Prove the reciprocal of a function is integrable if it is bounded

Let $f(x)$ be Riemann integrable on $[a,b]$ and satisfy $\inf\{\left|f(x)\right|:a\le x \le b\}=d>0$.

How can we prove that $g(x)=\frac{1}{f(x)}$ is Riemann integrable?

I know from the statement that $f(x)\neq 0$ since its $\inf$ is $d>0$, so $g(x)$ is bounded on $[a,b]$ since $f(x)$ is bounded and there is no division by $0$.

Is there a relationship like $M(f)=m(g)$ and $m(f)=M(g)$, since $g(x)$ is the reciprocal?

Solution 1:

A direct proof is what the OP needs here. Rather than giving a full proof I will give an outline. In order to show that some function $f$ is integrable in Riemann sense on $[a, b]$ we need two things:

1) show that $f$ is bounded on $[a, b]$ otherwise Riemann integral is not defined at all.

2) For every $\epsilon > 0$ there must be a partition $P = \{x_{0} = a, x_{1}, \ldots, x_{n} = b\}$ of $[a, b]$ such that $$\sum_{i = 1}^{n}(M_{i} - m_{i})(x_{i} - x_{i - 1}) < \epsilon$$ where $M_{i} = \sup\, \{f(x)| x \in [x_{i - 1}, x_{i}]\}, m_{i} = \inf\, \{f(x)| x \in [x_{i - 1}, x_{i}]\}$.

Now it is given that $f$ is Riemann integrable (so that condition mentioned in second point above is true) and also it is bounded away from zero so that $g = 1/f$ is bounded. Now let $G_{i} = \sup\, \{g(x)| x \in [x_{i - 1}, x_{i}]\}, g_{i} = \inf\, \{g(x)| x \in [x_{i - 1}, x_{i}]\}$. It should be easy to see that $G_{i}, g_{i}$ are $1/M_{i}, 1/m_{i}$ but not necessarily in that order. Now we can see that $$G_{i} - g_{i} = \frac{M_{i} - m_{i}}{|M_{i}m_{i}|} < \frac{M_{i} - m_{i}}{d^{2}}$$ This should help you to establish the second condition for integrability for $g$.