Uniqueness Theorem for integrable functions over a filtered probality spaces Let $(\Omega,\mathcal{F},\{\mathcal{F}_n\}_{n\in\mathbb{N}},\mu)$ be a filtered probability space and $f$ and $g$ be two integrable functions, such that for all $A\in\mathcal{F}_n$
$\int_Afd\mu=\int_Ag d\mu  -(*)$
Show that $f=g$ $a.e.$ on $\mathcal{F}$
I thought that this would be quite straightforward but am a bit stuck so any help will be greatly appreciated. My initial thoughts were to use the typical procedure and show that $\mathcal{A}:=\{A\in\mathcal{F} | (*) \text{holds}\}$ is equal to $\mathcal{F}$, but am stuck trying to show $\mathcal{A}$ is closed under complements and countable unions. I attempted to use the measurable set $\{\omega|f\geq g\}$ to help out which is standard in these types of questions and didnt get anywhere. So would really appreciate any help whether it is fixing my method or if this is wrong or there is a simpler strategy please let me know. Thanks in advance.
 A: Note that if you do not assume $f,g \in L^1$, then the conclusion is not true in general. 
Indeed it is not hard to construct a strictly positive  measurable function $f$  on $[0,1]$ which is finite everywhere yet its integral WRT Lebesge measure on any open interval is infinite. If ${\cal F}_n$ is the $\sigma$-algebra generated by $[0,2^{-n}),[2^{-n},2*2^{-n}),\dots,[(2^n-1)2^{-n},1]$, then ${\cal F}$ is the Borel-$\sigma$ algerba. Letting  $g=2f$, then the condition is satisfied yet, $g>f$ a.s. ($f$ was strictly positive).   
Therefore we'll assume $f,g \in L^1(P)$.  
To prove, you can use a monotone class argument. Let ${\cal M}=\{A \in {\cal F}: \int_A f  dP = \int_A g dP\}$. Then:


*

*By monotone convergence and fact that $f$ and $g$ in $L^1$, we have that ${\cal M}$ is a monotone class. That is if $(A_j:j=1,\dots)$ is an increasing, resp. decreasing, sequence of elements in ${\cal M}$ then their union, resp. intersection, is in ${\cal M}$. Note: we only needed the integrability to show that intersection of decreasing seqeucnes is in ${\cal M}$.  

*${\cal M}$ contains the algebra ${\cal A}= \cup_{n=1}^\infty {\cal F}_n$ (algebra is a nonempty collection of sets closed under (finite) unions and complement). 
By Dynkin's Monotone Class Theorem, ${\cal M}$ contains the $\sigma$-algebra generated by ${\cal A}$, ${\cal F}$. That is ${\cal M} \supseteq {\cal F}$.  
As a result, $\int _A f dP  = \int_A gdP$. for all $A\in {\cal F}$, in particular on the set $A=\{f>g\}$. Therefore, $f\le g$ a.s. Repeating the argument shows that $g\le f$, hence $f=g$ a.s. 
