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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.

Shay
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1 Answers1

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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..

José Carlos Santos
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  • Thanks a lot for the clarification in regards to the counter-example bit :) Much appreciated! In regards to the proof method: unfortunately, I'm not familiar with the concept of the Lebesgue measure of a set, and so was looking to formulate a proof that doesn't require this familiarity. If you have another direction, I'd really appreciate it! – Shay Apr 23 '21 at 13:27
  • I understand that. However, note that you do not need to learn all about Lebesgue measure in order to understande what I wrote. All that you need to know is that asserting that the Lebesgue measur of a set $M$ is $0$ means that, for every $\varepsilon>0$, there is some set ${(a_n,b_n)\mid n\in\Bbb N}$ of open intervals such that $M\subset\bigcup_{n\in\Bbb N}(a_n,b_n)$ and that $\sum_{n=1}^\infty(b_n-a_n)<\varepsilon$. – José Carlos Santos Apr 23 '21 at 13:36
  • I'll definitely take a look at that since it sounds fascinating (and useful, with your answer being direct evidence of that!). However, since I hadn't studied it yet, to use this in my proof I'd have to begin by proving as a "lemma" of sorts that a function is Riemann integrable iff the set of its points of discontinuity has a Lebesgue measure of 0, which would likely make the proof singificantly longer, as well. That's the main reason I was hoping for a proof that doesn't require it. – Shay Apr 23 '21 at 13:51
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    I found one, as you can see. – José Carlos Santos Apr 23 '21 at 13:55
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    I'm glad I could help. – José Carlos Santos Apr 23 '21 at 13:57
  • (I just accidentally deleted my previous comment saying thank you, so thank you!) – Shay Apr 23 '21 at 14:05