Question

Show that the function f defined on [1,0] by f(x) = (-1)^n-1 for 1/n+1 <x< = 1/n (for n= 1,2,3,...) is integrable on [0,1]

Answer 1

ANSWER

Since the set of value of the function is $\left \{ -1,1 \right \}$ . then the function is bounded

$|f(x)| = 1$ for all $x\in [-1,1]$ .

In each interval $(\frac {1}{n+1}, \frac {1}{n})$ $(n\in \N)$ the function is constant $(=-1$ or $1)$ , therefore it is continuous. At the points $x_{n}=\frac{1}{n}$ the function is discontinuous , because

$\lim_{x\rightarrow (\frac {1}{2n-1})^{-} }f(x) = \lim_{x\rightarrow (\frac{1}{2n-1})^{-} } (-1)^{ (2n-2) }=1 \\\lim_{x\rightarrow (\frac {1}{2n-1})^{+} }f(x) = \lim_{x\rightarrow (\frac{1}{2n-1})^{+} } (-1)^{ (2n-3) }=-1$

and vice versa for points $x_{2n}$ . Note, that $\lim_{n\rightarrow \infty }x_{n}=0,$ and $\lim_{n\rightarrow \infty}f(x_{n})=\lim_{n\rightarrow \infty}(-1)^{n}$ does not exist. Therefore $x=0$ is also a point of discontinuity of function.

Denote $B$ the set of discontinuity of function. $B= \left \{ 0,1,\frac{1}{2} ,..., \frac{1}{n},...\right \}$ and use

Definition .A set $A\subset \R$ has measure (Lebesgue) zero $(m(A)=0)$ , if for all $\epsilon>0,$ there is a countable collection of open intervals $\left \{ I_{0} .I_{1},..., I_{n},...\right \}$ such that

$A\subset\bigcup_{n=0}^{\infty} I_{n}$ and . $\sum_{n=0}^{\infty} m(I_{n} )<\epsilon$ .

If $I=(\alpha,\beta)$ , then $m(I)=\beta-\alpha.$

Let $A=B, I_{0}=(-\frac{\epsilon}{4}, \frac{\epsilon}{4}), I_{n}=(\frac{1}{n}-\frac{\epsilon}{2^{n+2}}, \frac{1}{n}+\frac{\epsilon}{2^{n+2}})$ . $0\in I_{0}, \frac{1}{n}\in I_{n}.$ So $B\subset\bigcup_{n=0}^{\infty} I_{n}$ ,

$m(I_{n})=\frac{\epsilon}{2^{n+1}} (n=0,1,2,... )$ and $\sum_{n=0}^{\infty} m(I_{n} )=\sum_{n=0}^{\infty} \frac{\epsilon}{2^{n+1}} =\frac{\frac{\epsilon}{2}}{1-\frac{1}{2}} =\epsilon$ . Hence , $m(B)=0.$

Therefore, by the Riemann -Lebesgue Theorem, $f$ is Riemann integrable on $[0,1]$.