209: Honors nalysis in R n Review of measure theory 1 Outer measure, measure, measurable sets Definition 1 Let X be a set. nonempty family R of subsets of X is a ring if, B R B R and, B R B R hold. bove, B = ( \ B) (B \ ). Note that B = ( B) ( B) and \ B = ( B). Definition 2 n algebra is a ring such that X R. Proposition 1 The intersection of a family of rings is a ring. Definition 3 The intersection of the family of rings that contain a family of sets F is called the ring generated by F and denoted R(F). Definition 4 semiring S is a family such that S, B S B S 1, S, 1 j S such that = N j=1 j j k =, j k Exercise 1 If S is a semiring, then R(S) = { N i=1 i i S} Consider a cube with sides parallel to the axis, i.e. a Cartesian product of intervals n Q = {x R n ; x = (x 1,..., x n ), x i I i } = 1 i=1 I i
where I i is a nonempty interval (a i, b i ) I i [a i, b i ]. The family S of such cubes is a semiring. We set n l(q) = (b i a i ). i=1 Proposition 2 The ring generated by the family of cubes is called the set of elementary sets, E = R(S). The function l defined above extends uniquely as a finitely additive function to the ring of elementary sets, i.e. to the family of sets that are finite unions of cubes: Finite additivity means m( N i=1p i ) = m : E R +. N m(p i ), if P i E, P i P j =, i j. i=1 The extension is done via N m(p ) = l(q i ), if P = N i=1q i, Q i Q j =, i j. i=1 Proposition 3 If E and k E are such that then k=1 k m() m( k ). Definition 5 Let m : R R + be defined on a ring R 2 X and obey m( ) = 0, m( N i=1 i ) = N i=1 m( i), i R, i j =, k=1 k m() k=1 m( k). k=1 We will refer to m as a premeasure. We define by λ m (Y ) = λ m : 2 X [0, ] inf Y k=1 k k R m(k ) 2
Definition 6 function λ λ : 2 X [0, ] is an outer measure if (i) λ( ) = 0 (ii) λ ( i N i ) i N λ( i) The property (ii) is called σ-subadditivity. Proposition 4 Let λ m : 2 X [0, ] be as in Definition 5. Then λ m is an outer measure and λ m R = m Definition 7 Let R n. Define the Lebesgue outer measure of to be m () = inf j N l(q j ) where the infimum is taken over all countable collections of open nonempty cubes Q j such that j N (Q j ). Exercise 2 Prove that m = λ m where m : E R + is defined in Proposition 2. Definition 8 Let λ be an outer measure on a set X. subset of X is measurable, C(λ) if holds for all sets S. λ(s) = λ(s ) + λ (S (X \ )) Note: in order to check C(λ) we need to check only for all S. λ(s) λ(s ) + λ (S (X \ )) 3
Definition 9 family Σ of subsets of X is a σ-algebra if it is non-empty, and satisfies (i) Σ (X \ ) Σ, (ii) j Σ, j = 1, 2,... j=1 j Σ. It follows that a σ-algebra contains the empty set, Σ, the whole space X, and is closed to countable intersections as well. If a family of σ-algebras is given, then their intersection, if nonempty, is again a σ-algebra. In particular Definition 10 Let F be an arbitrary non-empty family of subsets of R n. The intersection of the family of all σ-algebras that contain F, σ(f) = F Σ Σ is called the σ-algebra generated by F. This is the smallest σ-algebra that contains F. Definition 11 The Borelian σ-algebra is the σ algebra generated by the family of open sets U = {U R n ; U open}, B(R n ) = σ(u) Exercise 3 Prove that if F = {F R n ; F closed}, then σ(f) = B(R n ). Definition 12 Let Σ be a σ-algebra in X. function is a measure if µ : Σ [0, ] (i) µ( ) = 0 (ii) i j =, i j i, j Σ µ ( i N i ) = i N µ( i) Note carefully that the value is allowed. The property (ii) is called σ- additivity. Exercise 4 Prove (i) B, µ() µ(b) (ii) j j+1 µ ( j j ) = lim j µ( j ) (iii) µ( 1 ) <, j+1 j µ ( j j ) = lim j µ( j ) 4
Proposition 5 Let λ be an outer measure on X. Then the collection of measurable sets C(λ) is a σ-algebra and the outer measure λ restricted to C(λ) is a complete measure. Complete means that any set satisfying λ() = 0 is measurable. Such sets are called null or negligible. Proposition 6 Let λ m be the outer measure associated to m as in Definition 5. Then R C(λ m ) Definition 13 If C(m ) associated to the outer measure m defined in Definition 7, we define the Lebesgue measure of to be = m (). Proposition 7 TFE (the following are equivalent): (i) C(m ). (ii) ɛ > 0 U open, U, m (U \ ) ɛ. (iii) ɛ > 0 F closed, F, m ( \ F ) ɛ. (iv) G, G G δ, G, m (G \ ) = 0. (v) F, F F σ, F, m ( \ F ) = 0. Here G δ is the collection of countable intersections of open sets and F σ is the collection of countable unions of closed sets. Note that this proposition implies that any measurable set is the union of a Borelian set and a null set. 2 Measurable functions Definition 14 Let Σ 2 X be a σ-algebra. function f : X R = R { } { } is measurable if for any c R the set f 1 ((, c)) = {x; f(x) < c} is measurable, i.e. belongs to Σ. 5
