For b For p the conclusion fails Take E R and f k \u03c7 kk f 1 Then f k f L R f k f

For b for p the conclusion fails take e r and f k χ

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For (b): For p = , the conclusion fails. Take E = R and f k = χ ( k,k ) , f = 1 . Then f k , f L ( R ) , f k f a.e. on R , and k f k k = k f k = 1 . But we do not have k f k f k 0 as k → ∞ . Another example is E = (0 , 1) , f k = χ (1 /k, 1) , f = 1 , k N . Then we have the same conclusion. ¤ 7. (15 points) Let E = (0 , ) × (0 , 1) R 2 and f ( x, y ) = ye xy sin x, ( x, y ) E. (a) (5 points) Show that f ( x, y ) is integrable on E, i.e., f L ( E ) . (b) (10 points) Evaluate the integral ZZ E f ( x, y ) dxdy. 3
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solution: For (a), note that ZZ E | f ( x, y ) | dxdy ZZ E ye xy dxdy and by Tonelli Theorem we have ZZ E ye xy dxdy = Z 1 0 Z 0 ye xy dx ¸ dy = Z 1 0 dy = 1 which implies that f L ( E ) . For (b), we apply Fubini Theorem and get ZZ E f ( x, y ) dxdy = Z 1 0 y μZ 0 e xy sin xdx dy. Recall the elementary formula Z e ax sin bxdx = e ax ( a sin bx b cos bx ) a 2 + b 2 + C, a, b are constants we see that for each y > 0 Z 0 e xy sin xdx = e xy ( y sin x cos x ) y 2 + 1 ¯ ¯ ¯ ¯ x = x =0 = 1 y 2 + 1 and so Z 1 0 y μZ 0 e xy sin xdx dy = Z 1 0 y y 2 + 1 dy = 1 2 log 2 . ¤ 8. (20 points) Answer Yes or No to each statement. Just provide the answers. (a) If x is a point of density of a measurable set E, then x E. ANS: _________ (b) Let 0 < p < . If k f k f k p,E 0 as k → ∞ , then there exists a subsequence f k j f almost everywhere in E as j → ∞ . ANS: _________ (c) Let f and g be two nonnegative measurable functions on measurable set E satisfying |{ x E : f ( x ) > y }| = |{ x E : g ( x ) > y }| for all y > 0 . Then we must have R E f = R E g. ANS: _________ (d) k f k f k ,E 0 as k → ∞ if and only if f k f uniformly on E except on a set Z E of measure zero. ANS: _________ solution: The answers to the five problems are: NO, YES, YES, YES. ¤ 4
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  • Fall '08
  • Akhmedov,A
  • dx, lim sup

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