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WannabeNewton

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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

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pwsnafu

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While this is true from theory, in practical applications we usually are interested in topologies.Finally, there is noa prioritopology 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].

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.

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WannabeNewton

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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!While this is true from theory, in practical applications we usually are interested in topologies.

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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.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.

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quasar987

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Have you tried asking the Führer?

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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?So even when there is a topology involved, measurable does not imply open or closed.

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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.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 always measurable by definition.

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

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.

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mathman

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Simple example: 0≤x<1 is measurable and neither open or closed.

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Bacle2

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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.

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Bacle2

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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 2measurableor [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 noa prioritopology 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].

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pwsnafu

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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.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 2^{Aleph_1}(using, e.g., that all subsets of the Cantor set are measurable). There are also more constructive arguments used to show this.

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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].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 2^{Aleph_1}(using, e.g., that all subsets of the Cantor set are measurable). There are also more constructive arguments used to show this.

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Bacle2

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Right, I miswrote. [itex]2^{\aleph_0}[/itex] is what I meant. Seems clear;given the stated power set of Cantor set.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].

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