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1. • CommentRowNumber1.
   • CommentAuthorIan_Durham
   • CommentTimeMay 31st 2010
   • (edited May 31st 2010)
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   Author: Ian_Durham
   Format: MarkdownItexHere are my rough thoughts on an entry for the Lebesgue measure as per Eric's suggestion of putting them here: The Lebesgue measure's origins can be traced to the broader theory of Lebesgue integration. The original purpose of the latter, in broad terms, was to expand the class of integrable functions in order to give meaning to functions that are _not_ Riemann integrable. In order to accomplish this, the basic properties of the concept of the _length_ of an interval must be understood. This then leads to the need to fully define the concept of _measure_, particularly in relation to sets. We begin with a lemma and a corollary. **Lemma:** _Let I be an interval, $I=I_{1} \cup I_{2} \cup \cdots$ where $I_{1}, _{2}, \ldots$ are disjoint intervals. Then $|I|=\sum_{j=1}^{\infty}|I_{j}|$ (interpreted to mean that if one side $+\infty$, then so is the other, where $\sum_{j=1}^{\infty}|I_{j}|$ can be $+\infty$, either because one of the summands is $+\infty$ or because the series diverges). **Corollary:** _If I is any interval, then_ $$ |I|=inf\left\{\sum_{j=1}^{\infty}|I_{j}|:I \subseteq \bigcup_{j=1}^{\infty} I_{j}\right. $$ _where_ $\{I_{j}\}$ _is any countable covering of I by intervals._ $\}$. Now suppose _B_ is an arbitrary set. In order for _B_ to be measurable, we must have $|B| \le \sum_{j=1}^{\infty}|I_{j}|$ where $\bigcup_{j=1}^{\infty} I_{j}$ is any countable covering of _B_ by intervals. We must also have $|B|\le \bigcup_{j=1}^{\infty} |I_{j}|$ where the infinum is taken over all countable coverings of _B_ by intervals. **Definition:** We may define the Lebesgue measure as being $$ |B|=inf\left\{\sum_{j=1}^{\infty}|I_{j}|:B \subseteq \bigcup_{j=1}^{\infty} I_{j}\right\}. $$ Note that once the Lebesgue measure for is known for open sets, the outer regularity property allows us to find it for all Borel sets.

   Here are my rough thoughts on an entry for the Lebesgue measure as per Eric’s suggestion of putting them here:

   The Lebesgue measure’s origins can be traced to the broader theory of Lebesgue integration. The original purpose of the latter, in broad terms, was to expand the class of integrable functions in order to give meaning to functions that are not Riemann integrable. In order to accomplish this, the basic properties of the concept of the length of an interval must be understood. This then leads to the need to fully define the concept of measure, particularly in relation to sets. We begin with a lemma and a corollary.

   Lemma: _Let I be an interval, $I=I_{1} \cup I_{2} \cup \cdots$ where $I_{1}, _{2}, \ldots$ are disjoint intervals. Then $|I|=\sum_{j=1}^{\infty}|I_{j}|$ (interpreted to mean that if one side $+\infty$, then so is the other, where $\sum_{j=1}^{\infty}|I_{j}|$ can be $+\infty$, either because one of the summands is $+\infty$ or because the series diverges).

   Corollary: If I is any interval, then

   $$|I|=inf\left\{\sum_{j=1}^{\infty}|I_{j}|:I \subseteq \bigcup_{j=1}^{\infty} I_{j}\right.$$

   where $\{I_{j}\}$ is any countable covering of I by intervals. $\}$.

   Now suppose B is an arbitrary set. In order for B to be measurable, we must have $|B| \le \sum_{j=1}^{\infty}|I_{j}|$ where $\bigcup_{j=1}^{\infty} I_{j}$ is any countable covering of B by intervals. We must also have $|B|\le \bigcup_{j=1}^{\infty} |I_{j}|$ where the infinum is taken over all countable coverings of B by intervals.

   Definition: We may define the Lebesgue measure as being

   $$|B|=inf\left\{\sum_{j=1}^{\infty}|I_{j}|:B \subseteq \bigcup_{j=1}^{\infty} I_{j}\right\}.$$

   Note that once the Lebesgue measure for is known for open sets, the outer regularity property allows us to find it for all Borel sets.

2. • CommentRowNumber2.
   • CommentAuthorEric
   • CommentTimeMay 31st 2010
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   Author: Eric
   Format: MarkdownItexHuh? That seems pretty polished to me :) See [[Lebesgue measure]]

   Huh? That seems pretty polished to me :)

   See Lebesgue measure

3. • CommentRowNumber3.
   • CommentAuthorEric
   • CommentTimeMay 31st 2010
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   Author: Eric
   Format: MarkdownItexIs there a reference?

