# Borel Measurable

1. Nov 15, 2006

### Oxymoron

QUESTION 1: "If $f\,:\,X \rightarrow \mathbb{R}^n$ is a function then if its coordinate functions are measurable then does this imply that f is measurable?"

Suppose that we have a $\sigma$-algebra, $\mathcal{A}$ defined on some set X. We say that a function $f\,:\,X \rightarrow \mathbb{R}^n$ is $\mathcal{A}$-measurable if $f^{-1}(B) \in \mathcal{A}$ for every $B \in \mathcal{B}^n$, 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 $f\,:\,X\rightarrow\mathbb{R}^n$ is measurable if and only if its coordinate functions $f_i\,:\,X\rightarrow\mathbb{R}$ are measurable? (Where the coordinate functions are characterised by $f(x) = (f_1(x),\dots,f_n(x))$ for all $x \in X$).

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 $\mathbb{R}$ then every open set is a countable union of open intervals. Hence the Borel sets are generated by the set of all open intervals $\{(a,b)\,:\,a < b\}$.

QUESTION 2: "If I have a function $f\,:\,\mathbb{R}^n \rightarrow \mathbb{R}^m$ 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?

Last edited: Nov 15, 2006
2. Nov 15, 2006

### Hurkyl

Staff Emeritus
For (1), it's either obviously yes, or obviously no. I haven't decided which, yet.
For (2), yes. In this case, you are presumably using the Borel sets as your sigma-algebra.

3. Nov 15, 2006

### Oxymoron

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. Nov 16, 2006

### Oxymoron

I want to prove that if $f\,:\,\mathbb{R}^n \rightarrow \mathbb{R}^m$ is continuous then $f$ is Borel measurable.

Since the function $f$ is continuous we know that it preserves open sets. Since every open set is Borel, continuity of $f$ implies that $f^{-1}(B) \in \mathcal{B}$ for all open (Borel) sets $B \in \mathcal{B}$. 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 $B$, $f^{-1}(B)$ is open in $\mathbb{R}^n$. But since all open sets are Borel. Hence continuity implies Borel measurability.

$$\square$$

How does this look?

Last edited: Nov 16, 2006
5. Nov 16, 2006

### NateTG

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

P.S. Hurkyl, I'm not sure if you were making a joke, but if coordinate function means $f$ restricted to one of the dimensions in $\mathbb{R}^n$ then (1) is true.

6. Nov 16, 2006

### Oxymoron

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. Nov 16, 2006

### NateTG

The clause 'for every open set $B$' 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 $\mathcal{B}$ to refer to two possibly different sigma-algebras.

You could try it this way:
Start by assuming that $B$ is an arbitrary Borel set, and try to show that $f^{-1}|_B$ is a Borel set.
Or, alternatively, you can assume that $f^{-1}|_X$ is not Borel, and show that $X$ 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. Nov 16, 2006

### Oxymoron

Ok, I'll try those first.

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

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.

Last edited: Nov 16, 2006
9. Nov 16, 2006

### AKG

Define $\pi _i : \mathbb{R}^n \to \mathbb{R}$ by $\pi _i (x_1, \dots , x_i , \dots , x_n) = x_i$. Then $f_i = \pi _i \circ f$. It should be clear, then, that if f is measurable, so are the co-ordinate functions. The converse does not hold. Let $X = \mathbb{R},\, \mathcal{A} = \{\emptyset , X\}$, 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 $\{(t, \sin (t)) : -\pi /2 \leq t \leq \pi /2\}$?

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

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

where each $Y_{j_1j_2j_3j_4\dots j_{2n-1}j_{2n}}$ 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:

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

If f is continuous, then each $f^{-1}(Y_{j_1j_2j_3j_4\dots j_{2n-1}j_{2n}})$ is either open or closed, hence the right hand side of the above equation is a Borel set, so f is measurable.

10. Nov 16, 2006

### StatusX

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.

Last edited: Nov 16, 2006
11. Nov 16, 2011

### math4tots

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?

12. Nov 16, 2011

### micromass

Staff Emeritus
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

$$\mathcal{A}=\{A\in \mathcal{B}^m~\vert~f^{-1}(A)\in \mathcal{B}^n\}$$

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

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

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