Talk:Measurable function

Borel function[edit]

Borel function - that para really needs turning round to be clear. Charles Matthews 12:48, 6 Sep 2004 (UTC)

Composition of measurable functions not necessarily measurable[edit]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

Here's what Folland has to say:

If

(X,{\mathcal {M}})

is a measurable space, a real- or complex-valued function f on X will be called

{\mathcal {M}}

-measurable, or just measurable, if it is

({\mathcal {M}},{\mathcal {B}}_{R})

or

({\mathcal {M}},{\mathcal {B}}_{C})

measurable.

{\mathcal {B}}_{R}

or

{\mathcal {B}}_{C}

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)[reply]

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)[reply]

- 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)[reply]

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)[reply]

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)[reply]

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 $h\circ g$ is not, because the inverse image of $(0,\infty )$ under that map is $g^{-1}[h^{-1}[(0,\infty )]]=g^{-1}[A]=B$ 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

g:X\rightarrow Y

and

h:Y\rightarrow Z

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

(h\circ g)^{-1}[D]=g^{-1}[h^{-1}[D]]=g^{-1}[E]

is measurable. —vivacissamamente 07:24, 5 November 2006 (UTC)[reply]

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)[reply]

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

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

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

Well, well. The SpringerLink Maths dictionary (http://eom.springer.de/M/m063210.htm) makes the useful distinction between a measurable function and a measurable mapping. With a measurable function, the preimage of an open (hence, Borel) set is measurable, whereas with a measurable "mapping between two measure spaces", the preimage of a measurable set is measurable. The latter definition is closed to compositions, while the former is not. Maybe we should do the same - have two interlinked pages. mousomer 23:13, 17 January 2007 (UTC)[reply]

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 21:42, 2 August 2007 (UTC) 68.35.224.53 04:33, 6 February 2007 (UTC)[reply]

I think that the statement "the composition of measurable functions is measurable" is misleading because it is not clear a priori what "measurable" means. (This fact should be apparent from the above comments on this page.) I have changed the statement to a more explicit statement which is quite definitely correct, using Billingsley's notation of "measurable Σ/Τ". I know that no one ever says "measurable Σ/Τ"--one just says "measurable"--but depending upon one's field of study, "measurable" means different things. (To someone working with measure in a rather abstract sense, it is probably understood that the σ-algebra of the intermediate space is invariant during the composition; however, to someone working with Lebesgue measure, it is probably understood that the σ-algebra of the intermediate space is different depending on whether it is the domain or the codomain.) I think it is better to be explicit because it seems likely that people with both definitions of "measurable" in mind might read this page, and the answer to the question of composition depends on which definition one uses. Suppressing mention of the underlying σ-algebras is not appropriate here, in my opinion. Oh, and I think it's clear after perusing Billingsley that the σ-algebra on the target space need not be a Borel algebra (although it often is), so I changed this as well. —Babcockd 16:13, 30 July 2007 (UTC)[reply]

I think that the previous statement made by Babcockd may be more correct, depending on the context. A composition of functions may be measurable or not depending on the domain and the codomain, in the proof presented by Billingsley I believe that everything is considered measurable. Jka02 21:42, 2 August 2007 (UTC)[reply]

What is mid{-g,f,g}?[edit]

From the article:
A useful characterisation of Lebesgue measurable functions is that f is measurable if and only if mid{-g,f,g} is integrable for all non-negative Lebesgue integrable functions g.
Why is it useful? What does mid{-g,f,g} mean? Timhoooey 04:02, 14 October 2007 (UTC)[reply]

Sum and product of real-valued measurable functions[edit]

From the article:
The sum and product of two real-valued measurable functions are measurable.
What about when we have functions f,g and f+g is of the form $\infty -\infty$ or $-\infty +\infty$ ? -Timhoooey 04:24, 14 October 2007 (UTC)[reply]

Well, those wouldn't be "real-valued"; they'd be "extended-real-valued", and the problem would be not that f+g might not be measurable, but that it might not be a function at all. But for any reasonable way of extending the definitions to partial functions, the result should still be fine, I think. --Trovatore 19:10, 14 October 2007 (UTC)[reply]

I think it would be illustrative to include in the article a proof sketch of why fg is measurable. I'm trying to learn this stuff and I find it difficult to make the jump from definitions to understanding without anything in between. Timhoooey 23:21, 14 October 2007 (UTC)[reply]

Here it is if someone wants to fill in the details:

Since

f\,

is measurable,

f^{k}\,

is measurable for integer powers k. Since

f,g\,

are measurable,

f+g\,

is measurable. Then note that

fg={\frac {1}{4}}((f+g)^{2}+(f-g)^{2})\,

. Timhoooey 01:17, 15 October 2007 (UTC)[reply]

To be honest I would call that a "trickological" proof. I don't know what your priority queue is like, but I'd suggest you try to find a proof that's a little more enlightening as to what's going on. --Trovatore 20:49, 15 October 2007 (UTC)[reply]

