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Purpose of sigma algebras

  1. Sep 8, 2009 #1
    What's the point of defining [tex]\sigma [/tex] algebras, i.e., why must we assign measures only to elements of a [tex]\sigma [/tex] algebra?

    Second, can you give an example of a set contained in the Lebesgue [tex]\sigma [/tex] algebra, but not the Borel [tex]\sigma [/tex] algebra? Also, is there a set not contained even in the Lebesgue [tex]\sigma [/tex] algebra?
  2. jcsd
  3. Sep 8, 2009 #2


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    You need sigma algebras in measure theory because measure is countably additive. The reason we use this structure (sigma algebras) is to be able to define measure.

    Borel sets are defined when you have a topological space (open sets, etc.). The Borel sigma algebra is the smallest sigma algebra containing the open sets. However, sigma algebra (general case) can be defined even though there is no topology. Probability spaces are an example.

    Although they are hard to describe there are non-measurable sets on the real line. These are therefore not in the sigma algebra.
  4. Sep 8, 2009 #3
    I just saw a demonstration of a subset on the real line that is non-measurable.

    The proof relied on the fact that the measure of the union of a countable number of disjoint sets is additive, and also that the measure is unchanged by translation.

    But I thought the Banach-Tarski paradox said that measure could be changed by translation? So why did the proof assume that measure is unchanged by translation?

    So a sigma algebra guards against unmeasurable sets while maintaining the property that a countable number of disjoint set adds and that translation does not change the measure?
    So in effect, we are saying that instead of all these strange things happening with these strange sets, we just ignore these strange sets by restricting ourselves to sets contained in a sigma algebra?

    Most of these strange sets seem to be very rare, and it seems a lot of work is done just to guard against them. Wouldn't it be better to just ignore sigma algebras, and at the end of the book, give examples where the results don't hold for some very special sets?
  5. Sep 9, 2009 #4


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    Any Lebesque measure is, by definition, "translation invariant". The Banach-Tarski paradox does NOT say that measure can be changed by translation- it uses non-measurable sets.

    "Seem to be very rare" in what sense? They don't show up very often in analysis precisely because the have "strange things" happening but, in fact, it is the "good", measurable, sets that are "very rare". Almost all (in the Lebesque sense) sets are non-measurable.
  6. Sep 9, 2009 #5
    How do you define a measure on the power set of the reals?
  7. Sep 9, 2009 #6
    The idea of sigma algebra is designed to abstractly describe the properties for which our intuitive idea of measurement must hold. Arbitrary unions and complements seem necessary. For instance if the whole space has measure 5 and a subset has measure 2 we would think that its complement must be measurable and have measure 3. There is little else to it.

    I think the Borel sigma algebra is the smallest sigma algebra that contains all of the open sets.

    Such an algebra may have sets in it which we would like to be measurable but are not. These are subsets of sets of measure zero. We would like for these subsets to also have measure zero. Including them is called completing the sigma algebra.

    This definition is a little slick. What is really wanted is if A is a proper subset of B and has the same measure as B we would like any subset of B that contains A to also have the same measure. This works, meaning that the extended measure is still a measure and the extended collection of sets is still a sigma algebra. A little thought shows that by including all subsets of sets of measure zero and extending the collection of sets through arbitrary union and complementation, we get all of the sets we want.

    there are subsets of sets of measure zero - e.g. the Cantor set - that are not Borel measurable. I think Wikipedia should have something on this.

    I think that a basis for the reals over the rationals is not Lebesque measurable.
    The proof that I remember is to show that a linear map from the reals to the reals that is linear over the rationals but not over the reals is not only discontinuous but also non-measurable. One solves this by showing that if it were measurable then it would be continuous. I could be remembering this wrong. You should check it. I would be interested to see your proof.

    There is a cool construction of a non-Lebesque measurable subset of the square which shows that the idea of measurability is related to the axioms of set theory. Specifically, if one assumes that the continuum hypothesis is true i.e. that the cardinality of the reals is the same as that of the first uncountable ordinal, then one can find a non-measurable subset of the square.

    Take a well ordering of the reals assuming the Continuum Hypothesis is true.

    Consider the set of points, (x,y) in the square such that x precedes y in this ordering.
    This set is non-measurable. Why?

    Consider it's characteristic function. If the set is measurable then the integral of this function will equal its measure. Fubini's Theorem says that we can find this integral by double integration and that it does not matter which axis, x or y, that we integrate across first. let's try it.

    If we hold y fixed and integrate along x then the integral is zero because there are only countably many x that precede y and a countable set has measure zero. So the double integral is also zero.

    If we hold x fixed, then x precedes all but countably many y so the integral equals one. This contradiction of Fubini's Theorem shows that the set is not measurable.
    Last edited: Sep 9, 2009
  8. Sep 9, 2009 #7
    Here is a problem that helped me get some feel for the implication of the axioms for a sigma algebra.

