Question

14. Show that if a set E has positive outer measure, then there
is a bounded subset of E that also

has positive outer measure.

Answer #1

Assume to the contrary that every bounded subset of E has a outer measure 0

i.e m(E) =0

Define I=[n,n+1], nz to be countable collection of disjoint bounded interval

Then we can write E as countable union of bounded subset of E

i.e E= (E￼ I)

Now by finite sub additivity of m

0<m(E) =m(EI) m(EI)=0

which is a contradiction to our assumption

Hence there must be some nZ such that m(EI)>0

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bounded if and only if every infinite subset of E has a point of
accumulation that belongs to E.
Use Theorem 4.21: [Bolzano-Weierstrass Property] A set of real
numbers is closed and bounded if and only if every sequence of
points chosen from the set has a subsequence that converges to a
point that belongs to E.
Must use Theorem 4.21 to prove Corollary 4.22 and there should...

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(4)
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