Let $\mu$, as usual, denote the one-dimensional Lebesgue measure.
For a real interval $[0,x]$ and a nonempty $S \subset \mathbb{R}_{\geq 0}$ define the asymptotic density of $S$ in $\mathbb{R}_{\geq 0}$ as
$$A(S) :=\limsup_{x \to + \infty} \frac{\mu(S \cap [0,x])}{x}$$
Now, suppose $f:\mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$ is an arbitrary nonnegative continuous function such that $$\int_0^{+\infty} f < +\infty$$
Let $\epsilon>0$. Define $S_{\epsilon}$ as the preimage of $[\epsilon, +\infty)$ under $f$.
Conjecture: For any $\epsilon>0$, $A(S_{ \epsilon}) = 0$
Motivation: we know that there are continuous real functions such that $\int_0^{+\infty}f$ converges but $f$ does not go to $0$. For example, the 'thin triangle' type examples. However, in between the triangles there are 'plateaus of zero'. The triangles get thinner as $x \to +\infty$ which implies they become more and more sparse as we go further into the positive real line (ie., the plateaus are more prevalent). In other words, the function is mostly $0$ or mostly within $\epsilon$ of $0$.
One idea to construct a counterexample is to have the triangles be strictly side by side (no plateaus). I can't quite get this to work. For example, if the $n$th triangle has height $h$ (the height is invariant) and we want the integral to equal $C$, the width of the $n$th triangle should be $2\frac{C}{h} 2^{-n}$. However, $\sum 2\frac{C}{h} 2^{-n} = \frac{2C}{h}$ so the total width of all of the triangles is finite, which implies we just have a zero-plateau for sufficiently large $x$ (ie., $x>\frac{2C}{h}$). This is problematic.
Hence, we want the total width to equal $+\infty$. Hence we need the heights to go to $0$ to compensate. But this would mean for sufficiently large $x$ the triangles have height $<\epsilon$, which is problematic as well (ie., this would mean $\sup(S_{\epsilon}) < +\infty$ which obviously means $A(S_{\epsilon}) = 0$, hence we have no counterexample).