Intuition behind using $\pi$-system in Measure Theory

Question

Many proofs in measure theory rely on the π-λ theorem: if a π-system P generates a σ-algebra, then it suffices to verify a property on P and extend it to the whole σ-algebra. Examples include Carathéodory’s extension theorem, proving independence of σ-algebras, and many others.

I understand the proofs themselves, but I’m struggling with the underlying idea: why does one even think to use π-systems in the first place? The choice of this particular class—with its very specific closure property—feels somewhat ad hoc or "out of the blue." How did this concept arise, and what is the intuition for why π-systems are the right objects to work with when extending properties to a generated σ-algebra?

Answer 1

The key idea here is that $\lambda$-systems are very similar to $\sigma$ algebras except that $\lambda$-systems need only be closed under a restricted class of countable unions, namely disjoint unions.

If a $\lambda$-system is also closed under intersections (i.e. a $\pi$-system), then we can use some set theory tricks to turn any countable union into a countable disjoint union. In some ways, the limitation of a $\lambda$-system is that it is unable to deal with sets that intersect, so adding the $\pi$ condition is precisely what is necessary to deal with this deficiency by "disjointifying" sets.

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To show how this works, I will prove that any family $\mathcal{F}$ of subsets of $X$ which is both a $\pi$- and $\lambda$-system is a $\sigma$-algebra. As being a $\lambda$-system makes $\mathcal{F}$ already closed under complements, we need only show that it is closed under arbitrary unions.

Let $A_1, A_2,\dots \in \mathcal{F}$. Then define $B_1 = A_1$, $B_2 = A_2 \cap A_1^c$, $B_n = A_n \cap A_1^c \cap \cdots \cap A_{n-1}^c$, which are all in $\mathcal{F}$ as it is closed under complement ($\lambda$-system) and finite intersection ($\pi$-system).

I claim $\bigcup_{n =1}^\infty A_n = \bigcup_{n=1}^\infty B_n$. Proof by Double Containment: If $x \in A_n$ for some $n$, then consider the least such $n$. Then $x \not\in A_k$ for any $k<n$, so $x \in B_n$. Likewise if $x \in B_n$ for some $n$, then $x \in A_n$ as well.

And the latter union is by construction disjoint.

The proof of the full $\pi-\lambda$ is harder but along a similar vein.

Answer 2

I will give a probability-centric answer; $\pi$-systems seem to be mostly used by probabilists. Since probability measures are extremely complicated when interpreted as functions on $\sigma$-algebras, it is useful to know when they can be specified on subclasses of measurable sets. For example, cumulative distribution functions work because one can specify a Borel probability measure on the line uniquely by the probability assigned to sets of the form $(-\infty,r]$, a family forming a $\pi$-system.

The relevant fact is that two probability measures agreeing on a $\pi$-system must also agree on the generated $\sigma$-algebra. And this is not true without this condion. To find a family of sets that are not a $\pi$-system, you need to have a set with at least three points. So let $X=\{0,1,2\}$. A probability measure assigning probability 1ドル/2$ to each of the sets $\{0,1\}$ and $\{1,2\}$ could assign any probability between 0ドル$ and 1ドル/2$ to the intersection $\{1\}$. So not being closed under intersections causes a problem. If one eliminates this problem, no other problems are left.