I've been taking a look at some proposed proofs for the following theorem:
If $f$ is Riemann integrable in $[a,b]$ and is bounded from zero ($\inf\{|f(x)|,x\in[a,b]\} > 0$), then $\frac{1}{f}$ is also Riemann integrable there.
For convenience, let's denote $g(x)=\frac{1}{f(x)}$, and also for some partition $P=\{a<x_1<x_2<\dots<x_{n-1}<b\}$: $M_i=\sup\{f(x)|x\in[x_{i-1},x_i]\}$ and $m_i=\inf\{f(x)|x\in[x_{i-1},x_i]\}$, and $G_i$ and $g_i$ are the same except for $\frac{1}{f}$.
The ways I've been coming up with to deal with the fact that $f$ could have discontinuities & hop between positive and negative values were becoming increasingly ugly and I came here to see smarter alternatives, but about 3 different answers I saw were sort of ignoring that problem altogether in a way that doesn't seem right to me.
The problem I had is this: I want to say that $M_i=\frac{1}{g_i}$ and $m_i=\frac{1}{G_i}$ or, really, anything in a similar vein, but it just doesn't seem true: if $f$ does hop between positive & negative value in a given interval created by $P$ then the infimum of $g$ in that interval will be the reciprocal of the supremum of the negative values of $f$ in that interval, unless I'm missing something, and that isn't guaranteed to be (or rather: isn't going to be, I think, in most cases in which $f$ does behave as described) $M_i$ or $m_i$. A similar deal going on with the supremum, of course.
To quote one of the aforementioned answers I mentioned which seem to ignore this:
https://math.stackexchange.com/a/1106308
Quote:
It should be easy to see that Gi,gi are 1/Mi,1/mi but not necessarily in that order.
Here's a counter-example (unless I'm doing something really silly here, which certainly is a possibility):
$f(x)=\begin{cases} x &\text{} x\in [1,2]\\-x &\text{} x\in(2,3]\end{cases}$
$\Rightarrow g(x)=\frac{1}{f(x)}=\begin{cases} \frac{1}{x} &\text{} x\in [1,2]\\\frac{-1}{x} &\text{} x\in(2,3]\end{cases}$
Clearly, $f$ is Riemann integrable in $[1,3]$, and it is also bounded from zero in this interval, so the conditions are fulfilled.
With respect to the partition $P=\{1<3\}$, for example, we have $M_1=2$, $m_1=-3$, but $G_1=1$, $g_1=-\frac{1}{2}$. $G_1$ for example isn't the reciprocal of either of the two.
So my main question here is: am I missing something? Are any of the things I described above off, somehow?
And in addition: how would you handle this proof? Things I had considered are:
- Define a refinement of $P$ which includes all discontinuity points of $f$ to prevent this kind of "hopping" behaviour I mentioned, but there's nothing to promise me the amount of discontinuities is finite, and the definition of a partition I've been working with requires a finite amount of points.
- Figure out some slightly more complex expression to describe the actual supremum/infimum ($G_i$/$g_i$) we're going to end up with, but then the transition from $G_i-g_i$ to $\frac{f(x)-f(y)}{c^2}$ for some nonzero $c$ becomes significantly less nice. I might be able to still move forward with this somehow but it feels messy.
There were several other methods I had considered, some of which might have even worked, but they all seemed incredibly tedious and messy and I feel like I've been overcomplicating this. I'd really appreciate some help with clearing up this mess going on in my mind right now.
EDIT: I'm not familiar with the concept of a set's Lebesgue measure, so I'd prefer a solution that doesn't make use of it, if possible.
Your example is fine.
I would prove the statment as follows: since $f$ is bounded away from $0$, $\frac1f$ is bounded. And, since $f$ is Riemann-integrable, the Lebesgue measure of the set of its points of discontinuity is $0$. But then the set of points of discontinuity of $\frac1f$ also has Lebesgue measure $0$, and therefore $\frac1f$ is Riemann-inegrable..