I'm having a bit of trouble understanding the connection between measurable maps and continuous functions. Namely, I know the following: Assume $f:(\Omega_{1}, \mathcal{B}_{1}) \rightarrow (\Omega_{2}, \mathcal{B}_{2})$

  1. $E \subset \Omega_{2} \quad \mbox{open} \Rightarrow f^{-1}(E) \quad \mbox{open} $ by continuity
  2. $f$ is a measurable map $\Rightarrow$ $( A \in \mathcal{B}_{2} \Rightarrow f^{-1}(A) \in \mathcal{B} _{1})$

What I'm uncertain of is the connection between the two. Say, for instance, that $\Omega_{2} = \mathbb{R}$ and $\mathcal{B}_{2} = \mathcal{B}$, the $\sigma$-algebra generated by the collection of open intervals of $\mathbb{R}$ and $\Omega_{2} = \mathbb{R}^{d}$ and $\mathcal{B}_{2} = \mathcal{B}^{d}$, the same for $\mathbb{R}^{d}$. Does knowledge of how $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ were generated help? And if so, how does it lead the ability to talk about the whole of $\mathcal{B}_{1}$?

On the other hand, I've heard it mentioned that I can use compliments and limits of sets to somehow say that $f^{-1}(A)$ is measurable for all intervals $A \subset \mathbb{R}$, regardless of whether it's open, closed, or half-open, just by knowing that 1. from above. How is that so?

  • $\begingroup$ A hint would be: $f^{-1}$ preserves complements, countable unions, countable intersections. $\endgroup$ – GEdgar Aug 28 '11 at 21:57
  • $\begingroup$ As in $f^{-1}(E^{C}) = f^{-1}(E)^{C}$, $f^{-1}(\bigcup_{i=1}^{\infty}A_{i}) = \bigcup_{i=1}^{\infty}f^{-1}(A_{i})$, and so on? $\endgroup$ – duckworthd Aug 28 '11 at 22:09
  • $\begingroup$ @duckworthd : yes, that is what GEdgar meant. $\endgroup$ – Patrick Da Silva Aug 28 '11 at 22:41

If $\Omega_1$ and $\Omega_2$ are topological spaces and $\mathcal{B}_1$ and $\mathcal{B}_2$ are their Borel sigma-algebras, then every continuous function is measurable (but, in general, not every measurable function is continuous).

Proof. Let $u:\Omega_1\to\Omega_2$ be a continuous function. That $u$ is $\mathcal{B}_1/\mathcal{B}_2$ measurable is equivalent to the assertion that $\mathcal{C}=\mathcal{B}_2$, where $$ \mathcal{C}=\{B\in\mathcal{B}_2\mid u^{-1}(B)\in\mathcal{B}_1\}. $$ Thus, one has to show that $\mathcal{C}=\mathcal{B}_2$. This is done in two steps.

  • First, $u$ is continuous hence one already knows that $\mathcal{T}_2\subseteq\mathcal{C}$, where $\mathcal{T}_2$ is the topology of $\Omega_2$ (in other words, $\mathcal{T}_2$ is the collection of its open sets). Since $\mathcal{C}\subseteq\mathcal{B}_2$ by definition, one gets the double inclusion $$ \mathcal{T}_2\subseteq\mathcal{C}\subseteq\mathcal{B}_2. $$ By definition of Borel sigma-algebras, $\mathcal{B}_2=\sigma(\mathcal{T}_2)$, that is, $\mathcal{B}_2$ is the sigma-algebra generated by $\mathcal{T}_2$ (or equivalently, the smallest sigma-algebra which contains $\mathcal{T}_2$). Hence, the double inclusion above implies $$\mathcal{B}_2=\sigma(\mathcal{T}_2)\subseteq\sigma(\mathcal{C})\subseteq\sigma(\mathcal{B}_2)=\mathcal{B}_2, $$ that is, $\sigma(\mathcal{C})=\mathcal{B}_2$.
  • Second, and this will be enough to conclude, one wants to show that $\mathcal{C}=\sigma(\mathcal{C})$, or equivalently, that $\mathcal{C}$ is a sigma-algebra. It happens that $\mathcal{C}$ fulfills all the axioms defining a sigma-algebra and that it is easy to show, so I will be lazy and leave you this part of the proof.
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  • $\begingroup$ Why is it that we know that $\mathcal{T}_{2} \subseteq \mathcal{C}$? $\endgroup$ – duckworthd Aug 28 '11 at 23:05
  • $\begingroup$ Let B in T_2. This means B is open, hence u^(-1)(B) is open (since u is continuous). Every open subset of Omega_1 is in B_1, hence u^(-1)(B) is in B_1. This proves B is in C. $\endgroup$ – Did Aug 29 '11 at 0:50
  • $\begingroup$ Thanks! The argument makes perfect sense. My last question is how we know that "every open subset of $\Omega_{1}$ is in $\mathcal{B}_{1}$". Is this just by definition of $\mathcal{B}_{1}$, being that it's generated by the collection of all open subsets of $\Omega_{1}$ (thus implying that we could generate other $\sigma$-algebras that do not contain all open subsets of $\Omega_{1}$)? $\endgroup$ – duckworthd Aug 29 '11 at 8:32
  • $\begingroup$ Yes, one usually defines the Borel sigma-algebra B on a topological space (Omega,T) as the smallest sigma-algebra containing the topology T. Then every open set is in B by definition. If (Omega,T) is R^d with the usual topology, B is also the smallest sigma-algebra containing the collection I of (products of) intervals. This follows from the fact (which one should prove) that sigma(I) contains T. Other sigma-algebras on R^d than the Borel sigma-algebra may not be related to T at all. $\endgroup$ – Did Aug 29 '11 at 11:30

