# 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!

#### Solutions Collecting From Web of "Product of two Lebesgue integrable functions not Lebesgue integrable"

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 $a=1$.

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?