# Continuous function $f:\mathbb R\to \mathbb R$ such that $f(B)$ is not measurable for a Borel set $B$.

1) I saw in a book that there are $\mathcal C^1(\mathbb R)$ function $f:\mathbb R\to \mathbb R$ such that $f(B)$ is not measurable for $B$ a Borel set. I don't really find such an example. Any idea ?

2) If $f$ is continuous, does $f(O)$ measurable for $O$ open or closed or that is even not correct ?

• To answer Part (2) partially, an open subset of $\mathbb{R}$ is a union of countably many disjoint open intervals. It is easily seen that $f(I)$ is an interval for any interval $I$. Therefore, $f(U)$ is a union of countably many intervals if $U$ is an open subset of $\mathbb{R}$. Thus, $f(U)$ is measurable for an open set $U$. – Batominovski Aug 21 '18 at 22:12
• To answer the other question of Part (2), write $C_n:=C\cap [-n,+n]$ for each positive integer $n$, where $C$ is an arbitrary closed subset of $\mathbb{R}$. We know that $C_n$ is compact, so $f(C_n)$ is also a compact subset of $\mathbb{R}$, whence $f(C_n)$ is measurable. Thus, $$f(C)=\bigcup_{n=1}^\infty\,f(C_n)$$ is a countable of measurable sets, so $f(C)$ is also measurable. This leaves Part (1) to be answered. – Batominovski Aug 21 '18 at 22:40
• And to answer Part (1), I think you can use the result from this thread: math.stackexchange.com/questions/415759/…. Although the function $f$ in that thread is defined only on $[0,1]$, you can extend it smoothly outside $[0,1]$ to the whole $\mathbb{R}$. – Batominovski Aug 21 '18 at 22:50