# Integrate this function

• Oct 30th 2008, 03:01 PM
Integrate this function
Let $\displaystyle f: \mathbb {R} \rightarrow \mathbb {R}$ be defined by $\displaystyle f(x) = 1$ if $\displaystyle x= \frac {1}{n} \ \ \ n \in \mathbb {N}$, and $\displaystyle f(x) = 0$ otherwise.

Prove that $\displaystyle \int ^1 _0 f$ exist and find its value.

Question so far:

Define a partition P on [0,1] with $\displaystyle P = \{ a_0=0 , a_1,a_2,...,a_{N-1},a_N=1 \}$

Define $\displaystyle M_j = sup \{ f(x) : x \in [x_j,x_{j+1}]$
$\displaystyle m_j = inf \{ f(x) : x \in [x_j,x_{j+1}]$

So I need to show that for any given $\displaystyle \epsilon > 0$, we will have $\displaystyle U(f,P)-L(f,P)< \epsilon$, but how should I process? Thanks.
• Oct 30th 2008, 07:50 PM
NonCommAlg
Quote:

Let $\displaystyle f: \mathbb {R} \rightarrow \mathbb {R}$ be defined by $\displaystyle f(x) = 1$ if $\displaystyle x= \frac {1}{n} \ \ \ n \in \mathbb {N}$, and $\displaystyle f(x) = 0$ otherwise.

Prove that $\displaystyle \int ^1 _0 f$ exist and find its value.

obviously $\displaystyle L(f,P)=0,$ for any partition $\displaystyle P$ of [0,1]. so: $\displaystyle \int_{-} f = 0.$ suppose $\displaystyle n \geq 2.$ let $\displaystyle r= \frac{1}{2n^2}$ and define the partition $\displaystyle P_n$ by:

$\displaystyle x_0=0, \ \ x_1=\frac{1}{n}, \ \ x_2=\frac{1}{n} \ + \ r, \ \ x_{2i+1}=\frac{1}{n-i} \ - \ r, \ \ x_{2i+2}= \frac{1}{n-i} \ + \ r, \ \ 1 \leq i \leq n-2,$ and $\displaystyle x_{2n-1}=1-r, \ \ x_{2n}=1.$ see that

$\displaystyle U(f,P_n)=\frac{2n-1}{n^2}$ and thus: $\displaystyle 0 \leq \int^{-} f - \int_{-} f = \int^{-} f \leq U(f,P_n) = \frac{2n-1}{n^2}.$ therefore $\displaystyle f$ is integrable and $\displaystyle \int_0^1 f = 0. \ \ \ \Box$
• Nov 2nd 2008, 06:18 AM
$\displaystyle U(f,P_n) = \sum ^{2n-1}_{j=0}[sup \{ f(x) : x \in [x_j,x_{j+1}](x_{j+1}-x_j),$ but how does this equals to $\displaystyle \frac{2n-1}{n^2}$?