Are measurable sets open or closed?

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I'm seeing the term "measurable sets" used in the definition of some concepts. But when comparing with other concepts that rely on "closed sets", I can't seem to easily find whether measureable sets are open or closed. Does anyone have any insight into that? Thanks.
 

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  • #2
WannabeNewton
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Let [itex]X[/itex] be a set and [itex]\Sigma [/itex] a collection of subsets of [itex]X[/itex] such that: (1) [itex]\varnothing ,X\in \Sigma [/itex]; (2) [itex]A\in \Sigma \Rightarrow X \setminus A\in \Sigma [/itex]; (3) If [itex](A_n)[/itex] is a sequence of sets in [itex]\Sigma [/itex] then [itex]\bigcup_{n = 1}^{\infty }A_n\in \Sigma [/itex]. Such a [itex]\Sigma [/itex] is called a [itex]\sigma [/itex] - algebra. Any [itex]A\in \Sigma [/itex] is said to be measurable or [itex]\Sigma [/itex] - measurable.

As an example, take [itex]X = \mathbb{R}[/itex] and let [itex]\mathfrak{B}[/itex] be the [itex]\sigma [/itex] - algebra generated by the collection of all open intervals [itex](a,b)\subseteq \mathbb{R}[/itex]; [itex]\mathfrak{B}[/itex] is called the Borel algebra. In particular, note that by this definition the Borel algebra also contains all closed intervals [itex][a,b]\subseteq \mathbb{R}[/itex]. So both open and closed intervals in [itex]\mathbb{R}[/itex] are [itex]\mathfrak{B}[/itex] - measurable sets.

Finally, there is no a priori topology on a measurable space [itex](X,\Sigma )[/itex] and as such the notion of a measurable set or [itex]\Sigma [/itex] - measurable set precedes the notion of open and closed sets, with regards to [itex]\Sigma [/itex].
 
  • #3
pwsnafu
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Finally, there is no a priori topology on a measurable space [itex](X,\Sigma )[/itex] and as such the notion of a measurable set or [itex]\Sigma [/itex] - measurable set precedes the notion of open and closed sets, with regards to [itex]\Sigma [/itex].
While this is true from theory, in practical applications we usually are interested in topologies.
The Borel sigma algebra is smallest sigma algebra which contains all open sets. Secondly, if the topology is derived from a Polish space, you have the "standard Borel space", which honestly, covers every important application of modern measure theory.
 
  • #4
WannabeNewton
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While this is true from theory, in practical applications we usually are interested in topologies.
Sure I don't disagree but I just wanted to point out to the OP that there is no a priori attachment of measurable sets to open and closed sets - they are independent concepts when being formulated. Of course in practice one would ideally like to work in a topological space. Cheers!
 
  • #5
jbunniii
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While this is true from theory, in practical applications we usually are interested in topologies.
The Borel sigma algebra is smallest sigma algebra which contains all open sets. Secondly, if the topology is derived from a Polish space, you have the "standard Borel space", which honestly, covers every important application of modern measure theory.
To address the OP's question, it is worth mentioning explicitly that although the Borel sigma algebra is is the smallest one containing all of the open sets, it also contains sets that are neither open nor closed: for example, it contains countable unions of closed sets and countable intersections of open sets, and these are not generally open or closed. So even when there is a topology involved, measurable does not imply open or closed.
 
  • #6
quasar987
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I'm seeing the term "measurable sets" used in the definition of some concepts. But when comparing with other concepts that rely on "closed sets", I can't seem to easily find whether measureable sets are open or closed. Does anyone have any insight into that? Thanks.
Have you tried asking the Führer?

 
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  • #7
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So even when there is a topology involved, measurable does not imply open or closed.
So a measurable set can be either open or closed no matter what size it is? What about the empty set - is that measurable?
 
  • #8
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So a measurable set can be either open or closed no matter what size it is? What about the empty set - is that measurable?
When you say "measurable", then you're being ambiguous. There is not something such as a "measurable set" unless you specify the [itex]\sigma[/itex]-algebra involved. So before we can go on, you need to say in what [itex]\sigma[/itex]-algebra you're working.

In general, measurable sets can be open or closed, they can be both and they can be neither open nor closed. But this depends on the specific topology and [itex]\sigma[/itex]-algebra you have in mind.

The empty set is always measurable by definition.
 
  • #9
jbunniii
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So a measurable set can be either open or closed no matter what size it is? What about the empty set - is that measurable?
The empty set is both open and closed in any topology, and it is measurable in any sigma-algebra.

