# Construct Two Lebesgue Integrable Functions whose Product is not Lebesgue Integrable for Arbitrary $|E|>0$

Question. Let $$E\subseteq\mathbb{R}^n$$ be a measurable set and $$|\cdot|$$ be the Lebesgue measure on $$\mathbb{R}^n$$. Prove that if $$|E|>0$$, then there exist two Lebesgue integrable functions $$f$$ and $$g$$ on $$E$$ whose product $$fg$$ is not Lebesgue integrable on it.

If $$E$$ contains an interior point, the proof would be easy. Without loss of generality, we may assume that $$E$$ contains a sufficiently long open interval $$I$$. Then we let $$(I_n)$$ be a sequence of disjoint open intervals contained in $$I$$ with $$|I_n|=n^{-3}$$ for each $$n$$. Define $$f=\sum_{n=1}^{\infty}{n\cdot\chi_{I_n}}.$$ Clearly, $$\int_E{f}=\sum_{n=1}^{\infty}{n|I_n|}=\sum_{n=1}^{\infty}{\frac{1}{n^2}}<\infty$$ but $$\int_E{f^2}=\sum_{n=1}^{\infty}{n^2|I_n|}=\sum_{n=1}^{\infty}{\frac{1}{n}}=\infty.$$

Here my question is how to construct such functions $$f$$ and $$g$$ when $$E$$ contains no interior point, such as a fat Cantor set. It would be nice to hear from you.

• The Lebesgue measure is continuous in the sense that if $E$ is measurable and $|E|>0$, then for any $0<t<|E|$, there is a. measurable set $F\subset E$ with $|F|=t$. Oct 25, 2022 at 3:23
• Because you can obtain pairwise disjoint sets $F_n\subset E$ of size $\frac1n|E|$ (w.l.g. assume $E$ has finite measure); then it is easy to manufacture the functions $f$ and $g$. Oct 25, 2022 at 3:28
• @OliverDíaz Thanks, I also realized immediately after I posted my previous comment, so I deleted it. Oct 25, 2022 at 3:29

Your approach is basically right. The point is that your $$I_k$$ don't need to be intervals — any measurable sets with measure $$O(k^{-3})$$ will do, and the fact that we are dealing with the Lebesgue measure guarantees the existence of such sets.
Let $$m$$ be the Lebesgue measure on $$\mathbf{R}^n$$. Consider the function $$g\colon [0, \infty) \rightarrow [0, \infty]$$ defined by $$g(x) = m(E \cap (-x, x)^n)$$.
This function is increasing from $$0$$ to $$m(E)$$. Moreover, by continuity of measure together with the fact that the boundary of each $$(-x, x)^n$$ has Lebesgue measure $$0$$, it is continuous.
Hence, by the intermediate value theorem, we can choose $$x_k$$ such that $$g(x_k) = k^{-3}(m(E) \wedge 1)$$ and so define $$E_k = E \cap (-x_k, x_k)^n$$.
Now, as you did, we can define $$f = \sum_{k=1}^\infty k \cdot \chi_{E_k}.$$
The $$E_k$$ are not disjoint, but we can use Tonelli's theorem to switch the integral with the sum and conclude that $$\begin{gather*} \int_E f\,\mathrm{d}m = \sum_{k=1}^\infty k \cdot m(E_k) = (m(E) \wedge 1)\sum_{k=1}^\infty k^{-2} < \infty, \\ \int_E f^2\,\mathrm{d}m = \sum_{k=1}^\infty k^2 \cdot m(E_k) = (m(E) \wedge 1)\sum_{k=1}^\infty k^{-1} = \infty. \end{gather*}$$