Suppose you have the measurable space $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$ and a function $\phi:\mathbb{R}^n\to\mathbb{R}_{\geq 0}$. I am given the set/region $$ A = \left\{x\in\mathbb{R}^n\,:\, \phi(x) \leq \epsilon\right\}. $$
- Does this set belong to the Borel sigma-algebra $\mathcal{B}(\mathbb{R}^n)$?
- If this is true only for some $\phi$ what conditions shall it satisfy?
I know how to prove that a set is a sigma algebra. Not sure about how to show that a set belongs to one.
Answer following the comments
Suppose $\phi$ is a measurable function. This means that for any set $E\in\mathcal{B}(\mathbb{R}_{\geq 0})$ we have that its pre-image under $\phi$ is an element of $\mathcal{B}(\mathbb{R}^n)$ $$ \phi^{-1}(E) = \{x\in\mathbb{R}^n\,:\, \phi(x) \in E\} \in\mathcal{B}(\mathbb{R}^n). $$ Now if we take $E = [0, \epsilon]$ then $\phi$ measurable means that $$ \phi^{-1}([0, \epsilon]) = \{x\in\mathbb{R}^n\,:\, \phi(x) \leq \epsilon\} $$ is an element of $\mathcal{B}(\mathbb{R}^n)$. But this set is exactly $A$, so $A$ is $\mathcal{B}(\mathbb{R}^n)$-measurable.