Continuity implies Borel-measurability?

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function. Is it necessary that $f$ is Borel-measurable?

I'm considering $A=f^{-1}((a,\infty))$ where $a\in\mathbb{R}$. Is $A$ necessarily a Borel set? It looks like it should be, but I'm not sure.

score 8 · Accepted Answer · answered Oct 30 '13 at 05:05

8

By definition of continuity, and since $(a, \infty)$ is open,

$$f^{-1}((a, \infty))$$ is an open subset of $\mathbb{R}$. Every open set is Borel.

answered Oct 30 '13 at 05:05

2

Indeed, I totally forgot that definition of continuity, and was thinking about $\delta$-$\epsilon$... – Mika H. Oct 30 '13 at 05:06

score 4 · Answer 2 · answered Oct 30 '13 at 05:05

4

Hint: $(\alpha,\infty)$ is open, and $f$ is continuous.

answered Oct 30 '13 at 05:05

detnvvp

2 Answers2