Chapter 4. Lebesgue Integration
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1 4.2. Lebesgue Integrtion 1 Chpter 4. Lebesgue Integrtion Section 4.2. Lebesgue Integrtion Note. Simple functions ply the sme role to Lebesgue integrls s step functions ply to Riemnn integrtion. Definition. For simple function ψ on set of finite mesure with cnonicl representtion ψ = n i=1 iχ i, define the (Lebesgue) integrl n ψ = i m( i ). i=1 Lemm 4.1. Let { i n i=1 be finite disjoint collection of mesurble subsets of set of finite mesure. For 1 i n, let i be rel number. If ϕ = n i=1 iχ i on then ϕ = n i=1 im( i ). Note. Lemm 4.1 is necessry becuse the representtion of ϕ my not be the cnonicl one. Note. The following is the first (of four) times we ll see linerity nd monotonicity result.
2 4.2. Lebesgue Integrtion 2 Proposition 4.2. Linerity nd Monotonicity of Integrtion. Let ϕ nd ψ be simple functions defined on set of finite mesure. Then for ny α,β (αϕ + βψ) = α nd if ϕ ψ on then ϕ ψ. ϕ + β ψ Definition. Let f be bounded function on set of finite mesure. Define the lower Lebesgue integrl { f = sup Define the upper Lebesgue integrl { f = inf ϕ ϕ is simple, ϕ f. ψ ψ is simple, ψ f. Note. By monotonicity, since f is bounded, both f nd f re finite. Also by monotonicity f f. Definition. A bounded function f on domin of finite mesure is sid to be Lebesgue integrble over provided f = f. The common vlue is the Lebesgue integrl of f over, denoted f. Theorem 4.3. Let f be bounded function defined on [,b]. If f is Riemnn integrble over [,b] then it is Lebesgue integrble over [,b] nd the two integrls re equl.
3 4.2. Lebesgue Integrtion 3 Proof. Recll (see the Riemnn-Lebesgue Theorem hndout) tht upper nd lower Riemnn integrls re defined in terms of step functions. Since step functions re lso simple functions, R b [,b] f(x) dx = f = sup s f s step function { b s { b sup ϕ = f ϕ f ϕ simple [,b] { b { b inf ψ inf S = R ψ f S f ψ simple S step function b f(x) dx. If f is Riemnn integrble, then the inequlities must be equlities nd the Riemnn integrl equls the Lebesgue integrl nd the two integrls re equl. Note. The following result estblishes the Lebesgue integrbility of the clss of functions studied in this section. Theorem 4.4. Let f be bounded mesurble function on set of finite mesure. Then f is integrble on. Note. The converse of Theorem 4.4 is lso true. Tht is, if bounded f on set of finite mesure is integrble (i.e., f exists), then f is mesurble (see Chpter 5, Theorem 5.7). Note. The following is similr to Proposition 4.2, but dels with the clss of bounded functions on sets of finite mesure.
4 4.2. Lebesgue Integrtion 4 Theorem 4.5. Linerity nd Monotonicity. Let f nd g be bounded mesurble functions on set of finite mesure. Then for ll α,β R (αf + βg) = α f + β g. Moreover, if f g on, then f g. Corollry 4.6. Let f be bounded mesurble function on set of finite mesure. Suppose A nd B re mesurble disjoint subsets of. Then f = f + f. A B A B Corollry 4.7. Let f be bounded mesurble function on set of finite mesure. Then f f. Note. The following is not surprising in light of the corresponding result for Riemnn integrls. Proposition 4.8. Let {f n be sequence of bounded mesurble functions on set of finite mesure on. If {f n f uniformly on, then ( ) ( ) lim f n = lim f n = f.
5 4.2. Lebesgue Integrtion 5 Note. We re interested in generl in when (lim f n) = lim( f n) for convergent sequence {f n on. When this result holds, we hve convergence theorem. We will hve three such min results. Note. Define f n on s: 0 if x = 0 or x [1/n, 1] f n (x) = n if x (0,1/n). Then f n f 0 (pointwise but not uniformly) nd However, ( ) ( ) lim f n = 1 lim f n = 0 = 0. f n = 1 for ll n N. Note. To prove our first convergence theorem, we need: goroff s Theorem. (pge 64) Assume hs finite mesure. Let {f n be sequence of mesurble functions on tht converges pointwise on to rel-vlued function f. Then for every ε > 0, there is closed set F for which {f n f uniformly on F nd m( \ F) < ε. Bounded Convergence Theorem. Let {f n be sequence of mesurble functions on set of finite mesure. Suppose {f n is uniformly pointwise bounded on, tht is, there is number M 0 for which f n M on for ll n. If {f n f pointwise on, then ( ) ( ) lim f n = lim f n = f. Revised: 11/7/2016
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