Let $(\Omega,\mathcal A)$ be a measurable space, $\mathcal F$ a sub-$\sigma$-algebra of $\mathcal A$, and $A\in \mathcal A$. Show that $f:\Omega \to \mathbb R$ is $\sigma(A,\mathcal F)$-measurable if and only if $f=1_A f_1 + 1_{A^c} f_2$ for some $\mathcal F$-measurable maps $f_1,f_2 \to \mathbb R$.
My attempt:
Sufficiency is clear so lets show necessity. We first check that $\sigma(A,\mathcal F)=\Big\{(A\cap B_1) \cup (A^c\cap B_2) : B_1,B_2\in\mathcal F \Big\}$. Suppose first that $f$ is an indicator function, i.e. $f=1_C$ for some $C\in \sigma(A,\mathcal F)$. Then $f=1_A 1_{B_1} + 1_{A^c} 1_{B_2}$ for some $B_1,B_2\in\mathcal F$ so the claim holds in this case. By linearity the claim also holds if $f$ is simple, i.e. $f=\sum_{i=1}^n c_i 1_{C_i}$ with $c_i\in\mathbb R$ and $C_i\in \sigma(A,\mathcal F)$ for each $i$.
Now let $f$ be arbitrary and choose a sequence $(f_n)=(1_A f_{1,n} + 1_{A^c} f_{2,n})$ of simple functions converging pointwise to $f$. Define
$$f_1(\omega):=\begin{cases} \limsup f_{1,n}(\omega) & \text{if } \limsup f_{1,n}(\omega) < \infty \\ 0 & \text{otherwise} \end{cases} $$
$$f_2(\omega):=\begin{cases} \limsup f_{2,n}(\omega) & \text{if } \limsup f_{2,n}(\omega) < \infty \\ 0 & \text{otherwise} \end{cases} $$
Then $f_1,f_2$ are real-valued, $\mathcal F$-measurable, and $f=1_A f_1 + 1_{A^c} f_2$.
Is this correct?