    Prove that a infinite sigma algebra must be uncountable. It can not be countably infinite.
  9. Sep 9, 2009 #8
    That's intuitive enough. The intersection of B with A-compliment, which is in the sigma algebra, must have zero measure if A and B have the same measure. So if you take all subsets of regions of zero measure, then you can build any subset inside B (but containing all of A).

    I'm not exactly sure why this is important though to have this property. Why can't we have regions of space that have zero measure but aren't "sliceable"? We like things that have zero measure to be sliceable down to points?

    I've gone through the proof that the Cantor set has zero Borel measure. It was presented as an example of an uncountable set that has zero measure. However, the teacher didn't prove whether or not the Cantor set is actually an element of a sigma algebra. Is the Cantor set an element of the Borel sigma algebra, or is it only subsets of the Cantor set that aren't Borel measurable?

    That sounds too hard for me. I'm learning measure theory through some free online videos in my spare time. But I invite everyone else here to prove this!

    I'm a little confused here, but I guess you're not picturing the unit square with an integral of a simple function over its upper-diagonal half, where x<y? Or at least not an integral over the reals, where that area would be 1/2?

    This I have trouble seeing intuitively. I did the exercise where you have one subset of a set, and find the smallest sigma algebra containing that one subset. You get 4 elements, the null set, the entire set, the one subset, and the compliment of the one subset. If you add another subset, then the smallest sigma algebra contains 16 elements. So it seems if you take an infinite countable number of subsets, then you can construct a much larger but countable infinite sigma algebra. Maybe it gets too large when you keep adding new subsets (that aren't required for what you already have to be a sigma algebra) and becomes uncountable.

    I looked at one particular nonmeasurable set that involved irrationals, and I figured that the reals were bigger than the irrationals, so there ought to be more intervals and other nice sets than nonmeasurable sets. I guess this is wrong because the irrationals are as big as the reals, so the ugly sets outnumber the nice ones.
  10. Sep 9, 2009 #9
    I meant to say that there are set of measure zero contained within the Cantor set that are not Borel. The Cantor set is Borel because it is a closed set.

    There are a lot of reasons that a sigma algebra should be complete. But forgetting the mathematical niceties in probability theory for instance it is a little strange to says that an event has probability one but that a larger event that contains it has no probability at all.

    In the example of the square I was just doing a double integral of a function, nothing fancy.

    The proof that an infinite sigma algebra must be uncountable is not hard. try it. You'll like it.
  11. Sep 9, 2009 #10
    Wait a second. How does that work? Some subsets of the Cantor set are not measurable. But the Cantor set is itself measurable (with zero measure). The problem is the fact that the Borel sigma algebra is not complete. So if you go to a complete one, like the Lebesgue sigma algebra, you introduce those subsets that were unmeasurable in the Borel sigma algebra. But don't you still have the problem of assigning measure to these unmeasurable sets? In both cases, whether dealing with the Borel or the Lebesgue sigma algebra, we used the Lebesgue measure. So how do those subsets of the Cantor set all of a sudden have zero measure now when the only thing we did was expand the sigma algebra but not change the measure?

    Haha, I couldn't figure it out, so I cheated and looked online. I would have never gotten it anyways. The proof was interesting, but relied on the power set of a set of countably infinite sets being uncountable. So if I can just figure out why that's true, then I understand the proof.
  12. Sep 10, 2009 #11
    that is a great question. A subset of a set of measure zero is just declared to be measurable and to have measure zero. This works. You get a sigma algebra this way and a new measure.

    If you examine the definition of a measure, you can prove that a subset of a set of measure zero, if it is measurable, must have measure zero as well.
  13. Sep 10, 2009 #12
    Can you send me the link?
  14. Sep 10, 2009 #13
    The link I'm watching from is:


    It requires free registration, but is available to anyone. The course for some reason is called "real analysis", but it's entirely a course on measure theory.

    It's kind of boring: basically the professor just writes down and proves theorems on the blackboard, but I think that's how most analysis courses are.

    Unfortunately, the video of the first class is missing, so it begins from the 2nd class onwards. The only thing you need to know from that 1st class are what the infimum and supremum mean, and what the lim sup and lim inf mean.

    A good book is probably better, but sometimes a book contains a little too much information and detail and I just want a core that I can expand later if I want.
  15. Sep 10, 2009 #14
    That's not a problem. I took a course on real analysis before (About 9 years ago). I know what you mean about a book containing too much information sometimes.
  16. Sep 10, 2009 #15
    Royden's book on Real Analysis is readable and standard.

    I strongly suggest that you bang your head against the wall trying to solve these problems on your own. Even if it takes you a month 10 hours a day of standing in front of a blackboard trying to find a way, it is well worth it.
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