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funksional analiz misol va masalalar yechish 1 qism
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' & & A n = n − 1 3 n , n + 1 2 n & ! μ (A) = ∞ n =1 μ (A n ) = ∞ n =1 1 2 n + 1 3 n = 3 2 5 μ (A) = ∞, !5 μ (A) = 1. . 5 μ (A) = 1, 5 μ (A) = 0, !5 μ (A) = ∞, 5 μ (A) = 3, 5, 5 μ (A) = 2 1 3 + 1 e (e − 1) , 5 μ (A) = 2 3 10 . / 5 A 2n−1 = [n − 1, n) , A 2n = [−n, 1 − n) 5 A n = [n, ∞) ∪ (0, 1) . !5 A n = (−∞, n) ∪ N 5 A n = ∞ k =1 k + 1 √ n + k , k + 1 √ n + k − 1 . 5 μ (A) = π, 5 μ (A) = π. μ (A) = 1 − a E (f) = [0, λ]. $ + % B ⊂ [0, 1] ! & , A = {(x, y) : x ∈ B, y = 1} ∪ {(x, y) : x = 1, y ∈ B} & ! pr x A = B, pr y A = B & A \A 1 = P 1 ∪ P 2 ∪ P 3 ∪ P 4 3a#"! 5 P 1 = [0, 4) × [0, 7) , P 2 = [4, 7) × [5, 7) P 3 = [6, 7) × [0, 5) , P 4 = [4, 6) × [0, 3) . μ (A\A 1 ) = 45. '"# ' A \A 1 = P 1 ∪ P 2 . a%"! P 1 = [0, 3) × [0, 7) , P 2 = [0, 7] × {7} . μ(A\A 1 ) = 21 '&# . A \A 1 = P 1 ∪ P 2 ∪ P 3 . a`"! P 1 = [0, 7] × [6, 7] , P 2 = [0, 7] × {0} , P 3 = {7} × [0, 6) μ (A\A 1 ) = 7 '$# %/ : μ (A ∪B) = μ(A)+μ(B)−μ(A∩B) A ∪B & - A ∪ B = (A \ (A ∩ B)) ∪ (B \ (A ∩ B)) ∪ (A ∩ B). ' ! μ (A ∪ B) = μ(A \ (A ∩ B)) + μ(B \ (A ∩ B)) + μ(A ∩ B) (5.1j) ( ! E A & ! A ⊂ E μ (E \ A) = μ(E) − μ(A) 2 μ (A \ (A ∩ B)) = μ (A) − μ(A ∩ B), μ(B \ (A ∩ B)) = μ(B) − μ(A ∩ B) ' 3a#+5 μ (A ∪ B) = μ(A) + μ(B) − μ(A ∩ B) : - μ (AΔB) = μ((A ∪ B) \ (A ∩ B)) = μ(A ∪ B) − μ(A ∩ B). (5.2j) 3a%+5 μ (A ∪ B) μ (A) + μ(B) − μ(A ∩ B) μ (AΔB) = μ (A) + μ(B) − 2μ(A ∩ B) %$ 5 μ (A) = 1. 5 μ (A) = 2. !5 μ (A) = 1. 5 μ (A) = 1. %% 5 μ (A) = 11 − 2e 5 μ (A) = 3 !5 μ (A) = 4 + 2e 5 μ (A) = 5 %& d ! & ! ' ! [−1, 1] ! - x y x − y ∈ Q ' 2 ! [−1, 1] K (x), x ∈ [−1, 1] : + ' : 2 [−1, 1] K (x), x ∈ [−1, 1] : + [ " : & A A & ! & %' . ! ! A ⊂ [0, 1) B ⊂ [1, 2) ! & , A ∩ B = ∅ ! & A ∩ A = A ! & %. . ! ! A 0 = [0, 1) ! & A = A 0 ∪ [1, 2), B = A 0 A \B = [1, 2) ! & * A = A 0 ∪[1, 2), B = [1, 2) A \B = A 0 ! & &/ . ! ! & & * A ∩ B = ∅ ! & * B ⊂ A A ∩ B = B ! & * A ⊂ B A ∩ B = A ! & & * A ⊂ B A \B = ∅ ! & * B ⊂ A A \B ! & & A ∩ B = ∅ B \A = B ! & B ⊂ A B \A = ∅ ! & A ⊂ B B \A ! & &$ A ∩ B = ∅ ! & B ⊂ A A ⊂ B A ΔB & ! '$ = & '% ab%" a = 0, 1 μ (A) = 0, 9 ' A & ! ! & '& + '' A = K ∪ ([0, 1] ∩ Q) ' K = & '. 