How is continuity defined for set-valued functions? There is an obvious way in which certain functions from the reals to the powerset of the reals (or the powerset of $\Bbb R^n$, in general) are continuous. For example the function $F_X:\Bbb R\to\mathcal P(\Bbb R)$ given by
$$F_X(t)=\{x+t:x\in X\}$$
is obviously continuous, as is the function given by
$$G_X(t)=\{xt:x\in X\}.$$
There are obvious examples of discontinuous functions as well. Consider $H:\Bbb R\to\mathcal P(\Bbb R)$ given by
$$H(t)=\begin{cases} (-\infty,0) & t<0\\ \{0\} & t=0 \\ (0,\infty) & t>0\end{cases}$$
I would like to formalize this intuition.

My attempt so far:
To the extent of my knowledge, there is no "especially reasonable" topology on $\mathcal P(\Bbb R^n)$, so I figure it's best to keep as close to analysis as possible, without relying too much on the topological notion of continuity. My first thought is to generalize the $\epsilon-\delta$ definition of continuity using measures, so that for a function $F:D\to\mathcal P(\Bbb R^n)$, where $D\subseteq\Bbb R$, and a point $t_0\in D$, $F$ is continuous at $t_0$ with respect to a measure $\mu$ if and only if
$$(\forall \epsilon>0)(\exists \delta>0)(\forall t\in D\setminus\{t_0\})\left(|t-t_0|<\delta\implies \max\left\{\mu(F(t)\setminus F(t_0)),\mu(F(t_0)\setminus F(t))\right\}<\epsilon\right)$$
I think that this definition is sufficient for most practical purposes and - provided it isn't horribly wrong - the only additional questions I have are (bonus points)

*

*how can we define a topology to match this definition of continuity?


*is this equivalent to $\lim_{t\to t_0}(F(t)\setminus F(t_0))=\lim_{t\to t_0} (F(t_0)\setminus F(t))=\emptyset$ (and $\lim_{t\to t_0}F(t)=F(t_0)$) under other, accepted definitions?
Still, this only works for measurable sets, and there are "obviously continuous" functions whose range includes non-measurable sets. For example, take the previous $F_X(t)=\{x+t:x\in X\}$ for a non measurable set $X$. Clearly, $F_X$ is continuous, since we are merely translating $X$ to the right at a constant "speed," but since $X$ is non-measurable the continuity of $F_X$ can't be determined using the above definition.
An alternative approach might be to consider a function $F:\Bbb R\to\mathcal P(\Bbb R^n)$ to be continuous iff their exists a set of continuous functions $f_x:\Bbb R\to\Bbb R^n$, $x\in\Bbb R^n$ such that $F(t)=\{f_x(t):x\in\Bbb R^n\}$, but this seems a bit general, and I'm worried about potential pathological counterexamples.
Some notes on properties of intuitively continuous functions:
Just to paint a better picture of what I'm thinking of as "obviously" continuous functions:

*

*any "continuous deformation" of one set into another, as in popular animations used to depict homeomorphism and homotopy (side note: please stop using these to explain homeomorphism, it's misleading) is "continuous in the obvious sense."

[Some "continuous" functions are homotopies, provided that we think of $F(t)$ as an embedded topological space, but there's a lot of general yuckiness here that makes the exact relation to homotopy unclear to me.]

*

*if $F,G:D\to\Bbb R^n$ are both "continuous in the obvious sense," then so is $(F|G)(t)=F(t)\cup G(t)$


*if $F$ is "continuous" and there exists an interval $I$ such that for all $t\in I$, $F(t)$ is measurable, and $0<\mu(F(t))<\infty$, then $\mu\circ F$ is continuous on $I$ ("massive" sets do not pop into or out of existence instantaneously)


*all "finite speed" translations and reflections are "continuous in the obvious sense"
 A: $\newcommand{\M}{\mathscr{M}}$I don’t know if this will satisfy you. However, I present a pre-existing concept which I think is relevant.
You might like the concept of the Nikodym metric space. Let $(X,\M,\mu)$ be some measure space of interest (e.g. $\Bbb R$). If $\triangle$ denotes symmetric difference, put on equivalence relation $\sim$ on $\M$ via $A\sim B\iff\mu(A\triangle B)=0$. You can verify this satisfies the properties of an equivalence relation. We then enforce a metric $\rho$ on the quotient set $\M/_{\sim}$ by $\rho([A],[B])=\mu(A\triangle B)$, which is a well-defined metric. You then have the Nikodym metric space $(\M/_{\sim},\rho)$.
All metric spaces come equipped with a very natural and convenient topology. In the case of $X=\Bbb R$ and $\M$ the Lebesgue measurable sets, $\mu$ the Lebesgue measure (it’s not quite $\mathcal{P}(\Bbb R)$, but it avoids pathological sets), this metric space supports scalar multiplication too, as well as translation by a countable set, and intuitively, it feels quite nice and aligned with what you would want.
Incidentally, if $X$ is a finite measure space, the Nikodym metric space will be complete.
EDIT: As a comment from von Dongen has reminded me, the measure space must be finite for the metric to even make sense... and infinite measure sets would indeed break continuity.
Anyway, there is then a topological structure in which functions $F:\Bbb R\to\M/_{\sim}$ can be thought of as continuous. You lose access to all subsets of $\Bbb R$ but that’s alright for this intuitive thought experiment, since the sets you lose are weird and probably don’t play nice with the intuitive continuity criteria. If you really dislike taking a quotient, and just want all sets in $\M$, you can abandon the equivalence relation at the cost of dealing with the pseudometric space, and the pseudometric topology - it’ll still have a topology defined in the nice way, though.
In this case, your function $F_X,G_X$ are continuous for nice sets. I suspect $G_X$ is continuous for any measurable choice of $X$ but I can’t justify that yet. On this note, you might want to further restrict to the Borel sigma algebra to avoid pathological, but still measurable, sets.
