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funksional analiz misol va masalalar yechish 1 qism
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⊂ A, μ (A) = μ ∗ (A) = 0 μ ∗ (A ) = 0 B = ∅ ∈ S m μ ∗ (A Δ∅) = μ ∗ (A ) = 0 ( A ! μ (A ) = 0 ! $' . ! & ! c ! "# . ! & ! ! A ⊂ [0, 1] ! & B = [0, 1]\A ! & , A ∪ B = [0, 1] ! & * A ⊂ [0, 1) ! & B ⊂ [1, 2) ! & C = A ∪ B & ! A 0 = C ∩ E 0 = A ! & a7" ( C = A ∪ B ! & $. F (x) = [x] μ F − / "2 ! ! ! "# F (x) = [x] & Z ' & & ! a#`" ( μ F ! () * +, * *+- + # - $/ 1 & m : M(S) → R ! $ 3a#5 m : M(S) → R " A & ! $ m : M(S) → R ! m : S → R ! $ & & $ * A ∈ M(S) & m (A) = m(A) $$ * A ∈ M(S) & ! A 1 , A 2 , . . . , A n & ( A = n k =1 A k " - m (A) = n k =1 m (A k ) . $% * A ∈ M(S) & {A n } − & ! A ⊂ ∪ n A n m (A) ≤ n m (A n ) $& A & A 1 , A 2 , . . . , A n , . . . & ( A = ∞ n =1 A n , - m (A) = ∞ n =1 m (A n ) . a#W"a%`" A ⊂ R & " , P 1 , P 2 , . . . , P n A = n ∪ k =1 P k A & ! & $' A = 5 n =1 3 2−n , 2 3−n $. A = 6 n =1 3 3−n , e 4−n $ / A = 4 n =1 n 2(n + 1)! , 6 − n 2n! $ A = 4 n =1 n 4 n +1 , 5 − n 4 n $ A = 8 n =1 1 n − 1 8 , 1 n + 1 24 $ A = 4 n =1 3 n − 1 5 , 3 n + 1 5 $ \ & " $ $ S × S = S 2 $ % 1 ! ! − S 2 ,( , R 2 E 2 = [0, 1] × [0, 1] " x = 1 3 , x = 2 3 , y = 1 3 , y = 2 3 ! $ ! P 1 = (x, y) : 1 3 < x < 2 3 , 1 3 < x < 2 3 3aa"! 5 = W $ 3aa"! 5 ' + ! 0 + & A ' & ,( , ''# $ & 2 & ! & $ ' A ⊂ [0, 1] − & ! % ` ! ' & ! & $ . A ⊂ [0, 1] − ! W " & , ! & $ / A ⊂ [0, 1] − & ! # $ ! ! , ! & $ = & K ⊂ [0, 1] X ( ! $ ,) , K − = & A = {(x, y) : x ∈ [0, 1], y ∈ K} & ,) , A & [0, 1] × [0, 1] ! ! & μ (A) = 0. ,( , R 2 [0, 1] × [0, 1] x = 1 3 , x = 2 3 , y = 1 3 , y = 2 3 ! $ * ! & b '(# P 1 = (x, y) : 0 ≤ x ≤ 1 3 , 0 ≤ y ≤ 1 3 , P 2 = (x, y) : 0 ≤ x ≤ 1 3 , 2 3 ≤ y ≤ 1 , P 3 = (x, y) : 2 3 ≤ x ≤ 1, 0 ≤ y ≤ 1 3 , P 4 = (x, y) : 2 3 ≤ x ≤ 1, 2 3 ≤ y ≤ 1 & 3a7"! 5 ' A 1 = P 1 ∪ P 2 ∪ P 3 ∪ P 4 " '! $ '! P 1 , P 2 , P 3 , P 4 ! & 3a7"! Q 1 − Q 16 5 ' #7 A 2 + ! 0 + ! "! + A 1 ⊃ A 2 ⊃ . . . ⊃ A n ⊃ . . . " , " A = ∞ ∩ n =1 A n A & ( $ 2 & A ! 3 μ (A) = 0 5 [0, 1] × [0, 1] ! ! $ A ⊂ R & ! μ ( A ) = 0 " c ' A − A & & $ $ A ⊂ R ! ! & " x, y ∈ A + x − y ∈ Q $ % A ⊂ R ! & x, y ∈ A + x − y ∈ R\Q $ & R ! & σ − $ ' A ⊂ R ' & & / ! &- 5 A = ∞ n =1 n − 1 3 n , n + 1 2 n , 5 A = ∞ n =1 n n , n n + 1 ln(n + 1) \Q, !5 A = [0, 1]\Q. $ . < A ⊂ R & + ! & 5 A = (R\Q) ∩ [0, 1], 5 A = x ∈ R : x 4 ∈ Q , !5 A = [a, ∞), 5 A = ∞ n =0 (n − 1 2 n , n + 1 2 n ), 5 A = ∞ n =0 n − 1 e n , n + 1 4 n , 5 ∞ n =0 n 3 − 1 5 n , n 3 + 1 5 n ∩ (R\Q). $ / 2 {A n } )' & & ) " + - 5 μ (A n ) = 1, n ≥ 1, ∞ ∪ n =1 A n = R, 5 μ (A n ) = +∞, A n ⊃ A n +1 , n ≥ 1, μ ∞ n =1 A n = 1, !5 μ (A n ) = +∞, n ≥ 1, ∞ ∩ n =1 A n = N, 5 μ (A n ) = +∞, A n A m = ∅, n = m, n, m ∈ N. $ A ⊂ R 2 )' & & ) ! &- 5 A = (x, y) ∈ R 2 : x ∈ R, 0 ≤ y ≤ 1 1 + x 2 , 5 A = (x, y) ∈ R 2 : −1 < x < 1, 0 ≤ y ≤ 1 √ 1 − x 2 . $ a ∈ (0, 1) [0, 1] a 2 A 1 = 2 − a 4 , 2 + a 4 ! A 1 ! a 8 ! ' " A 2 ,! ! a 32 " ! , A 3 ' + ! A [0, 1]\ ∞ ∪ n =1 A n & A ! ! & $ ab%" A & [0, 1] ! ! $ A ⊂ [a, b] ! & μ (A) = λ > 0 , f (x) = μ ([a, x) ∩ A) f : [a, b] → R & $ $ [0, 1] [0, 1] ! # & & + c $ % 1 ! A ⊂ R 2 & & ! ab8"ab$" A 1 ⊂ A ! A A 1 & A \A 1 & P 1 , P 2 , . . . , P n ! A \A 1 = n k =1 P k A Download 1.57 Mb. Do'stlaringiz bilan baham: 
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