Does $r(x) >0$ almost everywhere imply $r(x) > 1/k > 0$ almost everywhere on an open set, some $k \in \mathbb{N}$? Let $(a,b)\subseteq \mathbb{R}$ be an open interval, and suppose $r :(a,b) \to \mathbb{R}$ is Lebesgue measurable and positive almost everywhere on $(a,b)$.

Do there exist $k \in \mathbb{N}$ and $a < c < d < b$ so that $r(x) > 1/k$ for almost all $x \in (c,d)?$

Since $r$ is positive almost everywhere, by continuity of Lebesgue measure from below, we are guaranteed that $A_k = \{x \in (a,b) : r(x) > 1/k \}$ has positive Lebesgue measure for some $k$, but it's unclear to me whether there must be an open interval $(c,d)$ such that $(c,d) \setminus A_k$ has measure zero.
Hints or solutions are appreciated.
 A: Here is an outline of how to construct a counterexample: for each $n$, take some subset $A_n$ of $[0, 1]$ which is closed and nowhere dense, and such that $m(A_n) > 1 - \frac{1}{n}$ (look up "fat Cantor set" for a way to construct such a set).  Now, define $r : (0, 1) \to \mathbb{R}$ such that if $n$ is the minimum integer such that $x \in A_n$, then $r(x) = \frac{1}{n}$; and otherwise, if $x \notin \bigcup_{n=1}^\infty A_n$, then set $r(x) = 0$.
It should now be straightforward to show that for this function $r$, $r$ is positive almost everywhere on $(0, 1)$; but for every $k \in \mathbb{N}$ and every $(c, d) \subseteq (0, 1)$, we do not have $r(x) > \frac{1}{k}$ for almost every $x \in (c, d)$ (in fact there exists $(c', d') \subseteq (c, d)$ such that $r(x) \le \frac{1}{k}$ everywhere on $(c', d')$).
A: The equivalent statement is: if $B_1 \subset B_2 \subset \ldots$ measurable with the complement of their union of measure $0$, does there exist a non-void open subset $U$ and $n$ such that $U\backslash B_n$ of measure $0$ ?
Not necessarily true. Here is a counterexample.
We can pass to the equivalent space $\{0,1\}^{\mathbb{N}}$ with the standard measure.
We know ( see normal numbers) that for almost all
$$x= (x_1, x_2, \ldots,)$$ we have (Borel's theorem)
$$\lim_{k\to \infty} \frac{\sum_{i=1}^k x_i}{k} = \frac{1}{2}$$
Let us define the Borel set
$$B_n = \{ (x_i) \ | \frac{ \sum_{i=1}^k x_i}{k} \ge \frac{1}{4} \textrm{ for all } k \ge n\}$$
(a closed subset in fact)
We have $B_1 \subset B_2 \subset \ldots$ and
$\cup_{n\ge 1} B_n $ has zero measure complement in $\{0,1\}^{\mathbb{N}}$.
Now, a basis for the topology of $\{0,1\}$ is given by subsets $J$ of $x=(x_i)$'s  where finitely many of the components of $x$ are fixed. It is easy to see that for every such $J$, and $n\ge 0$, there exists $J'\subset J$ such that
$$J'\cap B_n= \emptyset$$
( say $J$ is all the $x$  with prescribed  first $N$ components. Let $n\ge 0$. Now let $J'\subset J$ with enough $0$ prescribed components such that for all $x = (x_i)$ in $J'$ we have $\frac{\sum_{i=1}^k x_i}{k}  \le \frac{1}{4}$ for some $k\ge n$. )
Therefore, no $B_n$ will contain an open subset (up to measure $0$).
Note: we showed that $\{0,1\}$ is the union of a meagre set and a set of measure $0$.
So our answer is very similar to the answer of @Daniel Schepler: . For an alternative answer, check this.
