SOLUTIONS TO ASSIGNMENT 2 - MATH 355. with c > 3. m(n c ) < δ. f(t) t. g(x)dx =

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1 SOLUTIONS TO ASSIGNMENT 2 - MATH 355 Problem. ecall ha, B n {ω [, ] : S n (ω) > nɛ n }, and S n (ω) N {ω [, ] : lim }, n n m(b n ) 3 n 2 ɛ 4. We wan o show ha m(n c ). Le δ >. We can pick ɛ 4 n c n wih c > 3 δ n n 3/2. Then, m(b n ) 3 c n n < δ. n3/2 Now, if ω N c hen here exiss µ > such ha for all n N, S n (ω) nµ. There exiss N N such ha ɛ N < µ; hence ω B N. Thus, N c n B n and Since δ > is arbirary, m(n c ).. m(n c ) < δ. Problems from he exbook. Pages Problem 4. Suppose f is inegrable on [, b] and g(x) b < x b. Prove ha g is inegrable on [, b] and b x g(x)dx d, b d. Soluion. We can assume ha f(x) (and hence g(x) ), since f f + f and he Lebesgue inegral is linear.

2 2 SOLUTIONS TO ASSIGNMENT 2 - MATH 355 b g(x)dx b b b b x ddx b d, b b χ {x } (x)dx d χ { x} ()dxd b where we used Tonelli s heorem o swich he order of inegraion. dx d Problem 5. Suppose F is a closed se in, whose complemen has finie measure, and le δ(x) inf{ x y : y F }. Le I(x) : x y 2 dy. (a) Show ha, for all x, x, δ(x) δ(x ) x x. Soluion. Le ɛ >. There exiss y F such ha x y δ(x) + ɛ. Then, δ(x ) x y x x + x y x x + δ(x) + ɛ so δ(x) δ(x ) x x +ɛ. Swiching he roles of x and x we ge δ(x ) δ(x) x x +ɛ, so δ(x) δ(x ) x x + ɛ, and he resul follows. (b) Show ha I(x) + for all x F. Soluion. Le x F. Since F is closed, we have δ(x) a >. From he Lipschiz condiion we have ha δ(x) a 2 whenever x y < a 2. Then, x+a/2 x y 2 dy a 2 x a/2 a/2 a/2 x+a/2 x y 2 dy x a/2 dy, y 2 a/2 x y 2 dy where we used ha δ(x) a 2 iff δ(x) a 2 a 2 + δ(x), so a 2. (c) Show ha I(x) < for a.e. x F. Soluion. If y F we have. Le y F. For all x F, we have x y. Thus,

3 SOLUTIONS TO ASSIGNMENT 2 - MATH Then, F y x y 2 dx I(x)dx x y 2 dx + y+ x 2 2. x y 2 χ F (x)dy dx F c F F c 2 dy 2m(F c ) <. x y 2 dx dx dy x y 2 Again, we could swich he order of inegraion hanks o Tonelli s heorem (I(x) ). Since I(x)dx <, I(x) < for almos all x. Problem 6. Inegrabiliy of f on does no necessarily imply he convergence of f(x) o as x. (a) There exiss a posiive coninuous funcion f on such ha f is inegrable on, ye lim sup x f(x). Soluion: One example is 2n 4 x 2n(n 4 ) x [n, n], n n 3 f(x) 2n 4 x + 2n(n 4 + ) x [n, n + ], n ; n 3 oherwise. Here, he graph of f consiss on riangles of area n. Then fdx 2 n <. n 2 2n 2 cenred a every posiive ineger (b) However, if we assume ha f is uniformly coninuous on and inegrable, hen lim x f(x). Soluion. Le ɛ >. Since f uniformly coninuous, here exiss δ > such ha f(x) f(y) < ɛ 2 whenever x y < δ. Take δ < 2 Suppose f(x) as x. Then, here exiss x, s.. f(x ) ɛ. hus, f(y) ɛ 2 for all y (x δ, x + δ). Now, here exiss x 2 > x + s.. f(x 2 ) ɛ and hus f(y) ɛ 2 for all y (x 2 δ, x 2 + δ) Noe ha (x δ, x + δ) (x 2 δ, x 2 + δ). Coninue his way so as o find a sequence x n, x n+ > x n +, such ha f(y) ɛ 2 whenever y (x n δ, x n + δ). Therefore, f(x) dx f(x) dx δɛ. n (x n δ,x n+δ) n

