Consider the measurable spaces $\left( E,\mathcal{E} \right)$ and $\left(\mathbb{R},\mathcal{B}\left(\mathbb{R}\right)\right)$.
Prove that $f:E\rightarrow \mathbb{R}$ is $\mathcal{E}$-measurable $\iff$ $f^{+}, f^{-}$ are $\mathcal{E}-measurable.
As usual, $f^{+}= \left(0 \vee f \right)$, and $f^{-}=-\left( 0 \wedge f\right)$.
My attempt
"$\Longrightarrow$" Assume $f$ is $\mathcal{E}$-measurable, i.e., $\forall \; b \in \mathbb{R}$, $f^{-1}\big( \left( -\infty, b \right) \big) \in \mathcal{E}$. Then, for any $b \in \mathbb{R_+}$
${(f^{+})}^{-1}\big( \left( -\infty, b \right) \big) = {(f^{+})}^{-1}\big( \left[ 0 , b \right) \big) = f^{-1}\big( \left[0, b \right) \big) \in \mathcal{E}$, where the belonging follows from the assumption. Similarly,
${(f^{-})}^{-1}\big( \left( -\infty, b \right) \big) = {(f^{-})}^{-1}\big( \left[ 0 , b \right) \big) = f^{-1}\big( \left(-b, 0 \right] \big) \in \mathcal{E}$, where the belonging follows from the assumption.
"$\Longleftarrow$" Assume $f^{+}, f^{-}$ are $\mathcal{E}$-measurable.
Then, when $b \in \mathbb{R_+}$
$f^{-1}\big( \left( -\infty, b \right) \big)=\left({f^{-}}\right)^{-1}\big( \left[ 0, \infty \right) \big) \cup \left({f^{+}}\right)^{-1}\big( \left( 0, b \right) \big) \in \mathcal{E}$
and when $b<0$
$f^{-1}\big( \left( -\infty, b \right) \big)=\left({f^{-}}\right)^{-1}\big( \left(b, \infty \right) \big) \in \mathcal{E}$
In my opinion, this is correct. But I have seen a more elaborated proof, involving monotone sequences of functions that converge to $f$. So, if the author of that proof chose a more elaborated one, I am wondering if there are mistakes in my proof.
Any help would be much appreciated!