real_analysis_II

# real_analysis_II - 2 Lebesgue integration 1 Let(Ω A,μ be...

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Unformatted text preview: 2 Lebesgue integration 1. Let (Ω , A ,μ ) be a measure space. We will always assume that μ is com- plete, otherwise we first take its completion. The example to have in mind is the Lebesgue measure on R n , ( R n , L n , | · | ) . We will build the inte- gration theory for A-measurable functions. We will consider measurable functions f : Ω-→ ¯ R , where ¯ R = R 1 ∪ {-∞} ∪ { + ∞} (and also functions F : Ω → ¯ C ∪ {∞} , where ¯ C = C ∪{∞} ). First we define integrals of real valued nonnegative functions, and then reduce the general case to this. We will follow Rudin very closely. 2. Let f : Ω → R be a simple function : f = N X j =1 c j χ E j , c 1 ,...,N ∈ R , E 1 ,...,N ∈ A , N < + ∞ . It is useful to know that any simple f can be written as a linear com- bination of χ E j , E j ∈ A , with distinct c 1 , ... , c N and disjoint E 1 , ... , E N . In fact, a simple f takes only a finite number of distinct values α 1 < α 2 < ··· < α M . Hence the desired representation is f = M X m =1 α m χ f- 1 ( { α m } ) . Adding a term with c j = 0 if necessary, we can always assume that N [ j =1 E j = Ω . 3. Let f ≥ be a simple function written as f = N X n =1 c n χ E n , E j ∩ E k = ∅ if j 6 = k. Define the Lebesgue integral of f with respect to μ over Ω by Z Ω f ( x ) dμ ( x ) = Z Ω f dμ def = N X j =1 c j μ ( E j ) agreeing that ∞ = 0 . 1 We do not assume here that c 1 ,...,N are distinct. (However, since E j s are disjoint, we have c 1 ,...,N ≥ .) The definition is correct. In fact, suppose we also have f = M X m =1 d m χ F m , F j ∩ F k = ∅ if j 6 = k. Notice that χ A ∪ B = χ A + χ B if A ∩ B = ∅ . But then f = N X n =1 M X m =1 e nm χ E n ∩ F m , e nm = c n = d m . Now, keeping in mind that all sets E n (and all F m ) are disjoint we derive N X n =1 c n μ ( E n ) = N X n =1 c n M X m =1 μ ( E n ∩ F m ) = N X n =1 M X m =1 e nm μ ( E n ∩ F m ) , and by the same argument M X m =1 d m μ ( F m ) = N X n =1 M X m =1 e nm μ ( E n ∩ F m ) . 4. For a general measurable F ≥ define Z Ω F ( x ) dμ ( x ) = Z Ω F dμ = sup f simple , ≤ f ≤ F Z Ω f dμ. The definition is correct when applied to simple functions. This means, that for any f ≥ , f = N X n =1 c n χ E n with disjoint E 1 , ... , E N , the equality sup φ simple , ≤ φ ≤ f Z Ω φdμ = N X n =1 c n μ ( E n ) holds. In fact, take any simple φ written as φ = M X m =1 α m χ F m 2 with disjoint F 1 , ... , F M , such that ≤ φ ≤ f . Then for all m,n α m χ F m ∩ E n ≤ c n χ F m ∩ E n . Keeping in mind that all sets E n (and all F m ) are disjoint we derive M X m =1 α m μ ( F m ) = M X m =1 N X n =1 α m μ ( F m ∩ E n ) ≤ M X m =1 N X n =1 c n μ ( F m ∩ E n ) = N X n =1 M X m =1 c n μ ( F m ∩ E n ) ≤ N X n =1 c n μ ( E n ) ....
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## This note was uploaded on 12/26/2011 for the course MATH 5A taught by Professor Rickrugangye during the Fall '07 term at UCSB.

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real_analysis_II - 2 Lebesgue integration 1 Let(Ω A,μ be...

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