Product of two Lebesgue integrable functions not Lebesgue integrable

Try $f(x)=g(x)=\dfrac1{\sqrt{x}}$ 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?