Let $f$ be defined on a measurable set $E$. Show that $f$ is measurable if and only if for every Borel set $A$, $f^{-1}(A)$ is measurable.
Hint: The collection of sets $A$ that have the property that $f^{-1}(A)$ is measurable is a $\sigma$-algebra.
I want to know if the following proof is correct:
$\mathcal{A}$ is the collection in question, $\mathcal{B}$ is the set of Borel sets, and $\mathcal{M}$ is the set of measurable sets.
- Let $c \in \mathbb{R}$ arbitrarily chosen. Then $(c, \infty)$ is open so it is in $\mathcal{B}$ and $f^{-1}((c,\infty)) \in \mathcal{M}$.
- Also since $f^{-1}(\{\infty\}) \in \mathcal{M}$ so $\{\infty\} \in \mathcal{A}$.
- By both of the above this shows that $(c, \infty] \in \mathcal{A}$.
- Let $a,b \in \mathbb{R}$. Since $f$ is measurable $f^{-1}([-\infty,b))$ and $f^{-1}((a,\infty])$ are both in $\mathcal{M}$ so $[-\infty,b)$, $(a,\infty]$ are both in $\mathcal{A}$ and $f^{-1}([-\infty,b)) \cap f^{-1}((a,\infty])=f^{-1}((a,b)) \in \mathcal{M}$. So $(a,b)\in A$. This shows that all open intervals are in $\mathcal{A}$.
- For $A \in \mathcal{B}$, $f^{-1}(A) \in \mathcal{M}$ so $A \in \mathcal{A}$.
- $f^{-1}(\emptyset) \in \mathcal{M}$, so $\emptyset \in \mathcal{A}$.
I've gotten $\LaTeX$ fatigue, but I think all that is left to show is that countable unions and complements are in $\mathcal{A}$ to complete the definition of $\sigma$-algebra.