no numerical calculations are needed. Thank you for these good comments. – mathos Jun 03 '15 at 20:14

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