Questions about construction of Lebesgue measure in Rudin's RCA I have questions about the construction of Lebesgue measure in Rudin's RCA (Theorem 2.20). 


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*On pg52, Rudin defines $\lambda(E) = m(E+x)$ for measurable $E$ and says that $\lambda$ is a measure. But, why is $E+x$ measurable?

*In the same paragraph on pg 52, Rudin first shows $\lambda(E) = m(E)$ for all boxes, then using the observation he makes in the preceding paragraph, he shows $m(E+x) = m(E)$ for all Borel sets $E$, then finally he says the same equality holds for every measurable $E$, by using 2.20(b). But, can't I make the same conclusion without using 2.20(b)?  Theorem 2.18 shows the regularity $\lambda$ and $m$, so the conclusion holds immediately. Doesn't it?
 A: *

*There are many ways to show this. Suppose $E$ is measurable, then by that we mean there exists $A\subset E \subset B$, such that $B$ is a $G_\delta$ set and $A$ is an $F_\sigma$ set,and $m (B \setminus A) = 0$. Now, denote  $K_x=K+x$ for any set $K$. We notice $B_x \setminus A_x = (B\setminus A)_x$. We would like to show $m (B_x \setminus A_x) =0$, and it's enough to show that translation maps null sets to null sets.
Let $N$ be a null set, then we notice that for any cover $\{ B_n^{\epsilon} \} $ of open cubes (or dyadic, or closed...), such that  $m( \bigcup B_n ^{\epsilon}) < \epsilon$. By Taking finite unions and intersections we may take disjoint cubes. Now; The collection $\{ (B_n ^\epsilon)_x \}$ is a cover of $N_x$ with the measure (or volume, since we're talking about cubes). This is easy to show, each cube in the collection retains its volume.Now, because $\epsilon$ is arbitrary - $N_x$ is null set.

*The corollary of 2.18 is relevant to Borel measures, meaning, this is true for any set $F$ in the Borel collection, but the Lebesgue collection is larger, and at first sight it is unclear if a completion of a regular measure is still regular (with respect to the, perhaps, greater collection of sets).

