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Kostya_I
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I think there's indeed a lot of confusion here, in that in many situations the $\pi-\lambda$ is applied where a weaker result would suffice. What one uses most of the time is not the $\pi-\lambda$ theorem itself, but the following corollary:

If $\mu_1$ and $\mu_2$ are two probability measures that agree on a $\pi$-system, then they agree on the $\sigma$-algebra generated by that $\pi$-system.

The example about independent events given above by Peter Luthy falls into this category.

In many textbook situations the $\pi$-system in question is actually a semi-ring. For ($\sigma$-)finite premeasures, Carathéodory extension is unique for very simple reasons: any extension of the measure is bounded from above by the outer measure, and by passing to the complement it is also bounded from below. Thus, for semi-rings the above corollary is trivial.

A typical example of a $\pi$-system that is not a semi-ring are closed sets in a topological space. But here, the fact that a Borel measure is determined by its values on closed sets follows from regularity, which does not require $\pi-\lambda$ theorem either.

One place where you need $\pi-\lambda$ theorem (or a similar result) in an essential way is Fubini. There, you have a $\pi$-system (which is also a semi-ring) of sets of the form $E_1\times E_2$, and two ways to extend the pre-measure: by product-measure (= Carathéodory extension), and by integrating the measures of the slices. What is tricky to show is that the collection of sets for which the latter is well defined is a $\sigma$-algebra, in particular, that it is closed under pairwise intersections. Indeed, it's not clear under such operation, the function under the integral remains measurable. The $\pi-\lambda$ theorem allows one to bypass this difficulty in a very neat way.

That said, once you have the $\pi-\lambda$ intheorem at hand, it is often just a bit shorter to write down the proof that something is a $\lambda$-system than that it is a $\sigma$-algebra. No wonder the authors of textbooks use it systematically even when it's an overkill.

I think there's indeed a lot of confusion here, in that in many situations the $\pi-\lambda$ is applied where a weaker result would suffice. What one uses most of the time is not the $\pi-\lambda$ theorem itself, but the following corollary:

If $\mu_1$ and $\mu_2$ are two probability measures that agree on a $\pi$-system, then they agree on the $\sigma$-algebra generated by that $\pi$-system.

The example about independent events given above by Peter Luthy falls into this category.

In many textbook situations the $\pi$-system in question is actually a semi-ring. For ($\sigma$-)finite premeasures, Carathéodory extension is unique for very simple reasons: any extension of the measure is bounded from above by the outer measure, and by passing to the complement it is also bounded from below. Thus, for semi-rings the above corollary is trivial.

A typical example of a $\pi$-system that is not a semi-ring are closed sets in a topological space. But here, the fact that a Borel measure is determined by its values on closed sets follows from regularity, which does not require $\pi-\lambda$ theorem either.

One place where you need $\pi-\lambda$ theorem (or a similar result) in an essential way is Fubini. There, you have a $\pi$-system (which is also a semi-ring) of sets of the form $E_1\times E_2$, and two ways to extend the pre-measure: by product-measure (= Carathéodory extension), and by integrating the measures of the slices. What is tricky to show is that the collection of sets for which the latter is well defined is a $\sigma$-algebra, in particular, that is closed under pairwise intersections. Indeed, it's not clear under such operation, the function under the integral remains measurable. The $\pi-\lambda$ theorem allows one to bypass this difficulty in a very neat way.

That said, once you have $\pi-\lambda$ in hand, it is often just a bit shorter to write down the proof that something is a $\lambda$-system than that it is a $\sigma$-algebra. No wonder the authors of textbooks use it systematically even when it's an overkill.

I think there's indeed a lot of confusion here, in that in many situations the $\pi-\lambda$ is applied where a weaker result would suffice. What one uses most of the time is not the $\pi-\lambda$ theorem itself, but the following corollary:

If $\mu_1$ and $\mu_2$ are two probability measures that agree on a $\pi$-system, then they agree on the $\sigma$-algebra generated by that $\pi$-system.

The example about independent events given above by Peter Luthy falls into this category.

In many textbook situations the $\pi$-system in question is actually a semi-ring. For ($\sigma$-)finite premeasures, Carathéodory extension is unique for very simple reasons: any extension of the measure is bounded from above by the outer measure, and by passing to the complement it is also bounded from below. Thus, for semi-rings the above corollary is trivial.

A typical example of a $\pi$-system that is not a semi-ring are closed sets in a topological space. But here, the fact that a Borel measure is determined by its values on closed sets follows from regularity, which does not require $\pi-\lambda$ theorem either.

One place where you need $\pi-\lambda$ theorem (or a similar result) in an essential way is Fubini. There, you have a $\pi$-system (which is also a semi-ring) of sets of the form $E_1\times E_2$, and two ways to extend the pre-measure: by product-measure (= Carathéodory extension), and by integrating the measures of the slices. What is tricky to show is that the collection of sets for which the latter is well defined is a $\sigma$-algebra, in particular, that it is closed under pairwise intersections. Indeed, it's not clear under such operation, the function under the integral remains measurable. The $\pi-\lambda$ theorem allows one to bypass this difficulty in a very neat way.

That said, once you have the $\pi-\lambda$ theorem at hand, it is often just shorter to write down the proof that something is a $\lambda$-system than that it is a $\sigma$-algebra. No wonder the authors of textbooks use it systematically even when it's an overkill.

Source Link
Kostya_I
  • 9k
  • 1
  • 32
  • 51

I think there's indeed a lot of confusion here, in that in many situations the $\pi-\lambda$ is applied where a weaker result would suffice. What one uses most of the time is not the $\pi-\lambda$ theorem itself, but the following corollary:

If $\mu_1$ and $\mu_2$ are two probability measures that agree on a $\pi$-system, then they agree on the $\sigma$-algebra generated by that $\pi$-system.

The example about independent events given above by Peter Luthy falls into this category.

In many textbook situations the $\pi$-system in question is actually a semi-ring. For ($\sigma$-)finite premeasures, Carathéodory extension is unique for very simple reasons: any extension of the measure is bounded from above by the outer measure, and by passing to the complement it is also bounded from below. Thus, for semi-rings the above corollary is trivial.

A typical example of a $\pi$-system that is not a semi-ring are closed sets in a topological space. But here, the fact that a Borel measure is determined by its values on closed sets follows from regularity, which does not require $\pi-\lambda$ theorem either.

One place where you need $\pi-\lambda$ theorem (or a similar result) in an essential way is Fubini. There, you have a $\pi$-system (which is also a semi-ring) of sets of the form $E_1\times E_2$, and two ways to extend the pre-measure: by product-measure (= Carathéodory extension), and by integrating the measures of the slices. What is tricky to show is that the collection of sets for which the latter is well defined is a $\sigma$-algebra, in particular, that is closed under pairwise intersections. Indeed, it's not clear under such operation, the function under the integral remains measurable. The $\pi-\lambda$ theorem allows one to bypass this difficulty in a very neat way.

That said, once you have $\pi-\lambda$ in hand, it is often just a bit shorter to write down the proof that something is a $\lambda$-system than that it is a $\sigma$-algebra. No wonder the authors of textbooks use it systematically even when it's an overkill.