Prove Borel Measurability of a Function f:\mathbb{R}^n->\mathbb{R}^m

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In summary: Y_{j_1j_2j_3j_4\dots j_{2n-1}j_{2n}}\right )But since each f_i is continuous, they are all Borel measurable, so that the inverse image of an open set is a Borel set. This should be enough to conclude that f is Borel measurable.In summary, we discussed the concept of a function being measurable in terms of a \sigma-algebra on
  • #1
Oxymoron
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QUESTION 1: "If [itex]f\,:\,X \rightarrow \mathbb{R}^n[/itex] is a function then if its coordinate functions are measurable then does this imply that f is measurable?"

Suppose that we have a [itex]\sigma[/itex]-algebra, [itex]\mathcal{A}[/itex] defined on some set X. We say that a function [itex]f\,:\,X \rightarrow \mathbb{R}^n[/itex] is [itex]\mathcal{A}[/itex]-measurable if [itex]f^{-1}(B) \in \mathcal{A}[/itex] for every [itex]B \in \mathcal{B}^n[/itex], that is, if the pre-image of every Borel set is in the sigma-algebra.

My question is this: Would it be possible to prove that the function [itex]f\,:\,X\rightarrow\mathbb{R}^n[/itex] is measurable if and only if its coordinate functions [itex]f_i\,:\,X\rightarrow\mathbb{R}[/itex] are measurable? (Where the coordinate functions are characterised by [itex]f(x) = (f_1(x),\dots,f_n(x))[/itex] for all [itex]x \in X[/itex]).

I mean, is f measurable if and only if all of its coordinate functions are measurable?

I know that if one regards the standard topology on [itex]\mathbb{R}[/itex] then every open set is a countable union of open intervals. Hence the Borel sets are generated by the set of all open intervals [itex]\{(a,b)\,:\,a < b\}[/itex].

QUESTION 2: "If I have a function [itex]f\,:\,\mathbb{R}^n \rightarrow \mathbb{R}^m[/itex] which is continuous then does this mean that it is automatically Borel measurable?"

Is it true that I should be basing my idea of "Borel measurable" on the following?:

A function f is Borel measurable if the pre-image of any Borel set is in the sigma-algebra

If so, then when any question asks to prove that a function is Borel measurable then do I simply have to prove that the pre-image of every Borel set is in the sigma-algebra?
 
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  • #2
For (1), it's either obviously yes, or obviously no. I haven't decided which, yet. :frown:
For (2), yes. In this case, you are presumably using the Borel sets as your sigma-algebra.
 
  • #3
I assume it only makes sense to talk about a function's coordinate functions when we have R^n as our range. And each f_i maps to a 1-dimensional slice of R^n. This kind of reminds me of when I covered coordinate functions (and charts etc...) in differential geometry.

and yes, my sigma-algebra is the set of Borel sets.
 
  • #4
I want to prove that if [itex]f\,:\,\mathbb{R}^n \rightarrow \mathbb{R}^m[/itex] is continuous then [itex]f[/itex] is Borel measurable.

Since the function [itex]f[/itex] is continuous we know that it preserves open sets. Since every open set is Borel, continuity of [itex]f[/itex] implies that [itex]f^{-1}(B) \in \mathcal{B}[/itex] for all open (Borel) sets [itex]B \in \mathcal{B}[/itex]. But this is precisely the definition for a Borel measurable function! Therefore, since a continuous function, by defintion, preserves open sets - that is for every open set [itex]B[/itex], [itex]f^{-1}(B)[/itex] is open in [itex]\mathbb{R}^n[/itex]. But since all open sets are Borel. Hence continuity implies Borel measurability.

[tex]\square[/tex]

How does this look?
 
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  • #5
That's not a valid proof, since Borel sets are not necessarily open. (How does your proof show that the inverse image of [itex][0,1)[/itex] for a continuous function [itex]f:\mathbb{R} \rightarrow \mathbb{R}[/itex] is Borel?)

P.S. Hurkyl, I'm not sure if you were making a joke, but if coordinate function means [itex]f[/itex] restricted to one of the dimensions in [itex]\mathbb{R}^n[/itex] then (1) is true.
 
  • #6
Hmmm, you are right. Borel sets include closed and neither open nor closed sets, not just the open ones. However, I should still be able to argue that every open set is Borel, no?
 
