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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.

hallo007
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  • I could write $$ E={x\in \Omega: g_1(x)-g_2(x)\neq 0}\=\bigcup_{n \in \mathbb{N_{geq 1}}} \left{x\in \Omega: g_1(x)-g_2(x)\geq \frac{1}{n}\right} \cup \left{x\in \Omega: g_1(x)-g_2(x)\leq \frac{1}{n}\right} $$ But then I have still the same question. Why there is an index $ n\in \mathbb{N_{\geq 1}} $ with $ \mu\left(\left{x\in \Omega: g_1(x)-g_2(x)\geq \frac{1}{n}\right} \cup \left{x\in \Omega: g_1(x)-g_2(x)\leq \frac{1}{n}\right}\right)>0 $? – hallo007 Dec 22 '23 at 11:05
  • Then I would also get $ \mu(E)=0 $. – hallo007 Dec 22 '23 at 11:18
  • Apologies I should have wrote $>$ rather than $\geq$, I will amend my comment – Derek H. Dec 22 '23 at 11:21
  • Try writing $E={x\in\Omega:g_1(x)-g_2(x)\ne0}=\bigcup\limits_{n\in\mathbb{N}}$ ${x\in\Omega:g_1(x)-g_2(x)$ $>\frac{1}{n}}\cup{x\in\Omega:g_1(x)-g_2(x)>\frac{1}{n}}$. – Derek H. Dec 22 '23 at 11:22
  • If there is no set $E_n$ of positive measure with $E_n={x\in\Omega:g_1(x)-g_2(x)>\frac{1}{n}}$ and no set $G_n$ of positive measure with $G_n={x\in\Omega:g_2(x)-g_1(x)>\frac{1}{n}}$ then we obtain $\mu(E)=0$ which shows that $g_1=g_2$ almost everywhere on $E$. This is the same as saying if we assume that $g_1=g_2$ is false on some set of positive measure (false on $E$) then at least one of the sets $E_n$ or $G_n$ must have positive measure. What happens when we take the integral over such a set? – Derek H. Dec 22 '23 at 11:26
  • And just a heads up it is standard to write $\mathbb{N}={1,2,3,\dots}$ so that you don't have to specify that you mean natural numbers greater than 0; 0 is usually not considered a natural number unless we write, e.g., $\mathbb{N}_0$ – Derek H. Dec 22 '23 at 11:29
  • I could do a lower estimation by $$ \int_{E_n}(g_1-g_2)\text{ d}\mu\geq \frac{1}{n}\cdot \mu(E_n)>0 $$ and $$ \int_{E_n}(g_2-g_1)\text{ d}\mu\geq \frac{1}{n}\cdot \mu(G_n)>0 $$ which is a contradiction to $ \int_{G}(g_1-g_2)\text{ d}\mu=0 $ for all $ G\in \mathcal{G} $. – hallo007 Dec 22 '23 at 11:30
  • Correct, and this contradicts your original assumption. So, if your original assumption holds then we must have $\mu(E_n)=\mu(G_n)=0$ for all $n\in\mathbb{N}$, i.e., $\mu(E)=0$. – Derek H. Dec 22 '23 at 11:40
  • Also, you stated that $g_1,g_2\in L^2(\Omega,\mathcal{F},\mu)$. Do you know the inclusion result which shows that $g_1,g_2\in L^1(\Omega,\mathcal{F},\mu)$, given that $\mu(\Omega)<\infty$? Or are you supposed to assume that the integrals of $f,g$ exist as they are stated? – Derek H. Dec 22 '23 at 11:43

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