MATH 202 B - Problem Set 7 Walid
نویسنده
چکیده
where En are measurable sets. Let > 0. Since μ is outer regular, and inner regular on measurable sets of finite measure, there exist an open set On and a compact set Kn such that Kn ⊂ En ⊂ On μ(En)− 1 2N < μ(Kn) ≤ μ(En) ≤ μ(On) < μ(En) + 1 2N thus Kn ⊂ On and μ(On)− μ(Kn) < 1 N , and since X is of finite measure, we have μ(On \Kn) < 1 N consider the closed sets Kn and O c n. By Urysohn’s Lemma, there exists a continuous function gn : X → [0, 1] such that gn|Kn ≡ 1, and gn|Oc n ≡ 0. Therefore we have for all x ∈ Kn ∪O c n, gn(x) = 1En(x). Since X is compact and O n is closed, it is also compact, and so is Kn∪O n. Now let K = ∩n=1(Kn∪O n). K is compact as the finite intersection of compact sets. Consider the function g = ∑N n=1 cngn. It is continuous as the finite sum of continuous functions, and by definition of fn, we have for all x ∈ K,
منابع مشابه
MATH 202 B - Problem Set 10
(10.1) Let k ∈ N, and let X = C k ([0, 1] d) to be the set of all functions f ∈ C 0 ([0, 1] d) such that for all α ∈ ∆ k , f (α) = ∂f ∂x α 1 1 ...∂x α d d exists and is continuous on (0, 1) d , and extends continuously on [0, 1] d. Here
متن کاملMATH 202 A - Problem Set 13 Walid
1. Let z ∈ R, and consider the function fz : R→ R x 7→ x+ z We have fz is Borel-measurable since for all c ∈ R, {x ∈ R|fz(x) ≤ c} = {x ∈ R|x+ z ≤ c} = (−∞, c− z) ∈ B. We also observe that if E ⊆ R, then f−1 z (E) = {x ∈ R|x + z ∈ E} = {−z + (x + z)|x + z ∈ E} = −z + E. Thus we have: if E is a Borel set, then f−1 −y (E) = y + E is a Borel set. Conversely, if y + E is a Borel set, then −y + (y + ...
متن کاملMATH 202 A - Problem Set 4
(4.1) Let E,F be nonempty subsets of R, E is compact and F is closed. Then there exist (e, f) ∈ E ×F such that d(E,F ) = d(e, f). proof Let α = d(E,F ) = inf(x,y)∈E×F d(x, y) be the distance between the two sets. Let > 0, and let x0 be a given element of E. Since E is compact, it is in particular bounded, and there exists r > 0 such that E ⊂ B(r, x0). Now consider the closed ball B(r + α+ , x0)...
متن کاملMATH 202 B - Problem Set 2
we have by construction μ(En) = an. An important observation is that a subset of consecutive intervals En, En+1, . . . , En+k are either pairwise disjoint, or they cover the whole interval [0, 1]. Now for all n, let fn = 1En . Then we have • ‖fn‖1 = ∫ X |1En |dμ = μ(En) = an. • for all x ∈ [0, 1], (fn(x))n does not converge. Indeed, if we suppose by contradiction that (fn(x))n converges for som...
متن کامل