Outer measure of an open interval ....

In summary: Suppose $x$ is in the set of values for $\{\sum_{k}l(I_{k})\,:\,\ldots\}$ ... ...Then $x = \sum_{k}l(I_{k})$ for some covering of $(a,b)$ ...But ... $\vert(a,b)\vert = \sum_{k}l(I_{k})$ for the covering $(a,b), \emptyset, \emptyset, ...$ ...So ... $\vert(a,b)\vert$ is a member of the set of values for $\{\sum_{k}l(I_{k})\,:\,\ldots\}$ ... ...... so ... we have that $\vert(a,b)\vert \le
  • #1
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I am reading Sheldon Axler's book: Measure, Integration & Real Analysis ... and I am focused on Chapter 2: Measures ...

I need help with proving that the outer measure of an open interval, \mid (a, b) \mid = b - a

Axler's definitions of length and outer measure are as follows:
Axler - Defn 2.1 & 2.2 .png

Can someone demonstrate rigorously that \(\displaystyle \mid (a, b) \mid = b - a\) ...

I know it seems intuitively obvious but how would you express a convincing and rigorous proof of the above result ...
Help will be much appreciated ... ...

Peter
 
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  • #2
Hi Peter,

I agree with you here. For something that seems so intuitively clear, the proof isn't a one-liner using only the definition. My suggestion would be to try and use the result you asked about in another post, namely that $\vert[a,b]\vert = b-a$. Also think about approximating $(a,b)$ using closed intervals contained in $(a,b)$. These two ideas together are enough to formulate a short argument proving $\vert(a,b)\vert = b-a.$ I'll leave it at this for now, but feel free to let me know if you have additional questions.
 
  • #3
Thanks for the help GJA ...

BUT ... still struggling with constructing a proof ...

Can you help further ...

Peter
 
  • #4
Certainly. Let's approximate $(a,b)$ via the closed intervals $[a+1/2n, b-1/2n].$ We then have $b - a - 1/n = \vert[a+1/2n, b - 1/2n]\vert\leq \vert (a,b)\vert$ for all $n$ (large enough), because $[a+1/2n, b-1/2n]\subset (a,b)$ for all $n$ (large enough). Since this holds for all $n$ (large enough), it follows that $b-a\leq \vert(a,b)\vert.$ Can you think of an argument that would show $\vert(a,b)\vert \leq b-a$? Hint: $(a,b)$ itself is an open interval that contains $(a,b)$. Think about this and the definition of outer measure.

Edit: I needed to add the qualifier "large enough" because $[a+1/2n, b-1/2n]$ may not be contained in $(a,b)$ for all $n$. However, since $a<b$, there is $N$ such that $[a+1/2n, b-1/2n]\subset (a,b)$ for all $n\geq N$; i.e., for all $n$ "large enough."
 
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  • #5
GJA said:
Certainly. Let's approximate $(a,b)$ via the closed intervals $[a+1/2n, b-1/2n].$ We then have $b - a - 1/n = \vert[a+1/2n, b - 1/2n]\vert\leq \vert (a,b)\vert$ for all $n$ (large enough), because $[a+1/2n, b-1/2n]\subset (a,b)$ for all $n$ (large enough). Since this holds for all $n$ (large enough), it follows that $b-a\leq \vert(a,b)\vert.$ Can you think of an argument that would show $\vert(a,b)\vert \leq b-a$? Hint: $(a,b)$ itself is an open interval that contains $(a,b)$. Think about this and the definition of outer measure.

Edit: I needed to add the qualifier "large enough" because $[a+1/2n, b-1/2n]$ may not be contained in $(a,b)$ for all $n$. However, since $a<b$, there is $N$ such that $[a+1/2n, b-1/2n]\subset (a,b)$ for all $n\geq N$; i.e., for all $n$ "large enough."
Thanks for the help GJA ...

To show that $\vert(a,b)\vert \leq b-a$ ... we proceed as follows ...

We have I_1, I_2, ... = $(a,b), \emptyset, \emptyset,$ ... ... is a covering of $(a,b)$ ... ...

Therefore $(a,b)$ is a lower bound for $\sum_{n=1}^{\infty} I_n = b - a$ ... ...

... so ... $\vert(a,b)\vert \leq b-a$ ...Is that correct?

