Talk:Measurable function

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Borel function - that para really needs turning round to be clear.

Charles Matthews 12:48, 6 Sep 2004 (UTC)

[edit] Composition of measurable functions not necessarily measurable

This page has at least a few errors. The composition of two Lebesgue measurable functions is not neccesarily Lebesgue measurable. Some types of integrals work just fine on non-measurable functions, e.g. a stieltjes integral.(This comment made June 13 2006 by 66.171.165.218)Rich 09:46, 4 November 2006 (UTC)

• The current version (Nov4 2006) seems to be contradictory. 1)It says that measurable functions are the morphisms and 2) says composition of measurable funtions needn't be measurable. I guess the measurable functions could be a generating set for the morphisms if (2) is true, in which case (1) should be modified. I don't see why (2) should true, but (Statement (2) was by 128.30.51.97 on Sep 15, 2006) the statement is apparently backed up by Mathworld by Todd Rowland and also by Trovatore in answer to a question on the HelpDesk late last month. My obvious reasoning follows, and I'd be an idiot to think I'm right, it's too easy to be missed by you guys:

It's parallel to the composition of cts functions being cts in topology--If f and g are measurable and f is from A to B and g is fr B to C, then g^-1(meas set in C) is measurable in B and f^-1(g^-1(meas set in C)) is measurable in A. Now I know I'm wrong, since I'm close to an idiot in top. and analysis, but it will improve the article to have an explanation and references for this. Thanks,Rich 09:57, 4 November 2006 (UTC)

Hi Rich,

You'd be right, if the definition of "measurable function" were "the preimage of every measurable set is measurable". But that's too restrictive a definition (doesn't even include all the continuous functions). The standard definition is "the preimage of every open set is measurable" (and thus automatically the preimage of every Borel set is measurable).

The key thing to keep in mind is that every subset of a measure-0 set is measurable. That means you can have very pathological sets that are still measurable, just because all their pathology is coded up inside the Cantor set or something. Then you can use a well-behaved function under which the preimage of the Cantor set has positive measure, and then the preimage of your pathological set has a chance to "show itself" as it were. --Trovatore 19:00, 4 November 2006 (UTC)

But in fact the article does define a measurable function to be one in which the pre-image of every measurable set is measurable (and that's the correct definition). So the composition of two measurable functions is measurable, as long as you are consistent about the σ-algebra being used. Of course, what you are saying is that in many common situations one uses the Lebesgue σ-algebra for the domain and the Borel σ-algebra for the codomain, and so this consistency is lacking. I think the article needs to explain this clearly, rather than just stating that the composition of two measurable functions may not be a measurable function, since this statement is clearly false when interpreted in the obvious way. --Zundark 19:36, 4 November 2006 (UTC)

The article suffers from excessive generality. The most common sense in which the term "measurable function" is used is to mean that the preimage of a Borel set is measurable; that should be the first definition, with generalizations treated later. --Trovatore 20:09, 4 November 2006 (UTC)

Here's what Folland has to say:

If is a measurable space, a real- or complex-valued function f on X will be called -measurable, or just measurable, if it is or measurable. or is always understood as the σ-algebra on the range space unless otherwise specified.

Folland, Gerald B. Real Analysis: Modern Techniques and Their Applications, 1984, p. 43. --Trovatore 20:35, 4 November 2006 (UTC)

• Then what I called statement (1) above needs to be corrected or eliminated. Even if eliminated from this article, a correct statement about the morphisms in the usual category of measure spaces should be inserted in that article.

The 'common' definition of measurable Zundark describes should replace the current one asap. We can use the Folland reference for that, I think.

A simple example f(x) if there is one, with f measurable and f(f(x)) not measurable, would be helpful.

There is some good stuff in this article but overall it is currently not very good. Thanks for the very helpful replies from both of you.Rich 00:21, 5 November 2006 (UTC)

• • I'll put in an expert needed template to get an expert and give innocent readers a heads up.Rich 00:39, 5 November 2006 (UTC)

Okay, but it is otherwise specified later. Here's the rub.

A function from a measure space (X,Σ) to the reals is defined to be measurable when the inverse image of every open set is measurable.

• The very last word in the preceding sentence is [open]. I'm sure you meant [measurable].Rich 02:15, 5 November 2006 (UTC)

Right. Whoops. I've changed it now.

A function from a measure space (X,Σ) to another measure space (Y,Τ) is defined to be measurable if the inverse image of every set in Τ is a set in Σ.

This second definition doesn't specialize to the first definition because the first definition only implies that the inverse image of every Borel set is measurable, not every Lebesgue-measurable set.

Confusing matters worse is that usually, when unqualified, a "measurable" function on a space is a function from that space to the reals.

As pointed out, the composition of two measurable functions to the reals need not be measurable. First of all, in order to compose two such functions, the second function must be a function from the reals to the reals. But a measurable function f from R to R in the usual sense only has f^-1 of a Borel set Lebesgue-measurable; it doesn't usually have f^-1 of a Lebesgue-measurable set Lebesgue-measurable. I can find you an example if you want.

• If you can find one or tell me a promising place to look for one that would be great.Rich 02:15, 5 November 2006 (UTC)

The example I know is from the exercises in Royden's Real Analysis. It relies on the fact that every subset of the reals with positive outer measure contains a nonmeasurable set. Consider the function f : [0,1] → [0,2] given by f(x) = x + f₁(x), where f₁ is the Cantor function. Let C be the Cantor set.

• f is a homeomorphism, so g: = f ^{− 1} is continuous, and hence measurable.
• The image F := f[C] has measure 1, even though C has measure 0.
• There is a subset B of F that is nonmeasurable, but since A := g[B] is a subset of a set of measure zero (C), A is measurable.
• We know the characteristic function h of A is measurable, but is not, because the inverse image of under that map is which isn't measurable.
• A is a measurable set that is not Borel, because since f is a homeomorphism, B would be Borel if A was (and it's not even measurable).

On the other hand, the proof that the composition of two measurable functions in the other sense is measurable is quite easy: if we have and both measurable (using the same σ-algebra of sets on Y for each map, I mean), then if a subset D of Z is measurable, we know E: = h ^{− 1}[D] is measurable, and so is measurable. —vivacissamamente 07:24, 5 November 2006 (UTC)

However, the composition (where defined) of any two functions that are measurable in the second sense is indeed measurable, and in that sense they form a sensible class of morphisms to set up a category of measurable spaces and measurable functions.

• That second sense needs to be defined explicitly and put in a separate section with the morphism claim, and attention called to the difference in senses(in a good article).Rich 02:15, 5 November 2006 (UTC)

Hope this helps. —vivacissamamente 00:51, 5 November 2006 (UTC)

• Yes it does, thanks.Rich 02:15, 5 November 2006 (UTC)

I removed the morphisms statement until it can be clarified and corrected, but forgot the edit summary.Rich 00:53, 5 November 2006 (UTC)

[edit] A composition of measurable functions is measurable?

Please see the following and perhaps this will help: page 182 Probability and Measure 3rd ed. , Patrick Billingsley, Wiley, 1995. ISBN:0-471-00710-2. Jka02 01:05, 14 December 2006 (UTC)