# For measurable $f: \mathbb{R} \rightarrow \mathbb{R}$ prove $f(x)$ and $\frac{1}{f(1/x)}$ cannot both be Lebesgue integrable.

First question on MSE! I'd appreciate hints, theorem suggestions, or method suggestions regarding the question in the title or below. Please avoid full solutions. I'm studying for an exam coming up and got stuck on this question:

Problem Let $$f: \mathbb{R}\rightarrow \mathbb{R}$$ be a measurable function. Prove that $$f(x)$$ and $$\frac{1}{f(1/x)}$$ cannot both be Lebesgue integrable.

I've taken courses based on and read from Royden & Fitzpatrick if that helps with suggestions.

My attempts so far have focused on trying to find contradictions assuming $$f$$ is integrable: i.e. $$\int_{\mathbb{R}} |f| < \infty$$ and defining $$S_0 := \{x \in \mathbb{R} | f(x) = 0 \}$$. I'm thinking that something is happening with zeros and infinities that destroys the measurability of the alternative function.

• Can that really be true? Aren't $f(x) \equiv c$ and $\frac{1}{f(1/x)} \equiv \frac{1}{c}$ both Lebesgue integrable as long as $c \neq 0$? Commented May 30, 2020 at 21:21
• @md2perpe Non-zero constant functions are locally integrable, but not integrable over all of $\mathbb R$. Commented May 30, 2020 at 21:24
• @kimchilover. Ah, right. I thought of Lebesgue measurable. Commented May 30, 2020 at 21:39
• Have you studied what happens to some functions like $\frac{1}{1+x^2}$ and $e^{-x^2}$? Have you studied simple functions like $\chi_{[a,b]},$ where $0<a<b<\infty$? Commented May 30, 2020 at 21:54
• @md2perpe Yes, but only for the indicator function of the above. You quickly run into problems with not well defined function values: i.e. $\frac{1}{f(1/x)} = \frac{1}{\chi_{[a,b]}} = \frac{1}{0}$ for all $x \not \in [a,b]$. I haven't looked into the probability distributions that you've listed. Commented May 31, 2020 at 1:40

Suppose both $$\int_{\mathbb R}|f(x)|dx<\infty\text{ and }\int_{\mathbb R}\frac{1}{|f(\frac 1 x)|}dx<\infty$$ hold. Note that this implies $$f(x)\ne0$$ for almost all $$x$$. Then a change of variables gives us $$\int_{\mathbb R}\frac{1}{|x|^2|f(x)|}dx<\infty$$ as well. Now by the Arithmetic-Geometric Mean inequality (that is, $$\sqrt{uv}\le(u+v)/2$$ if $$u,v\ge0$$), we have $$\frac 1 {|x|} = \sqrt{|f(x)|\times\frac{1}{|x|^2|f(x)|}}\le\frac 1 2\left(|f(x)|+\frac{1}{|x|^2|f(x)|}\right).$$ The right hand side is supposed to be integrable, so then $$1/|x|$$ is too. Which is well known to not be the case. So...