Completing the proof of a theorem My question is about the proof of the Theorem 1.3 which is on the page 9 of the book "Topics in Real Analysis", which is available electronically through the previous link.
I want to demonstrate that theorem in the case $\mu (X)=\infty$. In that book there's the following tip:

To extend our result to the general case observe that the finite case implies $\mu (A\cap X_j)=\tilde \mu (A\cap X_j)$ (just restrict $\mu,\tilde\mu$ to $X_j$). Hence
$$\mu (A)=\lim_{j\to\infty}\mu (A\cap X_j)=\lim_{j\to \infty}\tilde \mu(A\cap X_j)=\tilde \mu (A)$$

Below is my attempt to use that tip to prove that theorem in the case $\mu (X)=\infty$.
Let $j\in\mathbb{N}$ be any element and define $S_j:=\big\{E\cap X_j:E\in S\big\}$. It's trivial to verify that $S_j$ is a $\pi$-system (because $S$ is by hypothesis a $\pi$-system). Besides, since $X_j\in S$, then $S_j\subseteq S$ which implies that $\Sigma (S_j)\subseteq\Sigma (S)=\Sigma $ in which $\Sigma (S_j)$ means the $\sigma$-algebra generated by $S_j$.
Let $\mu _{S_j}$ and $\tilde \mu _{S_j}$, respectively, be the restrictions of $\mu$ and $\tilde \mu$ to $\Sigma(S_j)$. It's easy to see that $(X_j,\Sigma (S_j),\mu_{S_j})$ is a measure space.
The elements of $\{X_j\cap X_n\}_{n\in\mathbb{N}}$ belongs to $S_j$ and satisfy $(X_j\cap X_n)\nearrow X_j$ and $\mu_{X_j}(X_j\cap X_n)<\infty $ for all $n\in\mathbb{N}$.
Because of the above considerations and $\mu_{S_j}(X_j)<\infty$, we can use the first part of the proof of the Theorem 1.3 which is in the book mentioned above to prove that $\mu_{S_j}(E)=\tilde \mu_{S_j}(E)\,\,\color{red}{(1)}$ for all $E\in \Sigma (S_j)$.
Let $A\in\Sigma $ be any element.
If there's a $j\in\mathbb{N}$ such that $A\cap X_j\in \Sigma (S_j)$, then I can use $(1)$ and that tip above to prove that $\mu (A)=\tilde \mu (A)$. However, if $A\cap X_j\,\,{\color{red}{\notin}} \,\,\Sigma (S_j)$ for all $j\in\mathbb{N}$, then I can't use $(1)$ and, therefore, that tip is useless.
Please help me to finish the proof. At least indicate some reference that contains the proof of that theorem.
Thank you for your attention!
 A: Actually, the $S_j$ you've defined are the best they can be, in the sense that the condition you speak of as a problem, cannot happen! Basically , your work so far is correct, and what I'm going to show is that for every $A \in \Sigma$, we have $A \cap X_j \in \Sigma(S_j)$ for every $j$.

Why can't that condition happen? Well, $S$ generates $\Sigma$. $S_j$ is just every element of $S$ intersected with $X_j$.  Now because any intersections, unions etc. behave nicely with respect to just an intersection with some $X_j$, the sigma-algebra generated by $S_j$ creates every possible intersection of an element of $\Sigma$ , and $X_j$. In one line : $S$ approximates $\Sigma$, and intersecting with an $X_j$ (or any set in $\Sigma$) should still retain the approximation.

The precise statement is this :

$$
\Sigma = \{F \in \Sigma : F \cap X_j \in \Sigma(S_j) \ \ \forall j = 1,2,...\}
$$

That is, there's no set $A \in \Sigma$ such that $A \cap X_j \notin \Sigma(S_j)$ for any single (forget about all) $j$. Note that $\Sigma(S_j)$ is a sigma-algebra of subsets of $X_j$ by construction (you are using this because the probability space is set up that way). Therefore $X_j \in \Sigma(S_j)$.
To prove this, note that $S$ is a subset of the RHS by definition of $S_j$. So we only need to prove that the RHS is a sigma-algebra. Let's call the RHS as $T$.

*

*Note that $\emptyset \cap X_j = \emptyset \in \Sigma(S_j)$ for all $j$ and $X \cap X_j = X_j \in \Sigma(S_j)$. So $0,X \in T$.


*Let $F \in T$. We know that $F \cap X_j \in \Sigma(S_j)$ for all $j$. Now, remember that $\Sigma(S_j)$ are sigma-algebras of $X_j$, so when I take a complement, $(F \cap X_j)^c = F^c \cap X_j$ for each $j$, because I am taking the complement corresponding to each $X_j$. Therefore, $F^c \cap X_j \in \Sigma(S_j)$ for all $j$ and thus $F^c \in T$.


*Let $F_i \in T$ be countably many sets. Then , note that :
$$
\left(\cup_{i=1}^\infty F_i\right) \cap X_j = \cup_{i=1}^\infty(F_i \cap X_j)
$$
now if I fix a $j$, then $F_i \cap X_j \in \Sigma(S_j)$ for all $i$, so from the above we get $(\cup_{i=1}^\infty F_i) \cap X_j \in S_j$ for all $j$ so $\cup_{i=1}^\infty F_i \in T$.
Therefore, $T$ is a sigma-algebra, so $T$ contains $\Sigma(S) = \Sigma$. So $T = \Sigma$.

Now, let $A \in \Sigma$. Then $A \in T$. So $A \cap X_j \in \Sigma(S_j)$ for all $j$. From the result for $\Sigma(S_j)$, we know that $\mu(A \cap X_j) = \tilde{\mu}(A \cap X_j)$ for every $j$. Now $A \cap X_j \uparrow A$, you can conclude.
I actually ran into a very similar issue when I dealt with a similar problem. It turns out that this has to do with a far more general statement concerning pull-backs of sigma-algebras.
I state that as follows :

Let $f : (S,\mathcal S) \to (F , \mathcal F)$ be measurable. Let $\mathcal E \subset \mathcal F$. Then, we have :
$$
f^{-1}(\Sigma(E)) = \Sigma(f^{-1}(E))
$$

That is, pulling back every set in $\Sigma(E)$ to give a sigma-algebra is equivalent to pulling back every set of $E$ and generating a sigma-algebra out of that. We can use this result to prove the above equality of sigma-algebras very easily, using the inclusion map from $X_j \to X$ for each $j$. Use this to prove that $\Sigma(S_j) = \{A \cap X_j : A \in \Sigma\}$ for each $j$, which of course leads to the theorem I wrote down.
Being comfortable with simple lemmas and theorems around sigma-algebras is very important if you want to far into probability, especially when conditioning, martingales etc. come into play.
