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:
- 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"?
- 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"?
Best Answer
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
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.