Micaela Simon

2022-06-22

In most measure theory text books one derives the Lebesgue anh Borel-Lebesgue measure from Caratheodory's extension to outer measures by first proving that the set of ${\lambda}^{\ast}$ measurable sets is a sigma algebra (the Lebesgue sigma algebra) and that ${\lambda}^{\ast}$ restricted to that sigma algebra is a measure, the Lebesgue measure. This measure space is even complete. However, then still one continues to show that ${\lambda}^{\ast}$ restricted to the sigmal algebra generated by the ring (on which the pre-measure was defined that was used to obtain ${\lambda}^{\ast}$ via Caratheodorys extension) is also a measure, the Borel-Lebesgue measure, and that the generated sigma algebra is the Borel sigma algebra.

So my question is, why is the Borel sigma-algebra "better" than the Lebesgue sigma algebra, because most of the time text books continue to work only on the Borel sigma algebra, even though the Lebesgue sigma algebra is its completetion and has some other favorable properties? I.e., I am just missing an argument in all the lecture notes and text books why we continue to work on the Borel sigma-algebra after having shown that the Lebesgue sigma algebra is larger (and after all we do all the extension from a pre-measure to an outer measure and then restricting to a measure because we want to get a larger set than just the ring on which we initially defined the pre-measure).

So my question is, why is the Borel sigma-algebra "better" than the Lebesgue sigma algebra, because most of the time text books continue to work only on the Borel sigma algebra, even though the Lebesgue sigma algebra is its completetion and has some other favorable properties? I.e., I am just missing an argument in all the lecture notes and text books why we continue to work on the Borel sigma-algebra after having shown that the Lebesgue sigma algebra is larger (and after all we do all the extension from a pre-measure to an outer measure and then restricting to a measure because we want to get a larger set than just the ring on which we initially defined the pre-measure).

Sage Mcdowell

Beginner2022-06-23Added 19 answers

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I think the comments already mentioned striking points but here is an easy example that shows pathologies that can happen:

Consider the inclusion-like map

$f:\mathbb{R}\to {\mathbb{R}}^{n},x\mapsto (x,0,\dots ,0)$

for $n>1$.

This function is continuous and hence Borel-measurable. If you equip $\mathbb{R}$ and ${\mathbb{R}}^{n}$ with the Lebesgue-$\sigma $-algebra, this mapping suddenly is not measurable anymore. Take some $V\subseteq \mathbb{R}$ which is not Lebesgue-measurable (to argue that these exists, take some argument like in the proof of Vitali's theorem with the translation invariant Lebesgue-measure). By definition

$V\times \{0{\}}^{n-1}\subseteq \mathbb{R}\times \{0{\}}^{n-1}$

is Lebesgue measurable, as $\lambda (\mathbb{R}\times \{0{\}}^{n-1})=0$. But

math xmlns="http://www.w3.org/1998/Math/MathML" > f − 1 ( V × { 0 } n − 1 ) = V

which shows that $f$ is not measurable. We found a very simple continuous function, which is not measurable. That is very inconvenient.

Consider the inclusion-like map

$f:\mathbb{R}\to {\mathbb{R}}^{n},x\mapsto (x,0,\dots ,0)$

for $n>1$.

This function is continuous and hence Borel-measurable. If you equip $\mathbb{R}$ and ${\mathbb{R}}^{n}$ with the Lebesgue-$\sigma $-algebra, this mapping suddenly is not measurable anymore. Take some $V\subseteq \mathbb{R}$ which is not Lebesgue-measurable (to argue that these exists, take some argument like in the proof of Vitali's theorem with the translation invariant Lebesgue-measure). By definition

$V\times \{0{\}}^{n-1}\subseteq \mathbb{R}\times \{0{\}}^{n-1}$

is Lebesgue measurable, as $\lambda (\mathbb{R}\times \{0{\}}^{n-1})=0$. But

math xmlns="http://www.w3.org/1998/Math/MathML" >

which shows that $f$ is not measurable. We found a very simple continuous function, which is not measurable. That is very inconvenient.

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