Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a homework problem that says;

Give Borel functions $f,g: \mathbb{R} \to \mathbb{R}$ that are Lebesgue integrable, but are such that $fg$ is not Lebesgue integrable.

I saw this page too: Product of two Lebesgue integrable functions, but the question does not mention boundedness.

I also am not sure what to do with the fact that the functions are Borel. (Any help on this would be especially appreciated)

I know that if $fg$ were Lebesgue integrable then both $\int (fg)^+\,d\mu$ and $\int (fg)^-\,d\mu$ would be finite. This could lead to utilizing the finiteness of their difference (the function's integral) or their sum (the absolute value). I also know that $f+g$ are Lebesgue integrable if $f$ and $g$ are so I thought of using $$fg = \frac{1}{4}\,\big( (f+g)^2 - (f-g)^2 \big)\longrightarrow \int (fg)\,d\mu = \frac{1}{4}\,\int (f+g)^2\,d\mu - \frac{1}{4}\,\int (f-g)^2\,d\mu,$$ assuming linearity of the integral etc.

I also thought of the Hölder inequality, $$\int \mid fg \mid d\mu \leq \bigg( \int \mid f \mid^p d\mu \bigg)^{(1/p)}\,\bigg( \int \mid g \mid^q d\mu \bigg)^{(1/q)},$$ but there was no mention in the question of what $L^p$-space this was in. Maybe by the definition I gave it is such that $p=1$ and $q=1$? Then $$\int \mid fg \mid d\mu \leq \bigg( \int \mid f \mid d\mu \bigg)\,\bigg( \int \mid g \mid d\mu \bigg).$$

However, I still can't seem to think of an approach to show that $fg$ is not Lebesgue integrable, while $f$ and $g$ are.

Thanks for any guidance!

share|cite|improve this question
up vote 10 down vote accepted

Try $f(x)=g(x)=[0<x<1]\cdot\dfrac1{\sqrt{x}}$ for every $x$ in $\mathbb R$.

Edit Thus, $f(x)=g(x)=\dfrac1{\sqrt{x}}$ for every $x$ in $(0,1)$ and $f(x)=g(x)=0$ for every $x$ in $\mathbb R\setminus(0,1)$. The Borel measurability of $f=g$ stems from the fact that $f=g$ is continuous everywhere except at points $0$ and $1$. The integrability of $f=g$ over $\mathbb R$ stems from the fact that the Riemann integral $\int\limits_0^1\dfrac{\mathrm dx}{x^a}$ is finite for every $a<1$ and in particular for $a=1/2$. The non integrability of $f\cdot g$ over $\mathbb R$ stems from the fact that the Riemann integral $\int\limits_0^1\dfrac{\mathrm dx}{x^a}$ is infinite for every $a\geqslant1$ and in particular for $a=1$.

share|cite|improve this answer
Does $[0<x<1] $ denote the fractional part of $x$ ? – Ragib Zaman Nov 8 '11 at 2:18
@Didier, I realized you are the same person in the link I read. I was wondering if you could explain your suggestion more. I also notice that the domain of $f,g$ are $(0,+\infty)$. Thank you. – nate Nov 8 '11 at 4:34
@nate, my suggestion is to check that $f$ and $g$ are Lebesgue integrable while $h=fg$ (defined by $h(x)=[0<x<1]\cdot\frac1x$) is not. Both $f$ and $g$ are defined on $\mathbb R$, even if it happens that $f(x)\ne0$ or $g(x)\ne0$ if and only if $x$ is in $(0,1)$ – Did Nov 8 '11 at 5:04
@Ragib, the bracket is Iverson bracket. – Did Nov 8 '11 at 5:06

Well that link tells you how to do it: $f$ and $g$ must be unbounded.

Also, your computations show that you can reduce the problem to the case $f=g$, cause if it would be true in this case you can show it in general.

And since you use the Lebesgue integral, pick a step function $f= \sum n 1_{I_n}$, where $I_n$ is an interval... How can you make $f$ Lebesgue integrable but $f^2$ not?

share|cite|improve this answer
Hi. Thanks for the reply. I've been thinking of a way to write $f = \sum\,a_n\,1_{I_n}$ as have finite measure ($\lambda(I_n)<+\infty$) with finite coefficients $a_n$ and yet not satisfy $f^2$ as being finite. Still a little stuck. Could you elaborate on your last comment? – nate Nov 7 '11 at 23:28
@nate Hint: can you find an $\alpha$ so that $\sum n \frac{1}{n^\alpha}$ is convergent while $\sum n^2 \frac{1}{n^{\alpha}}$ is divergent? – N. S. Nov 8 '11 at 7:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.