Question

12. Let f be integrable on a closed interval [a, b]. Suppose that there is a real number C such that f(x) 2C for all E a, b (1) Prove that if C>0, then 7 is also integrable on la,b] (6 Marks) (2) If C 0, i, still integrable (assuming f(x)关0 for any x E [aM)? If yes, supply a short proof. If no, give a counterexample. (6 Marks)

Answer 1

1) Recall that a function is Riemann integrable on if for every $\epsilon>0$ there is a partition $P_\epsilon=\{a=x_0<x_1<\cdots<x_n=b\}$ such that $U({P_\epsilon},g)-L({P_\epsilon},g)<\epsilon$ ; here

UP, f)=〉' (x, - ri-1) sup g(x) 7l xi-xi-1

Fix $\epsilon>0$; since is integrable, there is a partition $P_\epsilon=\{a=x_0<x_1<\cdots<x_n=b\}$ such that

$U({P_\epsilon},f)-L({P_\epsilon},f)<{\frac{C^2\epsilon}{2}}\hspace{2cm}(\ast)$

Then

(zi-ri-1) re [inf TiTi-1sup f(x) 2ー 7l Ti-T 2-1

For each $i=1,\cdots,n$ let $\xi_i,\zeta_i\in [x_{i-1},x_i]$ be such that

sup inf

This is by definition of sup and inf. Now, we get

$\begin{align*}U({P_\epsilon},1/f)-L({P_\epsilon},1/f)&=\sum_{i=1}^n(x_i-x_{i-1})\begin{pmatrix}\sup_{x\in [x_{i-1},x_i]}{\frac 1{f(x)}}-\inf_{x\in [x_{i-1},x_i]}{\frac 1{f(x)}}\end{pmatrix}\\ &<\sum_{i=1}^n(x_i-x_{i-1})\begin{pmatrix}{\frac 1{f(\xi_i)}}+{\frac{\epsilon}{4(b-a)}}-{\frac 1{f(\zeta_i)}}-{\frac{\epsilon}{4(b-a)}}\end{pmatrix}\\ &=\sum_{i=1}^n(x_i-x_{i-1})\begin{pmatrix}{\frac 1{f(\xi_i)}}-{\frac 1{f(\zeta_i)}}\end{pmatrix}+\sum_{i=1}^n(x_i-x_{i-1}){\frac{\epsilon}{2(b-a)}}\\ &=\sum_{i=1}^n(x_i-x_{i-1})\begin{pmatrix}{\frac {f(\zeta_i)-f(\xi_i)}{f(\xi_i)f(\zeta_i)}}\end{pmatrix}+{\frac{\epsilon}{2(b-a)}}\sum_{i=1}^n(x_i-x_{i-1})\\ &<\sum_{i=1}^n(x_i-x_{i-1})\begin{pmatrix}{\frac {f(\zeta_i)-f(\xi_i)}{C^2}}\end{pmatrix}+{\frac{\epsilon}{2(b-a)}}(b-a)\\ &\leq {\frac 1{C^2}}\sum_{i=1}^n(x_i-x_{i-1})\begin{pmatrix}\sup_{x\in [x_{i-1},x_i]}f(x)-\inf_{x\in [x_{i-1},x_i]}f(x)\end{pmatrix}+{\frac{\epsilon}2}\end{align*}$

Using $(\ast)$ we get

f) - L(P 1 Cee C2 2 2

Thus, we have shown that for any $\epsilon>0$ there is a partition $P_\epsilon=\{a=x_0<x_1<\cdots<x_n=b\}$ such that

$U({P_\epsilon},1/f)-L({P_\epsilon},1/f)<\epsilon$

Therefore, is Riemann integrable.

2) We let $[a,b]=[0,1]$ and $C=-1$. Let be the following function on

-x for xメ0 2 for x=0 f(x) = f(x)=

Then is bounded and continuous on except at the point $x=0$. Therefore, it is Riemann integrable, and we have

$\begin{align*}\int_0^1f(x)~dx&=-\int_0^1x~dx\\ &=-{\frac 12}\end{align*}$

Notice that $\begin{align*}f(x)\geq -1\end{align*}$ for all $x\in [0,1]$. We claim that is not integrable. Notice that

${\frac 1f}(x)=\left\{ \begin{array}{rcl} -{\frac 1x} & \mbox{for} & x\neq 0 \\ {\frac 12} & \mbox{for} & x=0 \end{array}\right.$

Thus, is continuous on except at $x=0$. Therefore, if it was integrable, we would have

$\begin{align*}\int_0^1{\frac 1f}(x)~dx&=-\int_0^1{\frac 1x}~dx\\ &=-\ln x\mid_0^1\\ &=\ln(0)-\ln(1)\\ &=\ln(0)\end{align*}$

which is absurd since
In(0)
is not even a real number. Hence, is not integrable.

This is the counter-example that shows that the converse statement as in part 2) of question I FALSE.