Even non-Lebesgue measurable function are measurable.

Question

Let $f:\mathbb R\to \mathbb R$ a non Lebesgue measurable function. Refer to the definition $f:(\mathbb R,\mathcal F)\to (\mathbb R,\mathcal G)$ measurable iff $f^{-1}(G)\in \mathcal F$ for all $G\in\mathcal G $, can we create a $\sigma -$algebra s.t. $f$ is measurable ?

Answer 1

With $(\mathbb{R}, \mathcal{G})$ a fixed measurable space, every function $f: \mathbb{R} \to \mathbb{R}$ generates a sigma-algebra $$\sigma(f) = \{f^{-1}(G): G \in \mathcal{G}\},$$ and then $f: (\mathbb{R}, \sigma(f)) \to (\mathbb{R}, \mathcal{G})$ is $(\sigma(f), \mathcal{G})$-measurable.

It follows immediately from the definition of a measurable function that if $f$ is $(\mathcal{F}, \mathcal{G})$-measurable and $\mathcal{H} \supset \mathcal{F}$, then $f$ is $(\mathcal{H}, \mathcal{G})$-measurable. Hence, every $f$ is $(\mathscr{P}(\mathbb{R}), \mathcal{G})$-measurable.