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Adriana Ayers

Adriana Ayers

Answered question

2022-06-15

Let E R n and O i = { x R n d ( x , E ) < 1 i } , i N . Show that if E is compact, then m ( E ) = lim i m ( O i ). Does the statement hold if E is closed, but not bounded and if E is open, but bounded?
What I got was that since i O i = E and O 1 O 2 , then we have that
m ( E ) = m ( i O i ) = lim i m ( O i )
by the continuity of the Lebesgue measure. How is the fact needed that E is compact here?

Answer & Explanation

Amy Daniels

Amy Daniels

Beginner2022-06-16Added 20 answers

To use the result that you mention, you must have that at least one of you O i has finite measure. Here you need compactness, for example take an open ball B such that E is compactly contained in B. Hence, for i big enough E O i B, which implies that m ( O i ) < m ( B ) < . After this argument, you can invoke the result that you are using.
Edit: here is a counter-example that I always remember, consider E = R × { 0 }, O i = R × { 1 / i }. Then E = O i , but (as subsets of R 2 ) m ( E ) = 0 and m ( O i ) = (and hence the limit is infinite!).
xonycutieoxl1

xonycutieoxl1

Beginner2022-06-17Added 7 answers

If E is not compact then it is not closed or it is not bounded and for both cases there is a counterexample.
If E is a hyperplane then it is closed but not bounded.
It is evident that m ( E ) = 0 while m ( O i ) = for every i.
If E = { x Q n d ( x , 0 ) 1 } then it is bounded but not closed.
It is evident that m ( E ) = 0 and:
i = 1 O i = { x R n d ( x , 0 ) 1 }
and also:
lim i m ( O i ) = m ( i = 1 O i ) > 0
(here we apply the continuity of the Lebesgue measure).

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