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I have a problem to the following exercise:

Suppose that $f$ and $g$ are integrable functions that are $\mathcal F $ measurable. Suppose that we have a sequence of $\sigma$-fields {$\mathcal F_{n}$} such that $\mathcal F_{n} \subset \mathcal F_{n+1}$ and for which

(1) $\int_{A}f(x)dx = \int_{A}g(x)dx$ for all A $\in \mathcal F_{n}$

Show that if $\mathcal F$ is the smallest $\sigma$-field that contains $\mathcal F_{n}$ for all n, then

$f(x) = g(x)$ for all $x$ except a set of measure zero.

There is also a Hint: Note that $\{x: f \gt g\} \in \mathcal F$. Next, it may be useful to show that equation (1) holds for all $A \in \mathcal F $. Here one may want to consider the set $\mathcal G$ of all A $\in \mathcal F$ for which we have equation (1) and then show that $\mathcal G$ is a $\sigma$-field.

This was an Exercise given many years ago on my University but without a solution. But I'm interested in a proof of this problem. Can someone please give a nice proof. Thanks in advance

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Think in the set $B=\{ A\in F : \int_{A} f = \int _{A} g \}$. It is easy to see that $B$ is a $\sigma$-algebra that contains every $F_{n}$. We deduce $B=F$.

Then, if $\{ f>g\}=\bigcup_{n\in\mathbb{N}} \{x: f(x)>g(x)+1/n\}$ has measure not zero, there exists $m\in\mathbb{N}$ such that $\{x:f(x)>g(x)+1/n\}$ has measure not zero. Since $\{x:f(x)>g(x)+1/n\}\in F$, we have $\int_{\{x:f(x)>g(x)+1/n\}}f = \int_{\{x:f(x)>g(x)+1/n\}}g\leq\int_{\{x:f(x)>g(x)+1/n\}}(f-1/n)$. Then, $measure(\{x:f(x)>g(x)+1/n\})\leq 0$. $\rightarrow \leftarrow$

We deduce $f\leq g$ a.e.. The other inequality follows in a similar way.