Glitch in definition?[edit]

The article says that f : X → Y is measurable provided that for every subset Z of Y in T the preimage of Z is in Σ. Two questions... first, nobody said f was surjective, so in general, Z doesn't even have a preimage. Second, nobody said f was injective, so how can we talk about the preimage of Z? Am I missing something obvious?—PaulTanenbaum (talk) 01:26, 28 February 2008 (UTC)[reply]

The preimage of a set Q under a function f is the set of points x in the domain such that f(x) is in Q. This is well defined (and, in particular, unique) for every subset of the codomain of the function. It may be that the preimage of Q is the empty set, for example, if no point in the domain maps into Q. — Carl (CBM · talk) 01:56, 28 February 2008 (UTC)[reply]

Ah, I was unacquainted with the notion of preimage of a set so, in case it's not clear from my original question, I presumed that f^-1({y₁, y₂, ...}) was being used to mean {f^-1(y₁), f^-1(y₂), ...}.—PaulTanenbaum (talk) 04:23, 28 February 2008 (UTC)[reply]

For a countable set, that is exactly the right idea, with the caveats that if f^-1(y_k) doesn't exist for some k (because y_k isn't in the range) then there is no contribution to the preimage set, and if multiple points map to y_k then they all get put into the preimage set. — Carl (CBM · talk) 14:34, 28 February 2008 (UTC)[reply]

Deleted Section "Stationary transformations"[edit]

because it duplicates some of the material in measure-preserving dynamical system. (Moreover, there was an inconsistency, since the assumption that the measure is a probability measure was implicit, but not explicit.) —Preceding unsigned comment added by OdedSchramm (talk • contribs) 16:42, 8 April 2008 (UTC)[reply]

Separate section with formal definition?[edit]

I usually come to wikipediate maths pages looking for a specific result or definition. Therefore I think it best to include a section that contains a formal definition. Removing the definition from the introductory paragraph would also allow this paragraph to provide a more intuitive idea of the concept. BenWhale (talk) 22:44, 12 July 2010 (UTC)[reply]

Rewrite?[edit]

As one poster said earlier, this pages suffers from excessive generality. I am studying analysis in graduate school and this page is so convoluted it had me questioning what it meant to be measurable. Could we rewrite the page to focus on Lebesgue measurability and one or more of the definitions for that [e.g. $f^{-1}((x,\infty ))$ is measurable for all a or $f^{-1}(B)$ is measurable for all Borel sets B], then move on to definitions in full generality? Ckhenderson (talk) 21:58, 20 July 2010 (UTC)[reply]

Different Definition?[edit]

A function f:X->R is said to be measurable if, for every \alpha \in R, {x \in X : f(x) > \alpah} \in Sigma. Sigma is the sigma-algebra which contains X. from Linear Functional Analysis by B.P. Rynne. Brydustin (talk) 00:20, 8 February 2012 (UTC)[reply]

Measurable and μ-measurable functions[edit]

There is a notion of a function being $μ$ -measurable. At least I have seen it for outer measures. Let $X$ be a space with outer measure $μ$ and let $Y$ be a topological space. A function $f : X \to Y$ is $μ$ -measurable if the inverse image of every open set is $μ$ -measurable. ~~(Perhaps $μ$ is required to be Borel regular, don't remember.)~~ No it isn't. Then it is a theorem that $f$ is $μ$ -measurable if and only if the inverse image of every Borel set is $μ$ -measurable.

Now, as far as I can see, if $X = Y = ℝ$ and if $μ = Lebesquge$ , then the above reduces to the content of the section Caveat.

The notion of $μ$ -measurable function makes much sense, at least if $X$ is a topological space and $μ$ is Borel regular. I'm not too sure whether the distinction between measure and outer measure is devastating. At any rate, there is an outer measure bearing the name Lebesgue measure that reduces to (ordinary) Lebesgue measure when restricted to the $σ$ -algebra of measurable sets.

Provided my observations make sense, perhaps the article could introduce $μ$ -measurable and thereby put Lebesgue measurable function in a context. A possible reference is

Federer, H. (1996) [1969]. Geometric Measure Theory. Classics in Mathematics (1st ed reprint ed.). Berlin, Heidelberg, New York: Springer Verlag. ISBN 978-3540606567. {{cite book}}: Invalid |ref=harv (help)

YohanN7 (talk) 10:01, 11 July 2016 (UTC)[reply]

Re-writing the lede[edit]

I'm going to re-write the lede because the current one, in my view, doesn't serve its purpose. First, the opening sentence (i.e. "a measurable function is a structure-preserving function between measurable spaces") is misleading. Second, there is some extraneous fluff (e.g. "... functions that are not Lebesgue measurable are generally considered pathological"). Third, and most importantly, the lede in this particular article probably shouldn't be written in Scientific American style and try to dumb the subject down. In other articles ("Probability Theory" is a good example of that), having a SciA-like intro is great. But measurable function just seems too specialized a notion to be of interest to the non-geeky crowd. StrokeOfMidnight (talk) 21:53, 3 August 2017 (UTC)[reply]

Is the definition correct?[edit]

In textbooks such as Walter Rudin, Real and Complex Analysis(3e) or Johnsonbaugh and Pfaffenberger, Foundations of Mathematical Analysis, the definition of a measurable function goes like “a function f from a measurable space X to a topological space Y such that f⁻¹(V) is measurable in X for every open set V in Y.” Some other textbooks that only deal with real measurable ones require the same one with sets of the form (c, ∞] in place of every open set V, which form subbase for the standard topology.