(While the answer is not very useful to the OP, I am not deleting it since I do believe that someone will find it useful, and perhaps even the OP someday may find it useful.)

[This answer is using some amount of choice, in particular a countable union of countable sets is countable assumed in here]

The Borel measurable functions are constructed as iterated pointwise limits of continuous functions.

For this we need to elaborate a little bit on the construction of $\mathcal B$, that is the Borel $\sigma$-algebra.

  • $\Sigma^0_0=\Pi^0_0=$ a countable basis of open intervals (or balls),
  • For $\alpha$ a countable ordinal denote: $$ \Sigma^0_\alpha=\{\bigcup_{i\in\mathbb N} A_i\mid A_i\in\bigcup_{\beta<\alpha}\Pi^0_\beta\},\quad \Pi^0_\alpha = \{X\setminus A\mid A\in\Sigma^0_\alpha\},\quad \Delta^0_\alpha=\Sigma^0_\alpha\cap\Pi^0_\alpha $$

The $\Sigma^0_\alpha$ are closed under countable unions, the $\Pi^0_\alpha$ under countable intersections and $\Delta^0_\alpha$ are algebras (but not $\sigma$-algebra in general), and the whole hierarchy is closed under inclusion, that is: $\beta<\alpha$ then $\Sigma^0_\beta\subseteq\Delta^0_\alpha\subseteq\Sigma^0_\alpha$, and similarly for $\Pi^0_\alpha$.

We have that: $$\mathcal B=\bigcup_{\alpha<\aleph_1}\large\Delta^0_\alpha$$ (That is a union of all the countable levels of the construction). In the case of $\mathbb R^n$ (and in any case that the space has similar properties to these spaces) we cannot have this union any shorter. That is, on every stage there is a new set added to the construction.

We also have that $|\mathcal B|=|\mathbb R|=2^{\aleph_0}$. Which means that "most" subsets of $\mathbb R^n$ are not Borel sets. So it is not hard to show (or to believe if anything) that most mappings are not Borel measurable.

Now identify the continuous functions with $\Sigma^0_1$ sets. That is the preimage of an open set is an open set, this implies also that $\Pi^0_1$ sets are preserved (preimage of closed is closed). Furthermore, it is quite simple to show that if a function is continuous then the entire hierarchy is preserved. In particular, we only need to see what is the preimage of open sets, and from there the hierarchy will be preserved.

Now take the pointwise limit of continuous functions. This may not be continuous, but an open set will be at most $\Sigma^0_2$ or $\Pi^0_2$. We can iterate this, each time taking pointwise limits of anything previously defined. Since a pointwise limit of measurable functions is measurable we have that the union of $\aleph_1$ many steps of this construction yields only measurable functions.

On the other hand we can measure how high the basis for the topology is being transferred into the hierarchy of $\mathcal B_1$, and we can prove that applying this many pointwise limits iterations will give you this function.

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  • $\begingroup$ Thank you for your detailed response, but I must admit that it is quite over my head. Over the course of my studies I have yet to encounter this hierarchy and its meaning is lost to me. $\endgroup$ – duckworthd Aug 28 '11 at 22:21
  • $\begingroup$ @duckworld: Most people do not encounter this hierarchy in details. In measure theory one usually just refers to it as "$\sigma$-algebra generated by open sets". I wrote this answer, knowingly it may be over your head, to try and open a window to a new realm; and perhaps give a little bit of general knowledge to you and others whom are interested. $\endgroup$ – Asaf Karagila Aug 28 '11 at 22:25
  • $\begingroup$ Maybe it would be worth mentioning that you're talking about the Baire classes/Baire hierarchy of functions here. Also, the Lebesgue-Hausdorff-Banach theorem to the end that every real-valued Borel measurable function is in some Baire class should probably be mentioned explicitly. $\endgroup$ – t.b. Aug 29 '11 at 2:10
  • $\begingroup$ @Theo, that is indeed worth mentioning. I just didn't know it was called that way. :-) I'll learn from last time and edit only after drinking the coffee. $\endgroup$ – Asaf Karagila Aug 29 '11 at 6:56
  • $\begingroup$ @Theo: Also, Mr. Libraryman, did you see that I have added references to that answer about amorphous/Dedekind finite sets? $\endgroup$ – Asaf Karagila Aug 29 '11 at 6:57

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