   Is there a reference?

4. • CommentRowNumber4.
   • CommentAuthorEric
   • CommentTimeMay 31st 2010
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   Author: Eric
   Format: MarkdownItexIt would be cool to address this from the nPOV in conjunction with [Carathéodory's extension theorem](http://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_extension_theorem). I can smell some [[universal property]] somewhere in there.

   It would be cool to address this from the nPOV in conjunction with Carathéodory’s extension theorem. I can smell some universal property somewhere in there.

5. • CommentRowNumber5.
   • CommentAuthorIan_Durham
   • CommentTimeMay 31st 2010
   • (edited May 31st 2010)
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   Author: Ian_Durham
   Format: MarkdownItexPersonally I like the presentation in Strichartz's _The Way of Analysis_. I think it's a very intuitive and readable presentation. Edit: I added this to the reference list even though there isn't an online version. It's just such a great discussion I think everyone should read it.

   Personally I like the presentation in Strichartz’s The Way of Analysis. I think it’s a very intuitive and readable presentation.

   Edit: I added this to the reference list even though there isn’t an online version. It’s just such a great discussion I think everyone should read it.

6. • CommentRowNumber6.
   • CommentAuthorColin Tan
   • CommentTimeDec 16th 2013
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   Author: Colin Tan
   Format: TextLebesgue measure on Cartesian space is characterized by being translation invariant. Is there an intuitive reason why Lebesgue measure is also rotationally invariant?

   Lebesgue measure on Cartesian space is characterized by being translation invariant. Is there an intuitive reason why Lebesgue measure is also rotationally invariant?
7. • CommentRowNumber7.
   • CommentAuthorTodd_Trimble
   • CommentTimeDec 16th 2013
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   Author: Todd_Trimble
   Format: MarkdownItexMore generally it is invariant with respect to linear transformations $T$ of determinant $1$. Intuitively, the image of a parallelpiped $T([a_1, b_1] \times \ldots \times [a_n, b_n])$ can be chopped up into finitely many measurable pieces (polyhedra, in fact) and then rearranged in tangram fashion into another copy of $[a_1, b_1] \times \ldots \times [a_n, b_n]$. To see this, it is enough to consider the special case $T = E_{i, j, \lambda}$ of an elementary row operation (that adds $\lambda$ times the $j^{th}$ row to the $i^{th}$ row), since any $T$ of determinant $1$ is a composite of such. There it is not hard to specify a single chop-and-rearrange (consider the $2$-dimensional case to get the general idea).

   More generally it is invariant with respect to linear transformations $T$ of determinant $1$. Intuitively, the image of a parallelpiped $T([a_1, b_1] \times \ldots \times [a_n, b_n])$ can be chopped up into finitely many measurable pieces (polyhedra, in fact) and then rearranged in tangram fashion into another copy of $[a_1, b_1] \times \ldots \times [a_n, b_n]$. To see this, it is enough to consider the special case $T = E_{i, j, \lambda}$ of an elementary row operation (that adds $\lambda$ times the $j^{th}$ row to the $i^{th}$ row), since any $T$ of determinant $1$ is a composite of such. There it is not hard to specify a single chop-and-rearrange (consider the $2$-dimensional case to get the general idea).

8. • CommentRowNumber8.
   • CommentAuthorColin Tan
   • CommentTimeDec 19th 2013
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   Author: Colin Tan
   Format: TextAh, thanks, your answer really helped to clarify my intuition. Now I understand the slogan "under a linear transformation, the determinant measures the change in infinitesimal volume" better.

   Ah, thanks, your answer really helped to clarify my intuition. Now I understand the slogan "under a linear transformation, the determinant measures the change in infinitesimal volume" better.
9. • CommentRowNumber9.
   • CommentAuthorTodd_Trimble
   • CommentTimeDec 19th 2013
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   Author: Todd_Trimble
   Format: MarkdownItexOkay, good. (BTW, I did get your email; I just haven't quite figured out what I want to say in response.) I forgot to mention in my comment above that each "rearrangement" (after a "chop") is actually a translation, which reduces the argument to the translation-invariance of Lebesgue measure.

   Okay, good. (BTW, I did get your email; I just haven’t quite figured out what I want to say in response.) I forgot to mention in my comment above that each “rearrangement” (after a “chop”) is actually a translation, which reduces the argument to the translation-invariance of Lebesgue measure.