When does $f\in L^1$ vanish? I am asking about the sort-of converse to this question: under what additional conditions on $f:\mathbb{R}_+\rightarrow\mathbb{R}$ does the following hold?
$$
\int_{\mathbb{R}_+}f<\infty\implies \lim_{x\rightarrow\infty}f(x)=0
$$
Where the limit above is made in the topological sense: for every increasing diverging sequence $x_n$, $f(x_n)\rightarrow 0$.
Surely, by the linked question, the set of such functions includes uniformly continuous ones, but can we expand the set and completely characterize it? Is it the set of BV functions? Absolutely continuous?
 A: Other than a condition that is essentially a restatement of $\lim_{x \to \infty} f(x) = 0$, I don't see how you can expect to completely characterize it.  It's easy to construct some quite nasty functions 
(discontinuous and not BV) which have
limit $0$ at $\infty$ and finite integral.
A: I think that I have made some sort of generalization of your statement.
$$\int f= \sum_{i=0}^{\infty} \int_{[i,i+1]} f$$
Therefore $\lim_{i\to \infty} \int_{[i,i+1]} f=0$ or else the sum would diverge. This implies (from this fact here) that $\lim_{i\to \infty}  f|_{[i,i+1]}=0$ almost everywhere. This means that $f$ restricted to the set $[i,i+1]$ must be $0$ in the limit $i \to \infty$ except on a subset of measure zero.
Alternatively,
$$\int f = \lim_{a\to \infty} \int_{[0,a]} f = \lim_{a\to \infty} (\int f - \int_{[a,\infty]} f) = \int f - \lim_{a\to \infty} \int_{[a,\infty]} f$$
So $\lim_{a\to \infty} \int_{[a,\infty]} f=0$ if $\int f < \infty$
This like before implies that the $\lim_{a\to \infty} f|_{[a,\infty]} =0$ almost everwhere.
