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look at Problem14.12 of chapter3 of "Aliprantis-Burkinshaw-Principles of real analysis-3ed.1998" ; 12. Let A be the collection of all measurable subsets of X of finite measure. That is, A = {B in X: m(A) < oo}. a. Show that A is a semiring. b. Define a relation ~ on A by B ~ C if m(B \Delta C) = 0. Show that ~ is an equivalence relation on A. c. Let D denote the set of ail equivalence classes of A. For B in A let B* denote the equivalence class of B in D. Now for B*, C* in D define d(B*, C*) = m(B \Delta C). Show that d is well defined and that (D, d) is a complete metric space. (For this part, see also Exercise [3] of Section [31].)

Now let m be lebesgu measure on Real numbers and X be a subset of Real line, is (D,d) a connected space with topology induced with meter d?

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  • $\begingroup$ Not necessarily, for example $X=\Bbb N, the set of natural numbers, with counting measure. Then $\{\emptyset\}$ is open, and so is $\{A\subset \Bbb N,A\neq\emptyset\}=\{A,d(A,\emptyset)>0\}$. $\endgroup$ Nov 7 '12 at 17:05
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    $\begingroup$ Not necessarily, for example $X=\Bbb N$,the set of natural numbers, with counting measure. Then \{\emptyset\}$ is open, and so is $\{A\subset \Bbb N,A\neq\emptyset\}=\{A,d(A,\emptyset)>0\}$. $\endgroup$ Nov 7 '12 at 17:06
  • $\begingroup$ As far as I can see, the revised version of the question was already answered affirmatively (before the revision) by Pietro Majer, since Lebesgue measure on a (measurable) subset of the reals is always atomless. $\endgroup$ Nov 9 '12 at 13:13
  • $\begingroup$ by the way, the term "atomless" is better than the unclear "non-atomic" (="no atoms at all", or "not atoms only"?), which is also a funny litotes. I'll correct now. $\endgroup$ Nov 12 '12 at 14:03
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Note that to get a true distance, you have to restrict to measurable sets of finite measure, and to quotient over null sets. This way you get a closed subspace $M$ of $L^1(X)$ (the subset of binary functions). If the measure is atomless, then $M$ is even path-connected. A well-known result on atomless measures (essentially due to Sierpinski) is that for any measurable $A$, with measure $\alpha$, there exists a nested family of measurable sets $\{A_t\} _ {t\in[0,\alpha ]}$, such that $A_ 0 = \emptyset$,$A_\alpha=A$ and $m(A_t)=t$ for all $t$. On the other hand, if there is an atom $A\in M$, then $M$ is the union of its non-empty open sets $\{B\in M\, : \, B\supset A\}$ and $\{B\in M\, : \, B\cap A=\emptyset \}$, so $M$ is not connected.

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