Measurable function proof Suppose $\mathcal{F}$ and $\mathcal{G}$ are two $\sigma$-algebras on a prob. space and suppose that on some set $A$, for every $G \in \mathcal{G}$ there is $F \in \mathcal{F}$ such that 
$G \cap A = F \cap A$.
The claim is that for any $\mathcal{G}$ measurable function, $X$, there exists an $\mathcal F$ measurable function $Y$ with the property that
$1_{A} X = 1_{A} Y$ (almost surely).
This is clearly true for linear combinations of indicator functions of $\mathcal{G}$ measurable sets, i.e. simple functions by the assumed property.
To prove the statement more generally, I was thinking that I can take a sequence of simple functions $X_n$ converging (almost surely) to $X$. I get a corresponding sequence of $\mathcal{F}$ measurable (simple) functions $Y_n$ with
$1_A X_n = 1_A Y_n$.
Now I want to take the limit as $n \to \infty$, but I do not see how I can argue that $Y_n$ converges everywhere (i.e. on all of $\Omega$.) If this is true, then clearly the limit function is $\mathcal{F}$ measurable and the proof is complete.
Can anyone help with the final details?
Also, if we additionally assumed $X$ was integrable, can we arrange for $Y$ to be integrable?
 A: I assume that your $X$ and $Y$ are real valued, since you are mentioning integrability?
I don't think you have the convergence almost surely of $Y_n$ in the general case. But in a Bolzano-Weierstrass way, you could extract a subsequence of the one you constructed to get a converging one. In this case this won't work (or at least I don't know how). But here is what I think you can do :
Let $Y=\limsup Y_n$. With general theorems, $Y$ is still a $\mathcal{F}$-measurable function right? 
You also have $\limsup (1_{A} Y_n) = 1_{A} Y$. This $Y$ should answer you question.
If $X$ is integrable, then you could use a monotonic sequence $X_n$ of simple function to approximate $X$. If $Y_n$ was monotonic too you could use a monotonic convergence theorem.
But in this case I don't see how you can have monotonicity of $Y_n$ when you're not on $A$...
Edit : In fact because of the equality and the way you build $Y_n$ the sequence $Y_n$ is also monotonic, and you can use monotonic convergence theorem on $Y_n$.
Edit 2 : I forgot to mention this but if $X$ is integrable to do what I suggested you can assume positiveness of $X$ at first and then with integrability work on $X^+-X^-$. 
A: A cheap trick for handling sequences that may fail to converge: use the limsup or liminf.  If you take $Y = \liminf Y_n$, then $Y$ is $\mathcal{F}$-measurable, and since $Y_n = X_n \to X$ almost everywhere on $A$, $Y=X$ a.e. on $A$.
The integrability question is a little harder.  I'll think about it.
A: If $Y$ does the job, then taking the conditional expectation with respect to $\mathcal F$, we get
$$\mathbb E(\mathbf 1_AX\mid\mathcal F)=Y\mathbb E(\mathbf 1_A\mid\mathcal F),$$
hence 
$$Y=\frac{E(X\mathbf 1_A\mid\mathcal F)}{E(\mathbf 1_A\mid\mathcal F)}[E(\mathbf 1_A\mid\mathcal F)\neq 0].$$
In particular, when $X$ is bounded or $E(\mathbf 1_A\mid\mathcal F)$ is below bounded by a positive constant, $Y$ is integrable. According to Did's answer, it may be not the case when $X$ is only assumed to be integrable.
