If $f:[0,1]\to [0,1]$ is Riemann integrable on $[0,1]$, show that $g(x)=\frac{1}{f(x)-2}$ is bounded and Riemann integrable on $[0,1]$.

Question

If $f:[0,1]\to [0,1]$ is Riemann integrable on $[0,1]$, show that $g(x)=\frac{1}{f(x)-2}$ is bounded and Riemann integrable on $[0,1]$.

I can say that $g(x):[0,1]\to [-2,-1]$. How to show the boundedness and integrability of $g(x)$ without using continuity or monotonic property of functions? Please help me.

Answer 1

Assume that $f$ is Riemann integrable, then given $\epsilon>0$, there exists a partition $P=\{0=x_{0}<\cdots<x_{n}=1\}$ on $[0,1]$ such that \begin{align*} U(f,P)-L(f,P)=\sum_{i=1}^{n}(M_{i}-m_{i})\Delta x_{i}<\epsilon, \end{align*} where $\Delta x_{i}=(x_{i}-x_{i-1})$, $M_{i}=\sup_{x\in[x_{i-1},x_{i}]}f(x)$, $m_{i}=\inf_{x\in[x_{i-1},x_{i}]}f(x)$.

Note that $M_{i}-2<0$ and it is not hard to see that for all $x\in[x_{i-1},x_{i}]$, we have \begin{align*} \dfrac{1}{M_{i}-2}\leq\dfrac{1}{f(x)-2}\leq\dfrac{1}{m_{i}-2}, \end{align*} so with $\widetilde{M}_{i}=\sup_{x\in[x_{i-1},x_{i}]}g(x)$, $\widetilde{m}_{i}=\inf_{x\in[x_{i-1},x_{i}]}g(x)$, we have \begin{align*} \widetilde{M}_{i}-\widetilde{m}_{i}\leq\dfrac{1}{m_{i}-2}-\dfrac{1}{M_{i}-2}. \end{align*} And we have $-2\leq M_{i}-2,m_{i}-2\leq -1$, so \begin{align*} \dfrac{1}{(M_{i}-2)(m_{i}-2)}\leq 1, \end{align*} and hence \begin{align*} U(g,P)-L(g,P)&=\sum_{i=1}^{n}(\widetilde{M}_{i}-\widetilde{m}_{i})\Delta x_{i}\\ &\leq\sum_{i=1}^{n}\dfrac{M_{i}-m_{i}}{(M_{i}-2)(m_{i}-2)}\Delta x_{i}\\ &\leq\sum_{i=1}^{n}(M_{i}-m_{i})\Delta x_{i}\\ &<\epsilon. \end{align*}

Answer 2

Your function is bounded because if $f(x)\in[0,1]$ then $g(x)\in \left[-1,\dfrac{-1}{2}\right]$. Recall that $f$ is integrable if and only if $\Lambda(f)$ (the set where $f$ is discontinuous) have measure zero. Note that $g(x)=h\circ f$ where $h(x)=\dfrac{1}{x-2}$. Recall that $h\circ f$ is continuous if $h$ and $f$ are both continuous. Clearly $h$ is continuous, thus, $h\circ f$ is discontinuous iff $f$ is discontinuous. We conclude that $\Lambda(f)=\Lambda(h\circ f)$ and therefore, $g$ is integrable.

Answer 3

For a function $f:[0,1]\to [0,1]$ we have $ 2-f(x) \in [1,2]$ therefore $\frac {1}{2-f(x)} \in [1/2,1]$

Thus $g(x) = \frac {1}{2-f(x)} $ is bounded.

Note that the discontinuities of g(x) are the same as discontinuities as f(x), hence g(x) is also Riemann integrable.