In my thought, one defines measurable functions to use it to define (Lebesgue) integral. However, to define an integral, doesn't one need a limit process, and thus, a topology? If the codomain of a measurable function is not a topological space but merely a measurable one, is it possible to define a useful integral? Then why one defines measurable functions?

Does someone know the source of the current version of the definition? --Inactivate (talk) 22:24, 19 December 2017 (UTC)[reply]

The WP definition of LI is set-theoretic, so you don't need measurability or topology for the definition itself. But how useful this definition is without the two is a question. You at least need the so-called "continuity from below" property to prove the monotone convergence theorem. The MCT is then used to prove linearity.

In probability theory, you want to assign probabilities to individual events before you start talking about random variables, right? And it's perfectly meaningful to have several RVs attached to the same probability space. To me this means that measure spaces are primary, and measurable functions are secondary, not vice versa.

By the way, since we know that every non-negative measurable function is a pointwise limit of a non-decreasing sequence of simple non-negative functions, what prevents us from using this to define the LI? This is exactly what some authors do.

On the last question you asked, I looked at this article ("Measurable function") in several languages, and most, if not all, seem to use (to the best of my understanding) the definition we're using. StrokeOfMidnight (talk) 21:37, 20 December 2017 (UTC)[reply]

Thank you for kind response.

By the way, don't you mean by “The WP definition of LI” this section Lebesgue_integration#Towards_a_formal_definition? This definition becomes nonsense if f is not measurable in the sense that f⁻¹(t, ∞] is measurable for all t. So I don't understand this comment: “you don't need measurability or topology for the definition itself”.

Yes, if we want to define LI for non-negative measurable functions then the current definition of a measurable function works. However, if the current definition is justified only in that it fits for non-negative ones, then it is in no way generalized than requiring just f⁻¹(t, ∞] is measurable for all t. In my thought, in that comment you did not give a reason why we may extend the codomain in the definition of measurable functions from nonnegative reals to an arbitrary measurable space. I am wondering if such extension in the current definition has any meaningful result regarding integration(other than this: f: Σ/Τ-measurable, g non-negative Τ-measurable ⇒ g∘f non-negative Σ-measurable) or not.

On the last question, yes, I did the same thing, and I found out that any of the articles (which I looked at) did not give a source(I mean, any paper, any textbook or at least something outside this WP) of the definition. You know that articles in other language versions of WP tend to be a mere translation of this English version.

For instance, in Paul R. Halmos, Measure Theory, a function satisfying the current WP definition is called a “measurable transformation”, rather than merely a measurable function. The author adds: “We observe that this language is inconsistent with our earlier one concerning measurable functions; …… confusion can always be avoided by use of the proper one of the terms ‘function’ and ‘transformation.’” --Inactivate (talk) 05:36, 27 December 2017 (UTC)[reply]

1. The way I understood your question was: can

\textstyle \int _{X}f\,d\mu _{X}

be defined when

X

is not a measure space? Yes, it can be and has (sort of) already been. Use the cumulative distribution function in probability theory as an example. If

f:X\to Y

is a simple function, then, rather than defining

\textstyle \int _{X}f\,d\mu _{X},

we define

\int _{f(X)}f\,d\mu _{Y}=\sum _{i=1}^{n}c_{i}\mu _{Y}(\{c_{i}\})

and use

\sup

in the general case. Now, whether this definition is useful is a different question.

2. This might be too trivial, but you can always talk about measurable function pairs

(f^{+},f^{-})

instead of measurable functions

f^{+}-f^{-}

if you want to add, subtract them, etc. It's a terminology issue more than anything else. I, personally, find it easier to talk about arbitrary measurable functions, but, again, it's a matter of preference.

3. What makes you think everyone blindly translates math articles from English into their own language? StrokeOfMidnight (talk) 03:04, 2 January 2018 (UTC)[reply]

"Bimeasurable function" listed at Redirects for discussion[edit]

A discussion is taking place to address the redirect Bimeasurable function. The discussion will occur at Wikipedia:Redirects for discussion/Log/2021 January 11#Bimeasurable function until a consensus is reached, and readers of this page are welcome to contribute to the discussion. 𝟙𝟤𝟯𝟺𝐪𝑤𝒆𝓇𝟷𝟮𝟥𝟜𝓺𝔴𝕖𝖗𝟰 (𝗍𝗮𝘭𝙠) 13:09, 11 January 2021 (UTC)[reply]