- #1
e12514
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If E_1, E_2, ... is a sequence (of subsets of R^n) that decreases to E
(i.e. E_m+1 is a subset of E_m for all m, and E = intersection of all the E_m's)
and some E_k has finite (lebesgue) measure, i.e. lambda(E_k) is finite
it is a known result that the measure of E is equal to the limit of the measure of E_m.
But now if we are given some bounded set E
and we define E_m = { x : d(x,E) < 1/m }
where d(x,E) = minimum distance from x to any point in set E,
then howcome we have lambda(E) = lim_m->oo lambda( E_m ) when E is closed
but not when E is open?
Doesn't the fact that E is bounded imply some E_k has finite measure, and hence the above result applies, regardless whether E is open or closed or neither?
(i.e. E_m+1 is a subset of E_m for all m, and E = intersection of all the E_m's)
and some E_k has finite (lebesgue) measure, i.e. lambda(E_k) is finite
it is a known result that the measure of E is equal to the limit of the measure of E_m.
But now if we are given some bounded set E
and we define E_m = { x : d(x,E) < 1/m }
where d(x,E) = minimum distance from x to any point in set E,
then howcome we have lambda(E) = lim_m->oo lambda( E_m ) when E is closed
but not when E is open?
Doesn't the fact that E is bounded imply some E_k has finite measure, and hence the above result applies, regardless whether E is open or closed or neither?