Peter
 
  • #6
A few things to note. First, we wouldn't say $(a,b)$ is a lower bound because it's an interval, not a number. Next, we can only say something is a lower bound for a set when it's a value below all the numbers in the set. For example $3$ is a lower bound for the set $(5,10)$ but $9$ is not. Since you chose a specific covering of $(a,b)$ and not an arbitrary one, you cannot conclude that the value you came up with is a lower bound for $\{\sum_{k}l(I_{k})\, :\,\ldots\}$, because you don't know that the covering you chose produces a value that is smaller than all the values in the set. What you have shown is that $b-a$ belongs to the set of values $\{\sum_{k}l(I_{k})\,:\,\ldots\}$ and, therefore, $\vert[a,b]\vert\leq b-a.$
 
  • #7
GJA said:
A few things to note. First, we wouldn't say $(a,b)$ is a lower bound because it's an interval, not a number. Next, we can only say something is a lower bound for a set when it's a value below all the numbers in the set.
Below or equal to every number in the set.[/quote]

For example $3$ is a lower bound for the set $(5,10)$[//quote]
and so is 5.
but $9$ is not. Since you chose a specific covering of $(a,b)$ and not an arbitrary one, you cannot conclude that the value you came up with is a lower bound for $\{\sum_{k}l(I_{k})\, :\,\ldots\}$, because you don't know that the covering you chose produces a value that is smaller than all the values in the set. What you have shown is that $b-a$ belongs to the set of values $\{\sum_{k}l(I_{k})\,:\,\ldots\}$ and, therefore, $\vert[a,b]\vert\leq b-a.$
 
  • #8
GJA said:
A few things to note. First, we wouldn't say $(a,b)$ is a lower bound because it's an interval, not a number. Next, we can only say something is a lower bound for a set when it's a value below all the numbers in the set. For example $3$ is a lower bound for the set $(5,10)$ but $9$ is not. Since you chose a specific covering of $(a,b)$ and not an arbitrary one, you cannot conclude that the value you came up with is a lower bound for $\{\sum_{k}l(I_{k})\, :\,\ldots\}$, because you don't know that the covering you chose produces a value that is smaller than all the values in the set. What you have shown is that $b-a$ belongs to the set of values $\{\sum_{k}l(I_{k})\,:\,\ldots\}$ and, therefore, $\vert[a,b]\vert\leq b-a.$
Thanks GJA ...

Sorry ... I have confused things with a typo ... I wrote $(a,b)$ when I meant $\mid (a, b) \mid$ ... ...

So ... my post should have read as follows:

To show that $\vert(a,b)\vert \leq b-a$ ... we proceed as follows ...

We have $I_1, I_2, ... = (a,b), \emptyset, \emptyset,$ ... ... is a covering of $(a,b)$ ... ...

Therefore $\vert(a,b)\vert$ is a lower bound for $\sum_{n=1}^{\infty} I_n = b - a$ ... ...

... so ... $\vert(a,b)\vert \leq b-a$ ...
A word about the claim that " ... Therefore $\vert(a,b)\vert$ is a lower bound for $\sum_{n=1}^{\infty} I_n = b - a$ ... ..."

... $\vert(a,b)\vert$ is a lower bound for the set of all sums of the form $\sum_{n=1}^{\infty} I_n$ ... (indeed, it is the greatest lower bound ...) ... so it is less than the particular sum mentioned ...

Is the above correct ... apart from referring to a lower bound of a number rather than a set of numbers ...

Peter
 
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What is the definition of outer measure of an open interval?

The outer measure of an open interval is a measure of the length of the interval, which takes into account all possible coverings of the interval by smaller intervals.

How is the outer measure of an open interval calculated?

The outer measure of an open interval is calculated by taking the infimum (greatest lower bound) of the sums of the lengths of all possible coverings of the interval by smaller intervals.

What is the significance of the outer measure of an open interval?

The outer measure of an open interval is important because it allows us to measure the length of sets that are not necessarily measurable using traditional methods. It also helps us define the concept of Lebesgue measure, which is a more general and useful measure than the traditional length measure.

How does the outer measure of an open interval relate to Lebesgue measure?

The outer measure of an open interval is used to define Lebesgue measure. In fact, the Lebesgue measure of an interval is equal to its outer measure. This allows us to extend the concept of length to more complicated sets and spaces.

Can the outer measure of an open interval be infinite?

Yes, the outer measure of an open interval can be infinite. This can happen when the interval is unbounded and cannot be covered by a finite number of smaller intervals. In this case, the outer measure represents the "size" of the interval in an infinite sense.

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