Question

Let \(f:[0,1] \rightarrow \mathbb R\)
f(x) = 1if x =(1/n) n is an integer.
f(x) =0 otherwise.
a) Prove that f is integrable
b) Show that \[\int_{0}^{1} f(x) dx = 0\]
Please, help.

Answer 1

I am not sure.

Answer 2

Hmmmm I'm really curious about this one. The function is not continuous anywhere and the limit doesn't exist anywhere.

Answer 3

My attempt: \([0,1]=[0,1/2]\cup [1/2,1]\)
on [1/2,1] , we have f(x) =0 except {1}, but in every each interval, the value of f(x) =0,

Answer 4

hence on (1/2,1) , f(x) is a constant function, it is integrable everywhere.

Answer 5

on [0,1/2] , let \(P=\{0=x_0<\dfrac{1}{x_n}< \dfrac{1}{n-1}<.....< \dfrac{1}{3}< \dfrac{1}{2}\}\) be a partition of [0,1/2]
then for each subinterval \([x_{i-1}, x_i]\), there exists \(x_i^*\) such that \(f(x_i^*) =0\)
For example, on \([1/5,1/4]\) , we have \((1/5)<(7/25)<(1/4)\) and \(f(7/25)=0\)

Answer 6

Let \(m = inf \{f(x) |x \in [x_{i-1}, x_i]\}\) , then Lower sum of f with respect to P is
\(L(f,P)= \sum_{i=1}^n (x_{i-1}-x_i)*m =0\)

Answer 7

Let \(M= sup\{f(x)|x\in [x_{i-1},x_i]\}\) , then M =1, Upper sum of f with respect to P is
\(U(f,P) =\sum_{i=1}^n(x_{i-1}-x_i)\)

Answer 8

but \(||P|| = \dfrac{1}{n^2-n} \) as n--> infinitive, ||P||-->0
hence U(f,P) =0
combine the 2 , we have f is integrable on [0,1]