2 ./ A = K, B = K ' K = & . A = B = K. ' K = & K + K = [0, 2] #`a" . a##" ( μ F (A) = F (5) − F (1) = 2 · 5 + 1 − (2 · 1 + 1) = 11 − 3 = 8. . μ (A) = 6. .& : #: a$7" .' X = [0, 1] Q S m X (a, b) [a, b] [a, b), (a, b] &" S m * A ab = X (a, b) ( [a, b], (a, b], [a, b)) A ab & m (A ab ) = b−a & m : S m → R + σ − ! .. #5 m & R ! & ' %5 m (A) ≥ 0 m ! `5 A B & μ (A ∪ B) = n ∈N (A B ) 1 2 n = n ∈N A 1 2 n + n ∈N B 1 2 n = μ(A) + μ(B) ' ! ! μ (A) = 0 μ (B) = 0, 875. 0 "4- " 5 * #"d %"* `"d b"\ a"\ 7"* 8"* W"d $"' #>"d ##"d #%"\ #`"' #b"' #a"* #7"\ #8"d #W"* #$"' %>"\ %#"' %%"* %`"d %b"' %a"d %7"' %8"* %W"d %$"' `>"d `#"d `%"\ ``"* `b"\ `a"' `7"d `8"* `W"' `$"d b>"\ 6;6< =6< , ! / . ! / ! . ! (; " X ! & " & ! " " / (; * " ! ! % § # * ,-0) ' & ) ) ( " ! ! . ! / " ' E ⊂ R (E ⊂ R 2 ) / ( ! & f %7, ! c ∈ R {x ∈ E : f(x) < c} := E(f < c) f E % f : E → R, f(x) = a = const ! ( ! "# c ∈ R ! E (f < c) = {x ∈ E : f(x) < c} = ⎧ ⎨ ⎩ E, agar c > a, ∅, agar c ≤ a E ∅ & ! \ c ∈ R ! E (f < c) & ! 7#" ( f (x) = a E ! % ( ! 5 f (x) = [x], x ∈ [0, 2). 5 f (x) = {x}, x ∈ [0, 2). !5 f (x) = 2x + 3, x ∈ [0, 3] 5 f (x) = x 2 − 5, x ∈ [−2, 3]. 5 f (x) = 2 x − 1, x ∈ [0, 2]. 5 f (x) = ln(x + 1), x ∈ [0, 2). 5 f (x) = sin x + 5, x ∈ [0, π]. 5 f (x) = cos x + 5, x ∈ [−π, 0]. ! "# ' 5 ! c ∈ R ! E (f < c) = {x ∈ E : [x] < c} = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ ∅, agar c ≤ 0 [0, 1), agar 0 < c ≤ 1 [0, 2), agar c > 1 ∅ [0, 1) [0, 2) & ! \ c ∈ R ! E (f < c) & ! 7#" ( f (x) = [x] E e [0, 2) ! % . ! ! "# E ! & A ⊂ E ! & < - f (x) = ⎧ ⎨ ⎩ − 1, agar x ∈ A 1, agar x ∈ E\A . (6.1) ' ! E (f < 0) = A ! & \ f E ! % * A ⊂ E ! & g (x) = χ E \A (x) ! 1 & (; E (g < 0, 5) = A ! & \ g : E → R ! %$ * f E & ! a, b ∈ R ! & ! - 1) E(f ≥ a); 2) E(a ≤ f < b); 3) E(f = a); 4) E(f ≤ a); 5) E(f > a). * f ! ( " a ∈ R ! E (f < a) & ! #5 E (f ≥ a) = E\E(f < a) ! & ! ! E (f ≥ a) & ! ! %5 E (a ≤ f < b) = E(f ≥ a) ∩ E(f < b) ! & ! E (a ≤ f < b) & ! ! `5 E (f = a) & ! Download 1.57 Mb. Do'stlaringiz bilan baham: |
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