4 4 SOLUTIONS TO ASSIGNMENT 2 - MATH 355 Problem 8. If f is inegrable on, show ha is uniformly coninuous. F (x) x d Soluion. Le ɛ > and δ : ɛ ( f d) <, since f inegrable. Le x, y such ha x < y, x y < δ. Then, F (x) F (y) y Problem 5. Consider he funcion defined over by f(x) x d x y d < ɛ. { x /2, if < x < ;, oherwise. For a fixed enumeraion {r n } of he raionals Q, le F (x) 2 n f(x r n ). n Prove ha F is inegrable, hence he series defining F converges for almos every x. However, observe ha his series is unbounded on every inerval, and in fac, any funcion F ha agrees wih F a.e. is unbounded in any inerval. Soluion. We firs compue fdx. The improper iemann inegral of f is given by x dx 2 x 2. We know ha he proper iemann and Lebesgue inegrals coincide. Then, f(x) dx 2 n f(x)dx ( n+, n ] ( n n n n + ) 2. n n+ f(x)dx

5 SOLUTIONS TO ASSIGNMENT 2 - MATH Now, by Monoone convergence heorem and ranslaion invariance of he Lebesgue measure, N F (x) dx lim 2 n f(x r n ) dx 2 n f(x r n ) dx N n n 2 n f(x) dx 2 n+ 2. n Thus F is inegrable and herefore finie a.e.. Le I be an inerval. Since raionals are dense in here exiss N N such ha r N I. Since F (x) is a sum of posiive erms, n lim F (x) lim 2 N f(x r N ) +. x r N x r N Le F coincide wih F almos everywhere. Le M >. Since lim x rn F (x) +, There exiss ɛ > such ha F (x) > M for all x (r N ɛ, r N + ɛ). Then F (x) > M almos everywhere on (r N ɛ, r N + ɛ), so F (x ) > M for some x (r N ɛ, r N + ɛ). Since M is arbirarily large, F is unbounded on I. Problem 7. Suppose f is defined on 2 as follows a n, n x < n + and n y < n +, n ; f(x, y) a n n x < n + and n + y < n + 2, n ;, oherwise. Here, a n k n b k, wih {b k } a posiive sequence such ha k b k s. (a) Verify ha each slice f y (x) and f x (y) is inegrable. Also, for all x, f x (y)dy, and hence f(x, y)dy dx. Soluion. Le y. There exiss a unique n N such ha n y < n +. Then f y (x) dx [n,n) f y (x) dx + f y (x) dx a n + a n. [n,n+) Le x. There exiss a unique n N such ha n x < n +. Then Moreover, f x (y) dy [n,n+) f x (y) dy + f x (y) dy 2a n. [n+,n+2)

6 6 SOLUTIONS TO ASSIGNMENT 2 - MATH 355 I follows ha, (b) For y <, If y, Hence f x (y)dy (c) By Tonelli, [n,n+) f y (x)dx f x (y)dy + f x (y)dy a n a n. [n+,n+2) f(x, y)dy dx f y (x)dx [n,n) [,) f x (y)dy dx f y (x)dx a. f y (x)dx + f y (x)dx a n a n. [n,n+) ( f y (x)dx dy f y (x)dx + n [n,n+) [n,n) [a n a n ] b n s. since a n is increasing. n f(x, y) dxdy n [n,n+) n f x (y) dy dx [n,n+) f y (x)dx 2a n, Problem 9. Suppose f is inegrable in d. For each α > le E α {x : f(x) > α}. Prove ha + f(x) dx m(e α )dα. d n ) dy Soluion: Firs noe ha Then, by Tonelli, f(x) f(x) dα.

7 SOLUTIONS TO ASSIGNMENT 2 - MATH f(x) dx d d + + f(x) dα dx d + d χ {α< f(x) } (x)dx dα m(e α )dα. χ {α< f(x) } (α)dα dx