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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.

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  • $\begingroup$ 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 $|F|=t$. $\endgroup$
    – Mittens
    Oct 25, 2022 at 3:23
  • $\begingroup$ 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 $g$. $\endgroup$
    – Mittens
    Oct 25, 2022 at 3:28
  • $\begingroup$ @OliverDíaz Thanks, I also realized immediately after I posted my previous comment, so I deleted it. $\endgroup$ Oct 25, 2022 at 3:29

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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*}

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