  • #7
The clause 'for every open set [itex]B[/itex]' should be setting off alarms in your noggin.

It wouldn't hurt you to be a bit more careful in your notation as well. You seem to be using [itex]\mathcal{B}[/itex] to refer to two possibly different sigma-algebras.

You could try it this way:
Start by assuming that [itex]B[/itex] is an arbitrary Borel set, and try to show that [itex]f^{-1}|_B[/itex] is a Borel set.
Or, alternatively, you can assume that [itex]f^{-1}|_X[/itex] is not Borel, and show that [itex]X[/itex] is not Borel.

As warm-ups, you could show that:
1. The inverse images of closed sets under continuous functions are Borel.
and
2. Any Borel set can be decomposed into a countable union of open and closed sets.
 
  • #8
Ok, I'll try those first.

It wouldn't hurt you to be a bit more careful in your notation as well. You seem to be using [itex]\mathcal{B}[/itex] to refer to two possibly different sigma-algebras.

I was using [itex]\mathcal{B}[/itex] to be the Borel sigma algebra generated by all the open sets from [itex]\mathbb{R}[/itex].

Also, does anyone have a proof of 1. Or possibly direct me into creating one. I was hoping to find a similar thing that I did in diff. geometry, but its not working for me.
 
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  • #9
Define [itex]\pi _i : \mathbb{R}^n \to \mathbb{R}[/itex] by [itex]\pi _i (x_1, \dots , x_i , \dots , x_n) = x_i[/itex]. Then [itex]f_i = \pi _i \circ f[/itex]. It should be clear, then, that if f is measurable, so are the co-ordinate functions. The converse does not hold. Let [itex]X = \mathbb{R},\, \mathcal{A} = \{\emptyset , X\}[/itex], and define f by f(t) = (t, sin(t)). The image of f is the graph of the sine function. What is the pre-image of [itex]\{(t, \sin (t)) : -\pi /2 \leq t \leq \pi /2\}[/itex]?

For the second question, note that the Borel sets consist of sets of the form:

[tex]\bigcup _{j_1 = 1} ^{\infty}\bigcap _{j_2 = 1} ^{\infty}\bigcup _{j_3 = 1} ^{\infty}\bigcap _{j_4 = 1} ^{\infty}\dots\bigcup _{j_{2n-1} = 1} ^{\infty}\bigcap _{j_{2n} = 1} ^{\infty}Y_{j_1j_2j_3j_4\dots j_{2n-1}j_{2n}}[/tex]

where each [itex]Y_{j_1j_2j_3j_4\dots j_{2n-1}j_{2n}}[/itex] is either open or the complement of an open set. This is just what it means for the Borel sets to be generated by the open sets. Now inverse images commute with unions, intersections, and complements, then the sense that the inverse image of an arbitrary union of sets is the union of the inverse images of the sets, the inverse image of an arbitrary intersection of sets is the intersection of the inverse images of the sets, and the inverse image of a compelement of a set is the complement of the inverse image of the set. So:

[tex]f^{-1}\left (\bigcup _{j_1 = 1} ^{\infty}\bigcap _{j_2 = 1} ^{\infty}\bigcup _{j_3 = 1} ^{\infty}\bigcap _{j_4 = 1} ^{\infty}\dots\bigcup _{j_{2n-1} = 1} ^{\infty}\bigcap _{j_{2n} = 1} ^{\infty}Y_{j_1j_2j_3j_4\dots j_{2n-1}j_{2n}}\right ) = \bigcup _{j_1 = 1} ^{\infty}\bigcap _{j_2 = 1} ^{\infty}\bigcup _{j_3 = 1} ^{\infty}\bigcap _{j_4 = 1} ^{\infty}\dots\bigcup _{j_{2n-1} = 1} ^{\infty}\bigcap _{j_{2n} = 1} ^{\infty}f^{-1}(Y_{j_1j_2j_3j_4\dots j_{2n-1}j_{2n}})[/tex]

If f is continuous, then each [itex]f^{-1}(Y_{j_1j_2j_3j_4\dots j_{2n-1}j_{2n}})[/itex] is either open or closed, hence the right hand side of the above equation is a Borel set, so f is measurable.
 
