How is the product of two Lebesgue integrable functions?

Question

Let $(X, \cal{A}, \mu)$ be a measure space and $f, g : X → \mathbb{R}$. Determine if the following implications hold in general:
(i) both functions $f$ and $g$ are integrable $⇒ f \cdot g$ is integrable;
(ii) $f \cdot g$ is integrable $⇒$ at least one of the functions $f$ or $g$ is integrable;
For $(i)$, I think it follows immediately from the theory?
For $(ii)$, I was thinking to find a counter example, namely, $f$ $\cdot$ $g$ to be integrable, but neither $f$, nor $g$ to be integrable, but I don't know if it is possible...

EDIT: For $(i)$, if we take $X=(0,1)$ and $$f(x)=g(x)=x^{−1/2}.$$ Then $f,g∈\cal{L}$$(X)$, but $(fg)=1/x∉\cal{L}$$(X)$.

Answer 1

(i) It might well happen that $\int|f|d\mu<\infty$ and $\int f^2d\mu=+\infty$.

(ii) Let $A$ be a set with $\mu(A)=\infty=\mu(A^{\complement})$. Then let $f$ be the indicator function of $A$ and let $g$ be the indicator function of $A^{\complement}$.

Answer 2

Choose two non-measurable disjoint sets $S$ and $T$, and consider the indicator function $I_S$ and $I_G$. None of them are integrable, however, their product, the zero function, is integrable.

There are a number of ways to find a suitable pair $S$ and $T$. For example, remember a textbook example on the existence of non-measurable sets:

We may define a such equivalent relationship on $\mathbb R$: $x\sim y$ :iff $x-y\in \mathbb Q$. Denote the quotient set of $\mathbb R$ over this relationship as $\mathbb R/\mathbb Q$. By the axiom of choice, we may choose one element from each coset. These elements form a set $S$, and $S$ cannot be measurable.

Actually, each cosets of $\mathbb R/\mathbb Q$ has the cardinality of $\aleph_0>1$. So, it is possible to choose another element that is not in $S$ from each coset. Let set $G$ be the collection of them. For a similar reason, $G$ is not measurable.