Prove that the function fx 1 if 0 lessthanorequalto x Solut

Prove that the function f(x) = {-1, if 0 lessthanorequalto x

Solution

Given that

f(x) = -1 for 0<= x < 1/3

f(x) = 0 for 1/3 <= x < 2/3

   f(x) =1 for 2/3 <= x <= 1

f be bounded funtion on [0,1]

let p = {0=x₀<x₁<x₂<x₃=1} be partition on [0,1]

here x₀ =0, x₁ =1/3, x₂=2/3, x₃=1

let m_i , M_i   are infimum and suprimum of f on [x_i-1, x_i]

m₁= infimum of f on [0, 1/3] = -1

m₂= infimum of f on [1/3, 2/3] = 0

m₃ = infimum of f on [2/3, 1] =1

lower reman sum= S_p(f)= m₁ (x₁-x₀) + m₂ (x₂-x₁) + m₃ (x₃-x₂)

   =-1 (1/3 -0) + 0 (2/3 -1/3) + 1 (1- 2/3)

   =-1/3 + 1/3

   =0

lower reman integral = sup(S_p(f))

= sup(0)

   =0

M₁ = supremum of f on [0,1/3] =-1

M₂ = supremum of f on [1/3, 2/3] = 0

M₃ = supremum of f on [2/3, 1] = 1

I_p(f) =M₁ (x₁-x₀) + M₂ (x₂-x₁) + M₃ (x₃-x₂)

=-1 (1/3 - 0) + 0 (2/3-1/3) + 1 (1- 2/3)

   =-1/3 +1/3

=0

Upper reman intigral = Inf(I_p(f))

   =Inf(0)

   =0

lower reman integral = upper reman integral

so , f(x) is reman integrable on [0,1]