# Constructing a $\sigma$-finite measure using measures of its subsets

In Bogachev's 'Measure Theory', one first treats $(-\infty,\infty)$-valued measures, and then moves on to nonnegative measures, $[0,\infty]$-valued measures, and so on. Following this spirit, the book first shows how to construct a (finite) measure on any 'rectangle' of $\mathbb{R}^n$, where by a 'rectangle' we mean any subset of the form $[a_1,b_1]\times\cdots\times[a_n,b_n]$. The construction starts from an algebra of sets, and then uses the concept of an outer measure and a measurability criterion. The book denotes such a process by a 'Lebesgue extension'. (The resulting measure on a rectangle is of course nothing more than the restriction of the Lebesgue measure to the rectangle, assuming we know what 'the Lebesgue measure' is.)

The next step is to construct the Lebesgue measure of $\mathbb{R}^n$, and this is the part where I'm really confused. Without any proof or discussion whatsoever, the book throws the following example to the reader:

1.6.4. Example. Let $\frak{L}_n$ be the class of all sets $E\subset\mathbb{R}^n$ such that all the sets $E_k:=E\cap\{x:|x_i|\leq k,i=1,\ldots,n\}$ are Lebesgue measurable. Then $\frak{L}_n$ is a $\sigma$-algebra, on which the function $\lambda_n(E)=\lim_{k\rightarrow\infty}\lambda_n(E_k)$ is a $\sigma$-finite measure ... ... ...

The book places the above 'example' right after the following lemma, implying that the example is a result of the lemma. But I just don't see how. Also, it says that a different choice of the sequence of sets $X_n$ (in the lemma) does not change the final result, but I don't see why this is the case either.

1.6.2. Lemma. Let $\frak{R}$ be a ring of subsets of a space $X$ (i.e. $\frak{R}$ is closed with respect to finite intersections and unions, $\varnothing\in\frak{R}$ and $A\setminus B\in\frak{R}$ for all $A,B\in\frak{R}$). Let $\mu$ be a countably additive set function on $\frak{R}$ with values in $[0,\infty]$ such that there exists sets $X_n\in\frak{R}$ with $X=\bigcup_{n=1}^{\infty}X_n$ and $\mu(X_n)<\infty$. Denote by $\mu_n$ the Lebesgue extension of the measure $\mu$ regarded on the set $S_n=\bigcup_{j=1}^{n}X_j$ equipped with the algebra of sets consisting of the intersections of elements in $\frak{R}$ with $S_n$. Let $\frak{L}_{\mu_n}$ denote the class of all $\mu_n$-measurable sets. Let $\frak{A}$ be the class $$\{A\subset X:A\cap S_n\in\frak{L}_{\mu_n}\forall n\in\mathbb{N}, \bar{\mu}(A):=\lim_{n\rightarrow\infty}\mu_n(A\cap S_n)<\infty\}.$$ Then $\frak{A}$ is a ring closed with respect to countable intersections (i.e. a $\delta$-ring) and $\bar{\mu}$ is a $\sigma$-additive measure whose restriction to every set $S_n$ coincides with $\mu$.

Please help me clarify things up! (P.S. The TeX work in the definition of $\frak{A}$ is messed up with fraktur letters, but please understand.)