Consider the sequence $\{f_n\}_{n=1}^\infty$ of functions $f_n: (-\infty, n) \to \mathbb R$ given by $$f_n(t) = \frac{1}{n-t}.$$ On an intuitive level it feels like this sequence should have a limit, and that that limit should be $f: \mathbb R \to \mathbb R$, $f(t) = 0$. But we cannot use the usual notions of convergence of sequences of functions, because the domain is different for each function. So what is a good notion of convergence here?
This sort of thing appears naturally in the context of differential equations (see for example Example 4.33 in the ODE book by Logemann and Ryan), so it seems like there should be a somewhat standard notion. But I have been unable to find one.
One idea I had was to consider the graphs $\psi_n = \{(t, f_n(t)) \in \mathbb R^2: t \in \mathrm{dom}(f_n)\}$ and use the Hausdorff distance, but even this is tricky when the prospective limit function has an unbounded domain (or if the functions $f_n$ have such domains). Maybe it can be saved by considering only $\psi_n \cap U$ for all closed subsets $U \in \mathbb R$, or something like that.
The weak-L1 space $L^{1, \infty}(\mathbb R)$ has been introduced precisely to deal with the power-like function $g(t)=1/\lvert t \rvert$. Since it is translation-invariant, it contains the functions $$ f_n(t)=\begin{cases} \frac{1}{n-t}, & t<n, \\ 0&, t\ge n.\end{cases}$$ In $L^{1, \infty}(\mathbb R)$, the sequence $f_n$ is bounded and it has no limit.
Some details. The space $L^{1, \infty}(\mathbb R)$ consists of those measurable functions $f$ such that $$ \lVert f \rVert_{1, \infty}:=\sup_{\lambda >0} \lambda\lvert\{ t: \lvert f(t)\rvert >\lambda\}\rvert <\infty.$$
(Here $\lvert \{\ldots\}\rvert$ denotes the Lebesgue measure of a set. Also, $\lVert \cdot \rVert_{1, \infty}$ is not a norm but a quasi-norm, which is a minor technical point here). Letting $$ f_+(t)=\begin{cases} -\frac{1}{t}, & t<0, \\ 0,& t\ge 0,\end{cases}$$ it holds that $$\tag{1} \lVert f_n\rVert_{1, \infty}=\lVert f_+\rVert_{1, \infty}=1, $$ so $f_n$ is bounded. To see that $f_n$ has no limit, we refer to this Terry Tao blog on modes of convergence. First we note that convergence in $L^{1, \infty}(\mathbb R)$ implies convergence in measure, as it is immediate. Now, by Exercise 17 in the link, convergence in measure implies pointwise convergence at almost all points of a subsequence. Since $f_n(t)\to 0$ for all $t\in\mathbb R$, the candidate limit can only be the null function. But (1) shows that $f_n$ cannot converge to the null function, hence it cannot have a limit.
Remark. The space $L^{1, \infty}$ is a special case of a Lorentz space, which justifies the notation.