In general, "measurable" does not necessarily imply either open or closed. A countable union or intersection of measurable sets is measurable, but a countable union of closed sets need not generally be closed, and a countable intersection of open sets need not generally be open. (There are specific topologies such as the discrete topology that do guarantee this, however.)

In ##\mathbb{R}## with the usual topology and the Boreal sigma-algebra, for example, every singleton ##\{x\}## is measurable and closed, but the set of rational numbers, which is a countable union of singletons, is measurable but neither open nor closed.
 
  • #10
mathman
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Simple example: 0≤x<1 is measurable and neither open or closed.
 
  • #11
Bacle2
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Consider the fact that the measurable sets are/form a sigma-algebra, so that they are
closed under complement and under countable union and intersection. Then take,
e.g., [0,2] /\(1,3) in the standard Lebesgue measure on the real line. Extend this idea.

EDIT: I did not see Mathman's post when I made mine; still, hopefully this above can
help generalize.
 
  • #12
Bacle2
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Let [itex]X[/itex] be a set and [itex]\Sigma [/itex] a collection of subsets of [itex]X[/itex] such that: (1) [itex]\varnothing ,X\in \Sigma [/itex]; (2) [itex]A\in \Sigma \Rightarrow X \setminus A\in \Sigma [/itex]; (3) If [itex](A_n)[/itex] is a sequence of sets in [itex]\Sigma [/itex] then [itex]\bigcup_{n = 1}^{\infty }A_n\in \Sigma [/itex]. Such a [itex]\Sigma [/itex] is called a [itex]\sigma [/itex] - algebra. Any [itex]A\in \Sigma [/itex] is said to be measurable or [itex]\Sigma [/itex] - measurable.

As an example, take [itex]X = \mathbb{R}[/itex] and let [itex]\mathfrak{B}[/itex] be the [itex]\sigma [/itex] - algebra generated by the collection of all open intervals [itex](a,b)\subseteq \mathbb{R}[/itex]; [itex]\mathfrak{B}[/itex] is called the Borel algebra. In particular, note that by this definition the Borel algebra also contains all closed intervals [itex][a,b]\subseteq \mathbb{R}[/itex]. So both open and closed intervals in [itex]\mathbb{R}[/itex] are [itex]\mathfrak{B}[/itex] - measurable sets.

Finally, there is no a priori topology on a measurable space [itex](X,\Sigma )[/itex] and as such the notion of a measurable set or [itex]\Sigma [/itex] - measurable set precedes the notion of open and closed sets, with regards to [itex]\Sigma [/itex].
Nice. There is a cardinality argument showing not all measurable sets areBorel sets, i.e., they are not in the sigma-algebra generated by the open sets; comparing the process of generating combinations of F-sigma and G-delta sets with the fact that the cardinality of measurable sets is at least 2Aleph_1 (using, e.g., that all subsets of the Cantor set are measurable). There are also more constructive arguments used to show this.
 
  • #13
pwsnafu
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Nice. There is a cardinality argument showing not all measurable sets areBorel sets, i.e., they are not in the sigma-algebra generated by the open sets; comparing the process of generating combinations of F-sigma and G-delta sets with the fact that the cardinality of measurable sets is at least 2Aleph_1 (using, e.g., that all subsets of the Cantor set are measurable). There are also more constructive arguments used to show this.
Just to be specific, Barcle2 is using the term "measurable" in the sense of "subsets of ℝ which can be assigned Lebesgue measure" which forms a sigma algebra larger than the Borel sigma algebra.
 
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  • #14
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Nice. There is a cardinality argument showing not all measurable sets areBorel sets, i.e., they are not in the sigma-algebra generated by the open sets; comparing the process of generating combinations of F-sigma and G-delta sets with the fact that the cardinality of measurable sets is at least 2Aleph_1 (using, e.g., that all subsets of the Cantor set are measurable). There are also more constructive arguments used to show this.
Not [itex]2^{\aleph_1}[/itex]. The number of (Lebesgue)-measurable sets can be shown to have cardinality [itex]2^{2^{\aleph_0}}[/itex]. The number of Borel sets can be shown to have cardinality [itex]2^{\aleph_0}[/itex].
 
  • #15
Bacle2
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Not [itex]2^{\aleph_1}[/itex]. The number of (Lebesgue)-measurable sets can be shown to have cardinality [itex]2^{2^{\aleph_0}}[/itex]. The number of Borel sets can be shown to have cardinality [itex]2^{\aleph_0}[/itex].
Right, I miswrote. [itex]2^{\aleph_0}[/itex] is what I meant. Seems clear;given the stated power set of Cantor set.
 

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