  • #10
The Borel sets are the sigma algebra is generated by the open sets, ie, the open sets form a basis for it. Can you show that if the preimage of every basis set of a sigma algebra is measurable, then the preimage of every measurable set is measurable?

EDIT: sorry, I didn't read AKG's post. There's got to be a way that's easier on notation though.
 
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  • #11
I see this is an old thread, but I don't think it was well resolved,

For the second question, note that the Borel sets consist of sets of the form:

⋃j1=1∞⋂j2=1∞⋃j3=1∞⋂j4=1∞…⋃j2n−1=1∞⋂j2n=1∞Yj1j2j3j4…j2n−1j2n

I am not entirely sure that this is true. According to Real Analysis by Royden 2nd edition, pg. 50-51,

A set which is a countable union of closed sets is called Fσ (F for closed, σ for sum) ... We say that a set is a Gδ if it is the intersection of a countable collection of open sets (G for open, δ for durchschnitt). ... We could also consider sets of type Fσδ, which are the intersections of countable collections of sets each of which is an Fσ ... Thus the classes in the two sequences Fσ, Fσδ, Fσδσ, ... Gδ, Gδσ, Gδσδ, ... are all classes of Borel sets. However, not every Borel set belongs to one of these classes.

Quote frankly, I'm not sure how you might go about constructing Borel sets that isn't of that form, but it does raise concern right? It looks like the form that AKG put the Borel sets is equivalent to Gδσ..., so then according to Royden there is a Borel set that doesn't fit AKG's form?

So this question hasn't exactly been resolved?
 
  • #12
math4tots said:
I see this is an old thread, but I don't think it was well resolved,



I am not entirely sure that this is true. According to Real Analysis by Royden 2nd edition, pg. 50-51,



Quote frankly, I'm not sure how you might go about constructing Borel sets that isn't of that form, but it does raise concern right? It looks like the form that AKG put the Borel sets is equivalent to Gδσ..., so then according to Royden there is a Borel set that doesn't fit AKG's form?

So this question hasn't exactly been resolved?

You are entirely correct, there are Borek sets which are NOT of the form that AKG describes. In general we need more than countably many steps to obtain all Borel sets.

The question (2) can be resolved as follows: let

[tex]\mathcal{A}=\{A\in \mathcal{B}^m~\vert~f^{-1}(A)\in \mathcal{B}^n\}[/tex]

Our objective is to show that [itex]\mathcal{A}=\mathcal{B}^m[/itex]. This is quite easy. We just need to show that [itex]\mathcal{A}[/itex]

1) is a sigma-algebra (this is very easy)
2) contains the open sets (this follows from continuity

But [itex]\mathcal{B}^m[/itex] is the smallest sigma-algebra with that property. So this implies [itex]\mathcal{A}=\mathcal{B}^m[/itex].
 

Related to Prove Borel Measurability of a Function f:\mathbb{R}^n->\mathbb{R}^m

1. What is the definition of Borel measurability?

Borel measurability is a property of a function that describes its ability to be measured in terms of sets in a certain mathematical space. In the case of a function from Rn to Rm, Borel measurability means that the function can be measured in terms of Borel sets in Rn.

2. How is Borel measurability related to Lebesgue measurability?

Borel measurability is a stronger condition than Lebesgue measurability. A function that is Borel measurable is also Lebesgue measurable, but the converse is not always true. This means that a function that is Borel measurable is also measurable with respect to a more general measure, the Lebesgue measure.

3. What are the criteria for proving Borel measurability of a function?

In order to prove Borel measurability of a function f, there are two criteria that must be satisfied. First, the inverse image of every open set in the codomain Rm must be a Borel set in the domain Rn. Second, the function must be continuous.

4. Can Borel measurability be extended to functions with complex-valued outputs?

Yes, Borel measurability can be extended to functions with complex-valued outputs. In this case, the function must be measurable with respect to the Borel sets in Cm, the complex space of dimension m.

5. Are there any practical applications of proving Borel measurability of a function?

Yes, there are many practical applications of proving Borel measurability of a function. One example is in probability and statistics, where Borel measurability is used in defining and analyzing random variables. It is also important in the study of dynamical systems and differential equations, as well as in machine learning and signal processing.

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