MEASURE AND INTEGRATION: LECTURE 15. f p X. < }. Observe that f p

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1 L saes. Let 0 < < and let f : funtion. We define the L norm to be ( ) / f = f dµ, and the sae L to be C be a measurable L (µ) = {f : C f is measurable and f < }. Observe that f = 0 if and only if f = 0 a.e. Thus, if we make the equivalene relation f g f = g a.e, then makes L a normed sae (we will define this later). If µ is the ounting measure on a ountable set, then f dµ = f (x). x Then L is usually denoted l, the set of sequenes s n suh that ( ) / s n <. n= A funtion f is essentially bounded if there exists 0 M < suh that f (x) M for a.e. x. The sae L is defined as with the L norm L (µ) = {f : C f essentially bounded} f = inf{m f (x) M a.e. x }. Proosition 0.. If f L, then f (x) f a.e. Proof. By definition of inf, there exists M k f suh that f (x) < M k a.e, or, equivalently, there exists N k with µ(n k ) = 0 suh that f (x) M k for all x N k. Let N = k=n k. Then µ(n ) = 0. If x N = k= (N k ), then f (x) M k sine M k f. Thus, f (x) f for all x N. Date: Otober 23, 2003.

2 2 MEASURE AND INTEGRATION: LECTURE 5 Theorem 0.2. Let and / + /q =. Let f L (µ) and g L q (µ). Then fg L (µ) and fg f g q i.e., ( ) / ( ) /q fg dµ f g q. Proof. If < <, this is simly Hölder s inequality. If =, q =, then f(x)g(x) g f(x) a.e. Thus, fg g f. Theorem 0.3. Let. Let f, g L (µ). Then f + g L (µ) and f + g f + g. Proof. If < <, this is simly Minkowski s inequality. If =, then f + g f + g is true. If =, then f + g f + g f + g f + g. Normed sae and Banah saes. A normed sae is a vetor sae V together with a funtion : V R suh that (a) 0 x <. (b) x = 0 x = 0. () αx = α x for all α C. (d) x + y x + y. For examle, L (µ) is a normed sae if two funtions f, g are onsidered equal if and only if f = g a.e. Also, R n with the Eulidean norm is a normed sae. A metri sae is a set M together with a funtion d: M M R suh that (a) 0 d(x, y) <. (b) d(x, x) = 0. () d(x, y) > 0 if x = y. (d) d(x, y) = d(y, x). (e) d(x, y) d(x, z) + d(z, y). A normed sae is a metri sae with metri d(f, g) = f g. Reall that x i x M if lim n d(x n, x) = 0. A sequene {x i } is Cauhy if for every ɛ > 0 there exists N(ɛ) suh that d(x j, x k ) ɛ for all j, k N(ɛ). Claim: if x n x, then it is Cauhy. We know that lim n d(x n, x) = 0, so given ɛ > 0, there exists N suh that d(x k, x) < ɛ/2 for all k > N. for j, k > N, d(x k, x j ) d(x j, x) + d(x, x k ) < ɛ.

3 3 However, a Cauhy sequene does not have to onverge. For examle, onsider the sae R \ {0} (the untured real line) with the absolute value norm. The sequene x n = /n is Cauhy but it does not onverge to a oint in the sae. A metri sae is alled omlete if every Cauhy sequene onverges. By the Bolzano Weierstrass theorem, R n is omlete. (Every Cauhy sequene is bounded, so it has a onvergent subsequene and must onverge.) A normed sae (V, ) that is omlete under the indued metri d(f, g) = f g is alled a Banah sae. Riesz Fisher theorem. Lemma 0.4. If {f n } is Cauhy, then there exists a subsequene f nk suh that d(f, f nk ) 2 k. n k+ Theorem 0.5. For and for any measure sae (, M, µ), the sae L (µ) is a Banah sae. Proof. Let < and let {f n } L (µ) be a Cauhy sequene. By the lemma, there exists a subsequene nk with n < n2 < suh that f, f k nk < 2 k nk+. Let g k = f ni+ f and g = n i lim k g k = f ni+ f. By Minkowski s inequality, k k g f < 2 i k ni+ f n i <. Consider g k. By Fatou s lemma, lim inf g k lim inf g k, n i and so g g(x) < a.e. Thus, the series f n (x) + (f ni+ (x) f ni (x)) onverges absolutely a.e. Define { f n (f (x) + n i+ (x) f ni (x)) where it onverges; f(x) = 0 otherwise.

4 4 MEASURE AND INTEGRATION: LECTURE 5 The artial sum k f n (x) + (f ni+ (x) f ni (x)) = f nk (x), and so lim f nk (x) = f(x) a.e. k Thus we have shown that every Cauhy sequene has a onvergent subsequene, and we NTS that fn k f in L. Given ɛ > 0, there exists N suh that f n f m < ɛ for all n, m > N. We have that f f m = lim inf f nk f m sine f nk f a.e. Thus, f f m = lim inf f nk f m lim inf f nk f m < ɛ. This imlies that f f m < ɛ, and thus f = f f m + f + m + f f f m m <. We onlude that f L and f fm 0 as m. Now let = and let {f n } be a Cauhy sequene in L (µ). Let A k = {x f k (x) > f k } and B m,n = {x f n (x) f m (x) > f n f m }. These sets all have measure zero. Let ( ) ( ) N = A k B m,n. k= n,m= Then N has measure zero. For x N, f n is a Cauhy sequene of omlex numbers. Thus, f n f by omleteness of C uniformly. Sine fk is bounded, f k (x) < M for all x N. Thus, f(x) < M for all x N. Letting f = 0 on N, we have f < and fn f 0 as n. Theorem 0.6. Let and {f n } be a Cauhy sequene in L (µ) suh that f fn 0. Then f n has a subsequene whih onverges ointwise almost everywhere to f(x).

5 5 Proof. Sine f f n 0, f n f in measure. By the revious theorem, there exists a subsequene whih onverges a.e. Examles in R. () A sequene in L an onverge a.e. without onverging in L. Let f k = k 2 χ (0,/k). Then ( ) / f k = k 2 = k 2 (/k) / = k 2 / <. (0,/k) Thus f k L and f k 0 on R, but f k. (2) A sequene an onverge in L without onverging a.e. (HW roblem). (3) A sequene an belong to L L 2 and onverge in L without onverging in L 2. Let f k = k χ (k,2k). Then f k 0 ointwise and f k = k k / = k /. If >, then f k 0 as k, so f k 0 in L norm. But f k = so f k 0 in L.