From intermediate analysis course 1 Find a sequence xn such

From intermediate analysis course

1. Find a sequence {xn} such that S{xn} R = N.

2. Can you find a sequence {xn} such that S{xn} R = R?

Solution

For n = 1, 2, . . ., define An = {x X : 1/n |f(x)| < n}. Clearly, A1 A2 · · · A , [ n=1 An and each An is measurable (why?).3 Next, define A0 = {x X : f(x) = 0} and A = {x X : |f(x)| = }. Then X = A0 A A is a disjoint union, and Z X |f| = Z A0 |f| + Z A |f| + Z A |f| = Z A |f|. (2) The first term in the middle expression is zero since f is zero on A0, and the third term is zero since f L 1 (µ) implies µ(A) = 0. To prove the result, then, we must find a measurable set E such that R A\\E |f| < , and µ(E) < . Define fn = |f|An . Then {fn} is a sequence of non-negative measurable functions and, for each x X, limn fn(x) = |f(x)|A(x). Since An An+1, we have 0 f1(x) f2(x) · · · , so the monotone convergence theorem4 implies R X fn R A |f|, and, by (2), limn Z An |f| dµ = Z A |f| dµ = Z X |f| dµ. Therefore, there is some N > 0 for which Z X\\AN |f| dµ < . Finally, note that 1/N |f| < N on AN , so µ(AN ) N Z AN |f| dµ N Z X |f| dµ < . Therefore, the set E = AN meets the given crite