Why do sigma algebras define measurable sets?

Question

Since we can not measure all sets (e.g. vitali sets), it makes sense that measures need to be defined on a reduced set of sets. This reduced set of sets is the set of "measurable sets". But why is this reduced set of sets a sigma algebra?

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To illustrate the question I will suggest $d$-systems as an an alternative.

A $d$-system $\mathcal{D}$ (also known as $\lambda$-system or Dynkin system) is a set of sets, which has the following properties

1. $\emptyset \in \mathcal{D}$,
2. If $A, B\in \mathcal{D}$ and $A\subseteq B$, then $B\setminus A \in \mathcal{D}$,
3. If $A_1, A_2,\dots \in \mathcal{D}$ are disjoint, then $\biguplus_{i=1}^\infty A_i \in \mathcal{D}$.

Observe how these properties match the properties of a measure $\mu$:

1. $\mu(\emptyset) = 0$
2. If $A_1, A_2,\dots \in \mathcal{D}$ are disjoint, then $$ \mu\Bigl(\biguplus_{i=1}^\infty A_i\Bigr) = \sum_{i=1}^\infty \mu(A_i) $$

The empty set is always measurable with measure $0$, i.e. it should be measureable matching the first requirements of a $d$-system.

If you have a sequence of measurable disjoint sets then of course their union should be measurable because you can write their measure down by the second property. Thus their union should be "meausrable sets", matching the 3rd property of $d$-systems.

Finally, the second property of a $d$-system follows from the fact that $$ \mu(B\setminus A) + \mu(A) = \mu(B) $$ since $B\setminus A$ and $A$ are disjoint. So provided a measure of $A$ and $B$ we can define a measure of $B\setminus A$, thus it should be a measurable set.

So it is intuitive that measurable sets should have the properties of a $d$-system. Now the $\pi$-$\lambda$ Theorem essentially states that if the generator is stable under intersections $\cap$, i.e. a $\pi$-system, then the smallest $d$-system that contains the generator is a sigma algebra. So in most cases we are interested in, the two definitions of "measurable sets" coincide. But why do we use sigma algebras and not $d$-systems in the case where they do not coincide?

On wikipedia the $\pi$-$\lambda$ Theorem only states that any Dynkin system that contains the generator contains the generated sigma algebra. To obtain the statement above observe that the generated sigma algebra is always larger or equal to the generated $d$-system.

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Related questions:

• Why do we define measurable sets as a $\sigma$-algebra, not just an algebra?
• Why is $\sigma$-algebra necessary to define a measure?

Answer 1

My current hypothesis is

Historical & convenience reasons

One reason sigma algebras and not $d$-systems are used to define measurable sets might be that

1. in the cases we care about they coincide and
2. it is easier to show that two generators produce the same sigma algebra than it is to show that they produce the same $d$-system.

   To show they have the same sigma algebra, it is only necessary to show that both sigma algebras contain the other generator, which is easier to do with arbitrary unions than with disjoint unions.