# Continuous functions and Borel sets

1. Is every Borel set a preimage of some Borel set under some continuous and bounded function? That is, for a given topological space $$X$$ does it hold, that $$\forall_{\ B \ Borel \subset X} \ \exists_{\ C \ Borel \subset \mathbb{R}^n, \ \ f \in C_0(X, \ \mathbb{R}^n)} \ \ B=f^{-1}(C)$$?
2. Is Borel $$\sigma$$-field on $$X$$ the smallest $$\sigma$$-field that makes all continuous, compactly supported functions $$\in C_c(X, \ \mathbb{R}^n)$$ measurable (given the usual Borel $$\sigma$$-algebra on $$\mathbb{R}^n$$)?

If it's necessary, you can assume $$X = \mathbb{R}^m$$ or a metric space (my main interests), but I'm also interested in more genral cases.

Here is the answer to the 2. point (it's affirmative): lets take any bounded open ball $$B$$ in $$\mathbb{R}^n$$ and define $$f:\mathbb{R}^n \rightarrow \mathbb{R}$$, $$f(x) = dist(x, \ \partial B)$$ for $$x \in B$$ and $$f(x) = 0$$ for $$x \in \mathbb{R}^n / B$$. It's continuous and compactly supported.
Now lets define $$F: \mathbb{R}^n \rightarrow \mathbb{R}^m$$ componentwise so that $$F_i = f$$ for $$i \in \{1, 2, ..., m\}$$. Then $$F^{-1}((0, \infty]^m) = B$$.
Now the smallest $$\sigma$$-algebra containing all the bounded open balls is the Borel $$\sigma$$-algebra (since $$\mathbb{R}^n$$ is Lindelof and we have the countable union property of $$\sigma$$-algebras). I would still like to know the answer to the 1. question though.

• 1. is false for $m=2$, $n=1$, see my edit. Not sure about the general case $m > n$. @Jkbb
Feb 14 at 11:52

I'll prove these for $$X=\mathbb R^m$$, $$m=n$$.

1. Yes. Let $$f:X \to (0, 1)^m$$ be continuous with continuous inverse; we can take e.g. $$f(x)_i = 1/(1+e^{-x_i}).$$ Then for any Borel set $$B \subseteq X$$, the image $$C = f(B) = (f^{-1})^{-1}(B)$$ is Borel, and $$B = f^{-1}(C)$$.

2. Yes. Let $$A_n = [-n, n]^m$$, and choose continuous, compactly supported $$f_n:X \to \mathbb R^m$$ such that the image $$I_n = f_n(A_n)$$ is Borel, and the restriction $$f_n\mid_{A_n}:A_n \to I_n$$ has continuous inverse; we can take e.g. $$f_n(x) = f(x)g_n(x), \qquad g_n(x) = \max(0, \min(1, n + 1 - \max( \lvert x_i \rvert))).$$ Then for any Borel set $$B \subseteq A_n$$, we have $$B = g_n^{-1}(\{1\}) \cap f_n^{-1}(f_n(B))$$, so $$B \in \sigma(C_c(X, \mathbb R^m))$$. We deduce this $$\sigma$$-algebra contains all bounded Borel sets, and so by countable union all Borel sets.

Edit: for $$m < n$$, the same proofs hold, e.g. by setting the extra coordinates of $$f$$ to zero. For $$m > n$$, result 2. holds as in Jkbb's answer. For result 1., in the case $$m=2$$, $$n=1$$ the answer is no.

Let $$D \subset \mathbb R^2$$ be the open unit disc, and define the Borel set $$B = D \cup \{(1, 0)\}$$. Suppose we have continuous $$f$$ and Borel $$C$$ satisfying $$f^{-1}(C)=B$$.

By translation and scaling, we may assume $$f(1, 0) = 0$$, and $$f > 0$$ somewhere in $$\partial B$$. Applying the intermediate value theorem to a path along $$\partial B$$, we deduce that $$0 \in C$$, and $$C \cap (0,\epsilon) = \emptyset$$ for some $$\epsilon > 0$$. We thus have that $$f \le 0$$ on $$B$$, and $$f > 0$$ on $$\partial B \setminus B$$, contradicting the continuity of $$f$$.

• Thanks, that's a very good answer, but I would still like to prove it in a more general setting, so at least for functions $\mathbb{R}^m \rightarrow \mathbb{R}^n$ where $m \neq n$.
– Jkbb
Feb 13 at 22:43
• Yes, the case $n > m$ is easy, but for $n < m$ this seems like a harder problem. @Jkbb
• Please check my if reasoning for 2. is corrrect: for every compact set $K$ we can find a bounded open set $U, \ K \subset U$ and a function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ such that $supp \ f \subset U$, $f=1$ on $K$ and $f<1$ on $\mathbb{R}^n / K$. We combine these functions componentwise to get $F: \rightarrow \mathbb{R}^n \rightarrow \mathbb{R}^m$. Then $F^{-1}(\{1\}) = K$.
• Apply the intermediate value theorem to a path along $\partial B$. @Jkbb