Integrability of 1/f

Question

Let $f$ be integrable on $[a,b]$. Suppose there are constants $M$ and $N$ such that $0 < M \leq |f(x)| \leq N$, $\forall x \in [a,b]$. Prove $1/f$ is integrable on $[a,b]$. So far I know that $1/|f(x)|$ is not undefined at any points since $|f(x)| \geq M>0$. But I'm not sure of what's next. In terms of Riemann integrability, I was considering that the supremum of {$f(x):x \in [x_{i-1},x_i]$} is now the infimum, and vice versa, and the lower sum becomes the upper sum, but I am not confident in this idea. Any help is appreciated.

Answer 1

Since $f$ is integrable over $[a,b]$, for any $\varepsilon >0$ we can find a partition $P$ such that $$\sum_{i=1}^n \omega_i(f)\Delta x_i<\varepsilon$$ where $$\omega_i=\sup_{x,y\in [x_{i-1},x_i]}|f(x)-f(y)|=\sup_{x\in [x_{i-1},x_i]}f(x)-\inf_{x\in [x_{i-1},x_i]}f(x)$$

Now, note that from $M\leqslant f$ we get $1/f$ is defined and bounded on $[a,b]$, and

$$\left|\frac{1}{f(x)}-\frac 1{f(y)}\right|=\frac{|f(y)-f(x)|}{|f(x)f(y)|}<M^{-2}|f(y)-f(x)|$$

Thus under these hypothesis, if you know you can make $\displaystyle\sum_{i=1}^n \omega_i(f)\Delta x_i$ small, then you can make $\displaystyle\sum_{i=1}^n \omega_i(f^{-1})\Delta x_i$ small.