We say that a measurable function $g:\mathbb R \to \mathbb R$ has a Lebesgue point at $x \in \mathbb R$ if
$$ \frac1{2s}\int_{x-s}^{x+s} |g(y)-g(x)| \, dy \to 0 \quad \text{as } s \to 0.$$
If $g$ is continuous, then every $x \in \mathbb R$ is a Lebesgue point of $g$. The Lebesgue differentiation theorem states that a (locally) Lebesgue integrable function has Lebesgue points almost everywhere. The function given by
$$ g(x) = \sum_{k=0}^\infty (-1)^k\chi_{(2^{-k-1},2^{-k}]} $$
is an example of a function that does not have a Lebesgue point at $0$.
Now suppose that $g: \mathbb R \to \mathbb R$ and $h: \mathbb R \to \mathbb R$ have Lebesgue points at every $x \in \mathbb R$. Is it possible that the product $gh$ has non-Lebesgue points?
Best Answer
Take $$g(x)=\sum_{k=0}^\infty 2^{k/2} \chi_{[2^{-k}-4^{-k},2^{-k}]}.$$ It is easy to see that every $x \neq 0$ is a Lebesgue point of $g$. At $x=0$ we have $$\frac1{2^{-n+1}}\int_{-2^{-n}}^{2^{-n}} |g(y)-g(0)| \, dy = 2^{n-1}\sum_{k=n}^\infty 2^{-3k/2} = 2^{-n/2-1}\sum_{k=0}^\infty 2^{-3k/2}$$ which goes to $0$ as $n \to \infty$. This also implies $$\frac1{2s}\int_{-s}^{s} |g(y)-g(0)| \, dy \to 0 \quad \text{as $s \to 0$}$$ and therefore $0$ is also a Lebesgue point of $g$.
However, $$g^2(x)=\sum_{k=0}^\infty 2^k \chi_{[2^{-k}-4^{-k},2^{-k}]}$$ does not have a Lebesgue point at $0$, which can be seen by adapting the exponents in the above calculation. In fact, $g^2$ does not have a Lebesgue point at $0$ even after redefining $g^2(0)$. Also note that $g^2$ is integrable at $0$.