Inverse images of Borel sets by a continuous function are Borel

Question

Let $f\colon \Bbb R^p \to \Bbb R$ be a continuous function and $B\subseteq \Bbb R$ be a Borel set. Show that $f^{-1}(B)\subseteq \Bbb R^p$ is a Borel set.

Answer 1

There's an induction principle for Borel sets which you can use to prove your proposition. Stated generally:

Suppose $X$ is a topological space, and suppose $\phi(A)$ is a statement about subsets $A$ of $X$. If

1. $\phi(U)$ holds for every open set $U$ of $X$,
2. $\phi(X\setminus A)$ holds if $\phi(A)$ holds, for Borel $A\subseteq X$, and
3. $\phi(\bigcup_{i=0}^{\infty}A_i)$ holds if $\phi(A_i)$ holds for every $i\in \Bbb N$, where each $A_i\subseteq X$ is Borel,

then $\phi(A)$ holds for every Borel set $A$ of X.

So, given $X = \Bbb R$, and $f\colon \Bbb R^p\to \Bbb R$ continuous, proceed by "Borel induction", where $\phi(A)$ is the statement $$\phi(A) \iff \text{$f^{-1}(A)$ is Borel in $\Bbb R^p$}. $$

1. For $U$ open in $\Bbb R$, by continuity $f^{-1}(A)$ is open in $\Bbb R^p$, hence $f^{-1}(A)$ is Borel in $\Bbb R^p$. ($\phi(U)$ is true. We won't continue to mention $\phi$.)

2. Suppose $A\subseteq \Bbb R$ is Borel and $f^{-1}(A)$ is Borel. Then $\Bbb R\setminus A$ is Borel, and $f^{-1}(\Bbb R\setminus A) = \Bbb R^p\setminus f^{-1}(A)$ is Borel in $\Bbb R^p$, as complements of Borel sets are Borel sets.

3. Suppose $(A_i)_{i\in\Bbb N}$ is a countable family of Borel sets in $\Bbb R$ and for each $A_i$, $f^{-1}(A_i)$ is Borel in $\Bbb R^p$. Then $f^{-1}(\bigcup_i A_i) = \bigcup_i f^{-1}(A_i)$ is Borel in $\Bbb R^p$, as countable unions of Borel sets are Borel.

So we can conclude that for all Borel $A\subseteq \Bbb R$, $f^{-1}(A)$ is Borel in $\Bbb R^p$.