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The statement "Measurable functions are closed under addition and multiplication, but not composition." was generated by Wolfram Alpha from a source of mixed quality by turning all references (and hyperlinks) from the original text into a "related topics" section with no connection to the main text. I have two questions related to this:

  1. Is automatic removal of references from a mathematical text (as done by Wolfram Alpha) an "invalid" text transformation? For example, if I write a text, do I have to take care that it won't change its meaning dramatically if all references are "silently" removed? (Checking my last publication, I realized that some sentences in it would also "break" if the references would be "silently" removed.) Should Wolfram Research fix this "automatism"?
  2. Is there something wrong (or evil) with the "subsets of sets with measure zero" modification that make it especially "fragile" with respect to minor inaccuracies? Isn't such a "fragile" concept difficult to remember correctly, and hence would require a really good justification by a corresponding significant benefit with respect to more "canonical" concepts?

Context (original question)

In theory, I think the ideas behind Wolfram Alpha are very worthwhile. In practice, google seems to be good enough for me. But now I tried to use Wolfram Alpha for my math questions. It was surprisingly good at understanding my questions. Most answers were disappointingly "shallow", but not wrong. However, the answer to measurable function deeply worries me:

... When $X=\mathbb{R}$ with Lebesgue measure, or more generally any Borel measure, then all continuous functions are measurable. In fact, practically any function that can be described is measurable. Measurable functions are closed under addition and multiplication, but not composition.

Both the statement that the Lebesgue measure is a Borel measure and the statement that measurable functions are not closed under composition are seriously misleading for me. (I had to read the Wikipedia article to reassure me that my current understanding wasn't completely wrong). You may object that I quoted this out of context, but the removal of context done by Wolfram Alpha is exactly what worries me. The original article from MathWorld certainly wasn't great, but the removal of the hyperlinks and references by Wolfram Alpha really killed it.

However, this also made me think about the usefulness of the "subsets of sets with measure zero" modification to the Borel measurable sets. Is there any good practical reason that justifies the confusion created by the difference between "Lebesgue measurable" and "Borel measurable"? Are there any practically relevant functions that are "Lebesgue measurable" but not "Borel measurable"?

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This MO thread seems relevant –  t.b. Oct 15 '11 at 21:42
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I could be just me, but while reading the question, I was a bit lost if this is a question about Wolfram Alpha or about measures. Ironically, I think this question gives so much context that it is distracting. ;) Is it possible for you to highlight your actual question a bit more clearly? –  Srivatsan Oct 16 '11 at 3:24
    
@Srivatsan Narayanan: You are right, I mixed context and question in an unfortunate way. I guess my initial question was whether the answer provided by Wolfram Alpha can be considered as outright wrong. In contrast, I wouldn't consider the original MathWorld article as wrong, it's just sloppy. Despite all the shortcomings of my question, the provided answers were actually quite helpful for me, and helped me settle my open questions. I will think about how to rewrite this question, and how to avoid the same mistake in future questions. –  Thomas Klimpel Oct 16 '11 at 9:13
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2 Answers

up vote 6 down vote accepted

First of all, for many people, using the "Lebesgue measure" does not imply that you are dealing with all "Lebesgue measurable" sets. The measure in $\mathbb{R}$ that satisfies $\mu((a,b]) = b - a$ is usually called Lebesgue measure only when it is completed. But some may call the measure restricted to the Borel sets with the same name.

The statement

In fact, practically any function that can be described is measurable.

is not really meaningful. It depends a lot on one's ability to describe/construct a function.

About the "composition of functions", it is important to notice that if you are talking about measurable spaces, then the functions do not simply take one set to another... they take one space to another space. Let's write $\mathcal{B}$ for the Borel sets and $\mathcal{L}$ for the Lebesgue measurable sets. The problem is that very often probabilists say that a function $f: \mathbb{R} \to \mathbb{R}$ is measurable when $f: (\mathbb{R}, \mathcal{L}) \to (\mathbb{R}, \mathcal{B})$ is measurable. But, in general, when dealing with $f: (\Omega_1, \sigma_1) \to (\Omega_2, \sigma_2)$ and $g: (\Omega_2, \sigma_2) \to (\Omega_3, \sigma_3)$, it IS TRUE that $g \circ f: (\Omega_1, \sigma_1) \to (\Omega_3, \sigma_3)$ IS measurable. Probably, they do that because they are used to defining $f: (\Omega, \sigma) \to \mathbb{R}$ as measurable when $f^{-1}([a,b]) \in \sigma$ for all $a,b \in \mathbb{R}$.

About your last question, I guess it is clear that in many cases it is easier to deal with a complete space. Nevertheless, notice that the concept of "completion" does depend on a fixed measure. The Borel sets is a family that depends only on the topology, while you do need a measure in order to complete it. So, if you are talking about all measures that one may define on a certain topological space, or on a certain measurable space, you will not talk about the completion of the space. One example is the Riesz representation theorem.

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An important and very practical example of a function that is Lebesgue measurable (actually universally measurable), but not Borel measurable is the value function in dynamic optimization.

In general, if you have Borel subet of $\mathbb{R}^2$ onto one factor, the set is Lebesgue measurable but may not be a Borel subset of $\mathbb{R}$. The relation of this fact to optimization problems is explained in a beautiful way here (JSTOR needed).

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The link to "value function" doesn't work for me. –  Thomas Klimpel Feb 8 '12 at 1:20
    
It should work now. –  Michael Greinecker Feb 8 '12 at 1:25
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