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Let ($X, \tau_X$) and ($Y, \tau_Y$) be topological spaces. Let $\mathcal{B}(X)=\sigma(\tau_X)$ and $\mathcal{B}(Y)=\sigma(\tau_Y)$ be the corresponding Borel sigma algebras. Do we have that if $\mathcal{B} \subseteq \mathcal{P}(X)$ is a sigma algebra such that every continuous function $f:X\rightarrow Y$ is $\mathcal{B}/\mathcal{B}(Y)$ measurable, then $\mathcal{B}(X) \subseteq \mathcal{B}$?

If not, does it hold for the reals? i.e. when $X=\mathbb{R}^m,Y=\mathbb{R}^n$ for some $m,n$.

My attempt:

Let $f^{-1}(\tau_Y) \equiv \{f^{-1}(E)|E\in\tau_Y\}$. Then $f$ continuous iff $f^{-1}(\tau_Y) \subseteq \tau_X$, and $f$ Borel measurable iff $f^{-1}(\sigma(\tau_Y))=\sigma(f^{-1}(\tau_Y)) \subseteq \sigma(\tau_X)$. Assume that $f$ is $\mathcal{B}/\mathcal{B}(Y)$ measurable, then $f^{-1}(\sigma(\tau_Y))=\sigma(f^{-1}(\tau_Y)) \subseteq \mathcal{B}$. We now want to show that $\sigma(\tau_X) \subseteq \mathcal{B}$, and I don't know how to proceed.

For the real case, I found the following answer, but it doesn't seem to make sense for me, could someone please elaborate further?

Prove that Borel sigma field on R(d) is the smallest sigma-field that makes all continuous functions f:R(d)->R measurable.

uuddlrlrb4
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1 Answers1

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Part 1: In the general case, you cannot prove that $\mathcal{B}(X) \subseteq \mathcal{B}$.

Here is a counter-example.

Let $X=\{a,b\}$, $\tau_X$ be the discrete topology (that is $\tau_X=\mathcal{P}(X)$) and then $\mathcal{B}(X)=\mathcal{P}(X)$.

Let $Y=\{c\}$, $\tau_Y =\{\emptyset, Y\}$ and so $\mathcal{B}(Y)= \{\emptyset, Y\}$.

All functions $f:X\rightarrow Y$ are continuous and constant functions.

Let $\mathcal{B}=\{\emptyset, X\}$ be a $\sigma$-algebra in $X$.

All (continuous) functions $f:X\rightarrow Y$ are $\mathcal{B}/\mathcal{B}(Y)$ measurable, but $\mathcal{B}\subsetneq \mathcal{B}(X)$.

Part 2 For the case of $X=\mathbb{R}^m,Y=\mathbb{R}^n$, let us prove the following more general result.

Let ($X, d$) be a metric space, $\tau_X$ the topology induced by the metric and let $\mathcal{B}(X)=\sigma(\tau_X)$ the Borel sigma algebra of $X$. Let us consider $\mathbb{R}^n$ with its usual topology and its usual Borel sigma algebra. If $\mathcal{B} \subseteq \mathcal{P}(X)$ is a sigma algebra such that every continuous function $f:X\rightarrow \mathbb{R}^n$ is $\mathcal{B}/\mathcal{B}(\mathbb{R}^n)$ measurable, then $\mathcal{B}(X) \subseteq \mathcal{B}$.

Proof: Let us prove that for any $F$ closed set of $X$, $F \in \mathcal{B}$. It will follow immediately that $\mathcal{B}(X) \subseteq \mathcal{B}$.

Given any $F$ closed set of $X$, define the function $f:X\rightarrow \mathbb{R}^n$ such that for all $x \in X$, $f(x) = (d(x,F),0,\dots,0) \in \mathbb{R}^n$.

It is easy to see that $f$ is continuous, so $f$ is $\mathcal{B}/\mathcal{B}(\mathbb{R}^n)$ measurable. Since $\{(0,\dots,0)\}$ is a Borel set in $\mathbb{R}^n$, we have that $$F=f^{-1}(\{(0,\dots,0)\}) \in \mathcal{B}$$ So we have proved that for any $F$ closed set of $X$, $F \in \mathcal{B}$, and so $\mathcal{B}(X) \subseteq \mathcal{B}$.

Ramiro
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  • The proof is fantastic! Thank you. – uuddlrlrb4 Dec 17 '20 at 21:02
  • Can I ask how to prove that $\mathcal{B}\subset \mathcal{B}(X)$? – Hermi Sep 23 '21 at 19:37
  • @Bob , In the general case, it may not be true that $\mathcal{B}\subset \mathcal{B}(X)$, even if $X=\Bbb R^m$ and $Y=\Bbb R^n$.

    In fact, let ($X, \tau_X$) and ($Y, \tau_Y$) be topological spaces, such $\mathcal{B}(X) \subsetneq \mathcal{P}(X)$. Take $\mathcal{B} = \mathcal{P}(X)$. Then all functions from $X$ to $Y$ are $\mathcal{B}/\mathcal{B}(Y)$ measurable. So, all continuous functions from $X$ to $Y$ are $\mathcal{B}/\mathcal{B}(Y)$ measurable. But, it is not true that $\mathcal{B}= \mathcal{P}(X) \subseteq \mathcal{B}(X)$.

     – Ramiro Sep 24 '21 at 04:52