Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there any $\sigma$-algebra that is strictly between the Borel $\sigma$-algebra and the Lebesgue $\sigma$-algebra?

How about not in between the two, but in general, are there any other $\sigma$ algebra(s)?

What can be concluded about measure too, e.g. is Lebesgue measure the only measure for Lebesgue $\sigma$ algebra?

share|improve this question
1  
Please don't use the title as an integral part of the message; the body should be self-contained. –  Arturo Magidin Jan 24 '11 at 5:40

2 Answers 2

up vote 6 down vote accepted

First question (title): Sure. It is not hard to show that the sets of the form $B \cup S$, where $B$ is Borel and $S$ is a subset of the Cantor-set constitute a $\sigma$-algebra. There are $2^{\mathfrak{c}}$ subsets of the Cantor set but only $\mathfrak{c}$ Borel sets, hence this $\sigma$-algebra lies strictly between the Borel sets and the Lebesgue-measurable sets.

It's very rare that there is no $\sigma$-algebra strictly between two $\sigma$-algebras, it essentially means that there are atoms (sets that cannot be written non-trivially as union of proper subsets.

I'm not sure I understand the last question. You can take for instance Lebesgue measure plus a Dirac measure if you want something strictly different (i.e. not related via Radon-Nikodym).

share|improve this answer
3  
Let me add that there are natural examples of intermediate $\sigma$-algebras. For example, Suslin defined an operation that to infinitely many sets assigns another ("operation $A$"). The $\sigma$-algebra that results from applying $A$ to the Borel sets (and forming the smallest $\sigma$-algebra that contains all these sets) is strictly larger than the algebra of Borel sets but much smaller than the algebra of all Lebesgue sets. This is a natural class that appears in geometric measure theory, for example. Descriptive set theorists use it also with frequency. –  Andres Caicedo Feb 4 '11 at 20:29
    
@Andres: Thanks, this is of course a very good point but I think it goes way beyond the scope of the question. It's a deplorable fact that so few people outside set theory know about these things (and I myself only know some of the very basic facts from Kechris and Srivastava) –  t.b. Feb 4 '11 at 23:37

Your first question has been answered by Theo Buehler; for your second question, you can always take $\mathcal{P}(\mathbb{R})$ as a $\sigma$-algebra; a nontrivial example is the $\sigma$-algebra of all subsets of $\mathbb{R}$ that are either countable or co-countable (that is, all $X$ for which either $|X|\leq\aleph_0$ or $|\mathbb{R}-X|\leq\aleph_0$; it is straightforward to verify this is a $\sigma$-algebra).

For your last question: if you place no restrictions whatsoever on the measure, then the answer is trivial: yes, there are other measures. For a trivial example, just scale the Lebesgue measure by a factor different from $1$.

More generally, given a measurable positive function $f$, define the measure $\mu$ on the Lebesgue measurable sets by $\mu(X) = \int_X f d\lambda = \int f\chi_X d\lambda$, where $\lambda$ is the Lebesgue measure and $\chi_X$ is the characteristic function of $X$. This is a measure:

  1. If $X$ is any measurable set, then $\mu(X) = \int_X f\,d\lambda \geq 0$ because $f(x)\geq 0$ for all $x$, so $\mu$ is nonnegative.

  2. If $\{E_i\}$ is a countable collection of pairwise disjoint sets, then $$\mu(\cup E_i) = \int_{\cup E_i}f\,d\lambda = \sum_{i=1}^{\infty} \int _{E_i} f\,d\lambda = \sum_{i=1}^{\infty}\mu(E_i)$$ (with measures and sum possibly infinite); and

  3. $\mu(\emptyset) = \int_{\emptyset}f\,d\lambda = 0$.

Thus, $\mu$ is a measure on the Lebesgue $\sigma$-algebra; unless $f(x)=1$ for almost all $x$, you have $\mu\neq\lambda$. It may even be a finite measure, if $f\in\mathcal{L}^1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.