# $\{x \in \mathbb{R} : f$ is Lebesgue integrable in some open neighborhood of $x\}$ is dense in $\mathbb{R}$.

Suppose $$\{F_n\}_{n\in N}$$ are a sequence of closed sets with $$\cup_n F_n = \mathbb{R}$$. Suppose that $$f :\mathbb{R} \to \mathbb{R}$$ is measurable and $$f$$ is Lebesgue integrable on $$F_n$$ for each $$n$$. Prove that

$$\{x \in \mathbb{R} : f$$ is Lebesgue integrable in some open neighborhood of $$x\}$$ is dense in $$\mathbb{R}$$.

$$\textbf{my attempt}$$: is the following solution correct?:

suppose not, then $$\exists x \in \mathbb{R}$$ such that for some $$\epsilon >0 , f\notin L^1 (N_\epsilon(x))$$. Since $$\cup_n F_n = \mathbb{R}$$ , it must be that $$x\in F_n$$ for some $$n$$. This means that $$\exists N_\epsilon(x) \subset F_n$$ such that $$f$$ is not integrable. This is contradiction to the following lemma.

if $$f\in L^1 (A)$$ then $$f\in L^1 (B)$$ for any open (hence lebesgue measurable) $$B\subset A$$.

proof: since $$f$$ is measurable there is sequence of simple functions ,$$g_n$$, that converges pointwise to $$f$$. let $$f=\chi_B g_n$$. Since $$B$$ is measurable by fatu's lemma we have that $$f\in L^1 (B)$$.

$$\textbf{Question}$$: if above solution is right, what role does the closedness of $$F_n$$ plays?

• I don't know if the proof is right or not. But for me it isn't clear that $N_\epsilon(x)\subset F_n$. $F$ is closed, so it doesn't need to contain a neighbourhood of $x$. At most, $f$ wouldn't be integrable on $N_\epsilon(x)\cap F_n$. And then, the above lemma is contradicted only if you are assuming the relative topology,. Commented Oct 27, 2019 at 18:45
• Do you know Baire's Theorem? Commented Oct 27, 2019 at 18:47
• The closedness of $K_n$ is such that you can apply Baire's theorem. Commented Oct 27, 2019 at 18:47
• @conditionalMethod I was thinking of using Baire, but did not know how to apply that here. Commented Oct 27, 2019 at 18:50

Baire's theorem implies that one of the $$K_n$$ has non-empty interior, which obviously is not sufficient for proving the statement (since it will only provide you with the fact that your set contains some open set). However, you may intersect the $$K_n$$ with (say) dyadic intervals and apply Baire's theorem to show that your set contains an open subset of every dyadic interval, which implies density.