$\mathcal{B}(\mathbb{R})$-measurable vs. $\mathcal{B}(\overline{\mathbb{R}})$-measurable

Question

Let $(E,\mathcal{E})$ a measurable space and $f \colon E \to \mathbb{R}$ a function. Is it right, that $f$ is $\mathcal{E}$-$\mathcal{B}(\mathbb{R})$-measurable iff $i \circ f$ is $\mathcal{E}$-$\mathcal{B}(\overline{\mathbb{R}})$-measurable, where $i$ is the inclusion $\mathbb{R} \to \overline{\mathbb{R}}$? ($\mathcal{B}(\mathbb{R}) $ is the Borel $\sigma$-Algebra)

Answer 1

Yes, it is true.

If $f$ is $\mathcal{E}$-$\mathcal{B}(\mathbb{R})$-measurable, then, for all $E \in \mathcal{B}(\overline{\mathbb{R}})$, $$ (i \circ f)^{-1}(E) =f^{-1}(i^{-1}(E))= f^{-1}(E \cap \Bbb R) \in \mathcal{E}$$ (because $E \cap \Bbb R \in \mathcal{B}(\mathbb{R})$). So, $i \circ f$ is $\mathcal{E}$-$\mathcal{B}(\overline{\mathbb{R}})$-measurable

If $i \circ f$ is $\mathcal{E}$-$\mathcal{B}(\overline{\mathbb{R}})$-measurable, then, for all $A \in \mathcal{B}(\mathbb{R})$, we have that $A \in \mathcal{B}(\overline{\mathbb{R}})$ and $i^{-1}(A) =A$. So $$ f^{-1}(A) = f^{-1}(i^{-1}(A))= (i \circ f)^{-1}(A)\in \mathcal{E}$$ So, $f$ is $\mathcal{E}$-$\mathcal{B}(\mathbb{R})$-measurable