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I am learning probability theory now and our probability theory course is built upon the measure theory. In the first class, we talked about the concept of field and $\sigma$-field. One property of $\sigma$-field that we have been introduced is if $\mathcal{A}$ is a $\sigma$-field, it is a field. In my opinion, the field must be a subtype of the field. In other words, if we consider all types of field, a portion of them may be considered as $\sigma$-field. However, in later class when we were briefly introduced with the concept of measure, our teacher tried to give us a general picture of what we have done so far.

He said that $(\Omega, \mathcal{A})$ is a measurable space provided that $\mathcal{A}$ is a $\sigma$-field of subset of $\Omega$. Then he went on saying that when mathematician tries to find the proper class for measurable space they realize only a field would be inadequate. He then went on saying that $\sigma$-field is a larger class compared to a field.

I am very confused at this point as I thought $\sigma$-field must be a smaller class as compared to the field. I am not even sure if these two concepts can be compared using the langue of "large" or "small". I read another post today What are the foundations of probability and how are they dependent upon a σ -field? which gives an answer "No" to the relationship between the $\sigma$-field and a field, which made me more bewildered.

I understand that my understanding of a field is very rudimentary as I am not coming from a mathematical background. I hope my naïve way of understanding of the relationship between a field and $\sigma$-field can reveal the misconceptions I have in mind. In this post, I mainly expect someone to explain to me the relationship between a field and a $\sigma$-field in a broader sense (not that if a class is a $\sigma$-field it must also be a field). For example, why we give so much attention to $\sigma$-field. What are the implications of the difference between the condition "countably infinite union"? (What will happen if we don't make such distinction?)

JoZ
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  • If you read the comments under the answer, you’ll find that Qiaochu was assuming that the OP was using field in the abstract algebra sense, so that no is irrelevant. – Brian M. Scott Sep 02 '20 at 01:00

2 Answers2

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Indeed all $\sigma$-fields are fields. This was not what was meant by "larger class" by your teacher. Also, the answer "no" by Qiaochu Yuan was saying that $\sigma$-fields are not fields in the sense of algebra; they are indeed fields of sets.

Anyway, a $\sigma$-field "typically" will need to have "more sets" in it than a field would; in particular, you can ask for the minimal $\sigma$-field containing a given field, and typically it will be a proper superset of the field you started with.

Within probability theory you only really care about $\sigma$-fields and can essentially ignore the notion of fields. In real analysis there is a bit more going on because we want to build things like the Lebesgue measure on $\mathbb{R}^n$ "without doing too much work" so we come up with slick tricks of extending the Lebesgue measure restricted to "rather small" subsets of the Lebesgue $\sigma$-algebra (in real analysis the term "$\sigma$-algebra" is considerably more common).

Ian
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  • I cannot recall the exact wording my teacher said so it is more like how I understood what he said and reinterpreted in ways that make sense to myself. The second paragraph may be what my teacher meant. Thank you for your explanation! – JoZ Sep 02 '20 at 01:09
  • In set theory a field is defined as closed under finite unions and intersections. The point is that if it is closed only under these operations, it is not adequate for measure theory. – herb steinberg Sep 02 '20 at 01:37
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As in many other circumstances in mathematics, trying to list a couple of good examples helps you clarify two connected while different concepts. Let's consider the following example.

Let $\mathcal{A}$ consist of the finite and the cofinite sets (a set $A$ is cofinite if $A^c$ is finite). Then $\mathcal{A}$ is a field (straightforward to verify by definition). If $\Omega$ is finite, then $\mathcal{A}$ contains all the subsets of $\Omega$ hence is a $\sigma$-field as well. If $\Omega$ is infinite, however, then $\mathcal{A}$ is not a $\sigma$-field. For instance, choose a sequence $\omega_1, \omega_2, \ldots$ of distinct points in $\Omega$ and take $A = \{\omega_2, \omega_4, \ldots\}$. Then $A \notin \mathcal{A}$, even though $A$ is the union, necessarily countable, of the singletons it contains and each singleton is in $\mathcal{A}$. This example shows that the definition of $\sigma$-field is indeed more restrictive that of field.

On the other hand, a $\sigma$-field must be a field (again, check the definition). In this sense, it makes sense to say $\sigma$-field is always "larger" than a field.

To your second question, here is a perspective. In probability, we are interested in evaluating the uncertainty (i.e., probability) of certain "events". Probabilistically speaking, the collection of all those events made up of the "$\mathcal{A}$" in the probability space triple $(\Omega, \mathcal{A}, P)$. While it is OK to study many interesting problems with a field $\mathcal{A}$ (say, the classic/discrete probability space), it is of particular interest to study some event which is a countably infinite union of some basic events. As a famous example (the so-called Borel's normal number theorem), consider throwing a coin perpetually, the event of those outcomes for which the asymptotic relative frequency of head in the infinite sequence of heads and tails is $\frac{1}{2}$. While this event cannot be formulated as finite union/intersection of simpler events given $\mathcal{A}$ is a field, it is expressible as a countably infinite unions/intersections of simpler events given $\mathcal{A}$ is a $\sigma$-field. You might have heard many celebrated theorems in probability, such as weak/strong law of large numbers, central limit theorem, etc., which all rely on the assumption that $\mathcal{A}$ is a $\sigma$-field instead of merely being a field.

Zhanxiong
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