# Product of two Lebesgue integrable functions not Lebesgue integrable

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!

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## 2 Answers

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 $a=1$.

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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?

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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