measurable sets form a sigma algebra

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Theorem

The collection of measurable sets M is a sigma-algebra

Note

We want measurable sets to form a sigma algebra based on the desirable properties for measure:

Properties we want for this idea

  1. m(E) is defined for all ER
  2. if I is an interval then m(E)=(I) the "length" of I
  3. If {En} is a (countable) collection of disjoint subsets of E such that E=nEn, then we want m(nEn)=m(E)
  4. Translation invariance. ie, if ER and xR, then m(x+E)=m({x+y|yE})=m(E)

In particular, in order to get (3), we need to be able to define measure on any countable disjoint union.

Proof

We already know that M is an algebra, and we know since algebras have closure under finite disjoint countable unions that we just need to check countable disjoint unions are measurable.

So let {En} be a countable collection of disjoint measurable sets with E=nEn. Then since the other half is always satisfied via monotonicity, we just need to show m(AEc)+m(AE)m(A)

Let NN. Since M is an algebra, the union n=1NEnM so we have

m(A)=m(A[n=1NEn])+m(A[n=1NEn]c)

And since n=1NEnE, we have Ec[n=1NEn]c. Thus we have

m(A)=m(A[n=1NEn])+m(A[n=1NEn]c)m(A[n=1NEn])+m(AEc)

And since the measure of finite disjoint measurable sets is the sum of the measures, we get

m(A)n=1Nm(AEn)+m(AEc)(N)n=1m(AEn)+m(AEc)()m(nAEn)+m(AEc)=m(AE)+m(AEc)

Where () is from countable subadditivity.

References

References

See Also

Mentions

Mentions

File Last Modified
Functional Analysis Lecture 8 2025-07-14
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measurable sets form a sigma algebra 2025-07-08
measure satisfies countable additivity 2025-07-08

Created 2025-07-08 Last Modified 2025-07-08