Exercise 5 let f i : X R i = 1,... N are measurable and B(R N ) is a Borelian subset of R N then S = {x X (f 1 (x),... f N (x)) } Σ Hint The set of B R N such that S B Σ is a σ-algebra that contains all the generalized cubes. Proposition 8 If c is a constant f and g are measurable functions and F : R 2 R is continuous then x F (f(x), g(x)) is measurable. In paricular, cf, fg, f, max(f, g), min(f, g) and f + g are measurable. If f j is a sequence of measurable sets then sup j f j, inf j f j, lim sup j f j, lim inf j f j are all measurable. 3 Convergence Definition 15 Let (X, Σ) be a measurable space (i.e. a set and a σ-algebra on it), µ be a complete measure, f j be a sequence of measurable functions. We say that f n converge in measure to f if ɛ > 0 lim µ ({x; f n(x) f(x) ɛ}) = 0. j Definition 16 We say that f n converges almost everywhere (µ-a.e.) if the set where convergence fails is null. Proposition 9 If µ(x) is finite then convergence almost everywhere implies convergence in measure. Proposition 10 If f n converges in measure to f then f n has a subsequence that converges almost everywhere to f. Exercise 6 Give an example of a sequence of functions defined on [0, 1] so that it converges to zero in measure (w.r. to Lebesgue measure) but it does n ot converge almost everywhere. Definition 17 function f is simple if it is of the form m f = c k 1 k with c k R and k Σ. k=1 6
Here 1 (x) = 1 if x, and 1 (x) = 0 if x /. Without loss of generality we may assume that the values c k are nonzero and distinct, and that k = {x; f(x) = c k }. Proposition 11 Let f : X [0, ) be measurable. Then there exists a sequence of simple functions f n (x) such that and x X, x X, 0 f n (x) f n+1 (x) f(x) = lim n f n (x). Definition 18 measure µ on a σ-algebra Σ of subsets of R n (or, more generally, of a locally compact (Hausdorff) topological space X) is said to be a Radon measure if the measure is Borel regular, i.e. B(R n ) Σ, and S Σ, ɛ > 0, U open S U, µ(u \ S) ɛ, K, K compact µ(k) < S Σ, with µ(s) <, ɛ > 0, K compact K S, µ(s \ K) ɛ Proposition 12 (Egorov) Let µ be a Radon measure and let S Σ, µ(s) <. Let f n f be a sequence of measurable functions that converges almost everywhere on S. Then, for any ɛ > 0, there exists a compact K S such that µ(s \ K) ɛ and the convergence of f n K to f K is uniform in K. Proposition 13 (Lusin) Let µ be a Radon measure, let S Σ be such th at 0 < µ(s) < and let f : X R + be measurable. Then for any ɛ > 0 there exists a compact K S, such that µ(s \ K) ɛ and such that f K is continuous. 7
4 Integral Definition 19 If f 0 is simple and Σ we define fdµ = m c k µ( k ) k=1 If f : R n [0, ] is measurable and Σ we define fdµ = sup φdµ where the supremum is taken over all simple functions φ that satisfy 0 φ f µ-a.e. This is usually called an outer integral. Exercise 7 Let φ 0 be simple. Prove that φdµ is a measure. Proposition 14 (i) B, f 0 µ a.e. fdµ fdµ B (ii) 0 f g µ a.e. fdµ gdµ (iii) f = 0 µ a.e. fdµ = 0 (iv.) µ() = 0, f 0 µ a.e. fdµ = 0. (v.) fdµ + gdµ = (f + g)dµ. Definition 20 f 0 is said to be integrable if fdµ is finite. Theorem 1 (Lebesgue monotone) If 0 f j f j+1 µ-a.e. then f = lim j f j µ-a.e. is measurable and fdµ = lim f j dµ. j Proposition 15 (Chebyshev) Let f 0 µ-a.e. be measurable. Then µ({x; f(x) M}) fdµ M 8
Exercise 8 (Beppo-Levi). Let 0 f j f j+1 µ-a.e., and assume C, j fj dµ C. Let f = lim j f j µ-a.e. Then f is integrable and finite almost everywhere. Theorem 2 (Fatou). Let f j 0, µ-a.e. Then ( ) lim inf j dµ lim inf j j f j dµ Definition 21 function f is in L 1 (dµ) if it is measurable and f is integrable. In that case we define fdµ = f + dµ f dµ where f + = max(f, 0), f = max( f, 0). Theorem 3 (Lebesgue dominated convergence) Let g 0 be integrable and assume that f j g, µ-a.e., j. ssume that f = lim j f j µ-a.e. Then f j dµ = fdµ lim j and moreover 5 L p lim j f j f dµ = 0. Definition 22 Let µ be a complete measure and 1 p <. We say that f L p (dµ) is f is measurable and f p dµ <. The norm is ( f L p = ) 1 f p p dµ Definition 23 f L (dµ) if f is measurable and there exists a constant such that {x; f(x) C} is negligible. The norm is the infimum of such C f L = inf{c 0; µ({x; f(x) C}) = 0} 9
Proposition 16 Hölder inequality: Let p 1 + q 1 L q (dµ). Then fgdµ f L p g L q = 1, f L p (dµ), g is true for 1 p Proposition 17 (Minkowski inequality). Let 1 p. f + g L p f L p + g L p Exercise 9 Let µ = dx be Lebesgue measure in R n. Let f L 1, φ L 1. Then f φ(x) = f(y)φ(x y)dy is in L 1. If φ is continuous and bounded then f φ is continuous. 10