In his probability book Bauer proves the following
Theorem. Let $X$ be an extended real-valued integrable random variable on a probability space $(\Omega,\mathcal{A},P)$, and let $\mathcal{C}$ be a sub-$\sigma$-algebra of $\mathcal{A}$. Then there exist an extended real-valued integrable random variable $X_0$ on $(\Omega,\mathcal{A},P)$ which is $\mathcal{C}$ measurable, unique almost surely, and satisifies
$$\int_CX_0dP=\int_C XdP \hspace{0.5cm} \text{for all } C \in \mathcal{C}$$
He begins by proving the case where $X\geq 0$ (where $X$ and $X_0$ take values in $[0,\infty]$ but are not necessarily integrable) using the Radon-Nikodym theorem. He then writes:
If $X$ is integrable but of arbitrary sign, we decompose it into its positive part $X^+$ and negative part $X^-$. From what has just been proved follows the existence of $\mathcal{C}$ measurable extended real-valued random variables $X_0\geq0$ and $Y_0\geq0$ such that
$$\int_CX_0dP=\int_C XdP \hspace{0.5cm}\text{and}\hspace{0.5cm} \int_CY_0dP=\int_C XdP$$
for all $C \in \mathcal{C}$. Setting $C=\Omega$ we infer that, because $X^+$ and $X^-$ are integrable, $X_0$ and $Y_0$ are also integrable, and therewith almost surely finite. WLOG $X_0$ and $Y_0$ can therefore be assumed to be real-valued. But then $Z_0:=X_0-Y_0$ is an integrable solution to our problem.
Question: If am not mistaken a measurable alteration of $X_0$ and $Y_0$ on a $P$-null set (e.g. setting $X_0=Y_0=0$ on such set) dosen't ensure that they will remain $\mathcal{C}$ measurable.
However I think this issue can be avoided here: If we define $P_0$ as the restriction of $P$ to $\mathcal{C}$ then we have
$$\int X_0dP=\int X_0dP_0 \hspace{0.5cm}\text{and}\hspace{0.5cm} \int Y_0dP=\int Y_0dP_0$$
and so $X_0$ and $Y_0$ are actually $P_0$-a.s. finite. Hence we can alter them on a $P_0$-null set (which is also $P$-null) such that they become finite everywhere and remain $\mathcal{C}$ measurable.
In fact, when looking at the proof, it seems like the conditional expectation is actually unique $P_0$-a.s.
Is this correct?