Measure Theory?

Question

Show that the union of a countable collection of sets of measure zero is zero.

Find an uncountable collection of sets of measure zero whose union has positive measure.

Answer 1

Use additivity:



m(An) = 0 for all n => m(U An) ≤ ∑ m(An) = ∑ 0 = 0,



thus m(U An) = 0.



For the second part, how 'bout Ax = {x} for every real number x ∈ [0,1]. Then m(Ax) = 0 for all x but m(U{Ax : x ∈ [0,1]}) = m([0,1]) = 1. (m = Lebesgue measure)

Answer 2

i haven't the demonstration right now, but the example is very simple: Consider R (reals) with the Lebesgue measure, let A=[0,1] which has postive meaure = 1. Then A can be expresed as (in Latex code)

$$ A=\bigcup_{x\in A} \{ x \} $$

meaning: A is the uncountable union of its elements, each one having measure zero (a point has measure zero). The union is an uncountable one because A is uncountable.

Answer 3

I assume this is the Lebesgue measure on R^n. Note that a set S has measure zero if and only if for every ε>0 there is a countable set of open boxes A₁, A₂, A₃... such that S⊆[i=1, ∞]⋃A_i and [i=1, ∞]∑vol(A_i) < ε. So now suppose S₁, S₂, S₃... is a countable sequence of sets with measure zero. Let ε>0. For each S_i, find a countable sequence of open boxes A_(i1), A_(i2), A_(i3)... such that S_i ⊆ [j=1, ∞]⋃A_(ij) and [j=1, ∞]∑vol (A_(ij)) < ε/2^i. Then the collection of all the A_(ij) is a countable sequence of open boxes, and [i=1, ∞]⋃S_i ⊆ [i=1, ∞; j=1, ∞]⋃A_(ij), yet [i=1, ∞; j=1, ∞]∑vol (A_(ij)) = [i=1, ∞]∑[j=1, ∞]∑vol (A_(ij)) < [i=1, ∞]∑ε/2^i = ε. So the set of all the A_(ij) is a countable set of open boxes with total volume less than ε that covers [i=1, ∞]⋃S_i, and since we can find such a set for any ε>0, it follows that μ([i=1, ∞]⋃S_i) = 0.



The second part of this problem is easy: consider {{x}: x∈R}. Each set in this set has only one point, and thus measure zero, but the union of all of them is R, which of course has infinite measure.

Answer 4

stick to the definition of atom: formally, given a level area (X,?) and a finite degree ? on that area, a collection A in ? is declared as an atom if ? (A) >0 and for any measurable subset B of A with ?(A) > ? (B) one has ?(B) = 0. To show: " if ? is a non-atomic degree and A is a measurable set with ?(A) > 0, then for any actual variety b relaxing ?(A) > b > there exists a measurable subset B of A such that ?(B) = b" information: because of the fact that ? is a non-atomic degree, a collection A in it is no longer an atom, so with ?(A) > 0, the different of being an atom occurs, somewhat: for any b>0 ,right here exists measurable subset B of A with ?(A) > ? (B) =b > 0