Let be $ (\Omega, \mathcal{A},\mu) $ a measure space with $ \mu(\Omega)<\infty $, $ f\in L_2(\Omega, \mathcal{A},\mu) $ and $ \mathcal{G}\subseteq \mathcal{A} $ a $ \sigma $-Algebra. Further let $ g\in L_2(\Omega, \mathcal{G},\mu) $ with $$ \int_Gf\text{ d}\mu=\int_Gg\text{ d}\mu $$ for all $ G\in \mathcal{G} $.
Why is g almost everywhere unique?
My idea:
Let $ g_1,g_2\in L_2(\Omega, \mathcal{G},\mu) $ with
$$ \int_Gg_1\text{ d}\mu=\int_Gf\text{ d}\mu=\int_Gg_2\text{ d}\mu $$
for all $ G\in \mathcal{G} $. Then for all $ G\in \mathcal{G} $ it is
$$ \int_Gg_1-g_2\text{ d}\mu=\int_Gf-g_2\text{ d}\mu=0. $$
But how can I show $$ g_1-g_2=0 $$ for almost everywhere?
This would mean there is a zero set $ N\in \mathcal{G} $ such that $ g_1(x)-g_2(x)\neq 0 $ for all $ x\in N^c $.
So let's assume for all zero sets $ N\in \mathcal{G} $ there is at least one element $ x\in N^c $ with $ g_1(x)-g_2(x)\neq 0 $. Then define the new set $$ M:=\{x\in \Omega: |g_1(x)-g_2(x)|>0\}. $$
Further I can write $ M $ as $$ M=\bigcup_{n\in \mathbb{N_{\geq 1}}} \underbrace{\left\{x\in \Omega: |g_1(x)-g_2(x)|\geq \frac{1}{n}\right\}}_{=:E_n}$$
From here I get stuck because I don't see why there is an index $ n\in \mathbb{N_{\geq 1}} $ such that $ \mu(E_n)>0 $. Should that be the case then I can make a lower estimation by $$ \int_{E_n}|g_1-g_2|\text{ d}\mu\geq \frac{1}{n}\cdot \mu(E_n)>0 $$ and I would have a contradiction.