Let $\mu$ denote Lebesgue measure on the real line. In measure theory, the Borel-Cantelli Lemma states that if $\lbrace E_k \rbrace_{k\ge 1}$ is a collection of measurable sets such that $\sum_{k\ge 1}\mu(E_k)<\infty$, then almost every $x$ belongs to finitely many of the $E_k$'s. In simpler terms, there is a $K$ such that almost every $x$ belongs to one of $E_1,E_2,\ldots,E_K$.
I am asking if there is an analogous statement for a measurable function $f\in L^1(\mathbb{R}).$ If I had to guess, it would be the following:
If $\int _{\mathbb{R}}\vert f \vert <\infty$, then for every $\varepsilon>0$, almost every $x$ such that $\vert f(x) \vert >\varepsilon$ belongs to some interval of the form $[-R,R].$ I am confident that this is a true statement since if it fails, then there are sets of arbitrarily large measure for which $\vert f\vert >\varepsilon$, which cannot happen since $\int \vert f \vert <\infty $.
Can a stronger statement be made? Is there a precise analogue?
Follow-up: Can anything be said about the measure of the collection of intervals $[n,n+1]$, for arbitrarily large $n$ on which $\vert f \vert$ is large?
I don't think what you are saying is true. Consider taking $f_n(x) = 1_{[n,n+1/n^2]}$ and let $f(x)=\sum_n f_n(x)$. Then $\int |f(x)|dx = \sum \frac{1}{n^2} < \infty$ but $\{x:f(x)=1\}\cap (\mathbb{R}\setminus [-R,R])$ has positive measure for every $R$.
Of course if the interval is allowed to depend on $x$ then the problem is trivial since every $x$ is contained in $[-|x|-1,|x|+1]$.