MEASURE AND INTEGRATION: LECTURE 16 Cc dense in Lp for 1 ≤ p < ∞. Theorem 0.1. Let S = {s : X → C | s simple, measurable such that µ({x | s(x) = � 0})} . For 1 ≤ p < ∞, S is dense in Lp (µ), i.e., given f ∈ Lp (µ) there exists sequence sk ∈ S such that �sk − f �p → 0. Proof. Note that S ⊂ Lp (µ) since � � 0}) < ∞. sp dµ ≤ max sp µ({x | f (x) = X If f : X → R and f ≥ 0, then by the approximation theorem, there exists sk simple measurable functions � � such that 0 ≤ s1 ≤ · · · ≤ f and limk→∞ sk = f . Since sk ≤ f , spk ≤ f p < ∞. Thus, sk ∈ Lp ⇒ sk ∈ S. We have f − sn ≤⇒ |f − sn |p ≤ |f |p . So |f − sn |p ≤ f = |f |p ∈ L1 , and we can apply the dominated conver gence theorem. Thus, � � p lim |f − sn | dµ = lim(f − sn )p dµ − 0, X X and so �sn − f �p → 0. If f is not nonnegative, apply separately to f + and f − . � Corollary 0.2. If X is a locally compact Hausdorﬀ space, then for 1 ≤ p < ∞, Cc (X) is dense in Lp . Proof. Let S be as in the previous theorem. If s ∈ S and � > 0, there exists g ∈ Cc (X) such that µ({x | g(x) = � s(x)}) < � by Lusin’s 1 2 MEASURE AND INTEGRATION: LECTURE 16 theorem, and also |g| ≤ �s�∞ . Thus, �� 1/p �� p �g − s�p = = |g − s| p |g − s| + p �1/p |g − s| = �� ≤ � g=s |g − s| p �1/p g=s � g=s �� p 2 � g=s �s�p∞ �1/p ≤ 2 �s�∞ �1/p . Thus, Cc (X) is dense in S, and since S is dense in Lp , Cc (X) is dense � in Lp . Here is �an example. Let X = Rn and let f, g ∈ Cc (Rn ). Deﬁne ∞ d(f, g) = −∞ |f (t) − g(t)| dt. Note that Cc (Rn ) ⊂ L1 (Rn ) and L1 is complete. The space L1 (Rn ) is the completion of Cc (Rn ) under this metric, provided f ∼ g if f = g a.e. Any metric space has a unique completion under its metric. The case p = ∞. Let f, g ∈ Cc (X) and d(f, g) = sup |f (x) − g(x)| . x∈X ∞ Then L is not the completion of Cc (X) under d. A function f : X → C vanishes at inﬁnity if for every � > 0 there exists a compact subset K ⊂ X such that |f (x)| < � whenever x �∈ K. The set of all continuous function that vanish at inﬁnity is denoted by C0 (x). Cc dense in C0 . Theorem 0.3. The completion of Cc (X) under �·�∞ is C0 (X). Proof. We show that (a) Cc (X) is dense in C0 (X), and (b) C0 (X) is complete. Proof of (a). Let f ∈ C0 (X). For � > 0, there exists K compact such that |f (x)| < � for all x ∈ K c . By Urysohn’s lemma, there exists g ∈ Cc (X) such that K � g, 0 ≤ g ≤ 1, and g = 1 on K. Let h = f g. Then h ∈ Cc (X) and �f − h�∞ < �. (f = h on K and f < � on K c .) Proof of (b). Let fn be a Cauchy sequence in C0 (X), i.e., given � > 0, there exists N such that i, j > N , �fi (x) − fj (x)�∞ < �. In other words, fn converges uniformly. Thus, lim fn = f exists n→∞ and f is continuous. Given � > 0, there exists n such that �fn − f �∞ < �/2 and there exists K compact such that |fn (x)| < �/2 for all x ∈ K c . Then |f | = |f − fn + fn | ≤ �/2 + �/2 = � on K c . Thus, f ∈ C0 (X). �