The metric or topology between subsets of Euclidean space $\newcommand{\ps}{\mathcal{P}(\mathbb{R}^2)}$
Before asking a question, I would like to introduce a definition that motivated me. Consider the path-like function $f:[0, 1] \to \ps$ defined by
$$
\forall t \in [0,1]:f(t) = \{x \in \mathbb{R}^2~|~||x - (t,t)|| \le 1\}
$$
It is seemingly continuous in some sense (at least for me). The goal is to find a circumstance in which we can say that $f$ is continuous.
I tried to define a corresponding metric candidate $d: \ps \times \ps \to \mathbb{R}$ defined by
$$
\forall P,Q \in \ps:d(P,Q) = \left\Vert\frac{1}{A(P)}\int_{P}(x,y)dxdy - \frac{1}{A(Q)}\int_{Q}(x,y)dxdy\right\Vert
$$
where $A(P)$ is the area of $P$, and each term side the norm represents the geometric center. But the function $d$ fails to be a metric because there are $P \neq Q$ such that $d(P,Q) = 0$.
Question) Can we find some metric or topology on $\mathcal{P}(\mathbb{R}^n)$ by which $f$ is continuous?
 A: 
But the function $d$ fails to be a metric because there are $P \neq Q$ such that $d(P,Q) = 0$.

Note that you don't have to define a distance on all subsets of $\mathcal{P}(\mathbb{R}^n)$. In fact you already restricted yourself to measurable sets. So I assume you can restrict yourself furthermore, since $f(t)$ is a very well behaving set for any $t$, namely: a closed ball. A good way to do this is to take $F(\mathbb{R}^n)$ which is the set of all compact subsets of $\mathbb{R}^n$. This makes things easier, although your concrete definition will not work for compact subsets as well. But I think it does work if we restrict ourselves even further, by considering closures of open bounded subsets.
But the classical Hausdorff distance:
$$d(P,Q)=\max\bigg\{\sup_{p\in P}d(p,Q), \sup_{q\in Q}d(P,q)\bigg\}$$
does work for $F(\mathbb{R}^n)$. In fact it is a metric on $F(\mathbb{R}^n)$ (or $F(M)$ for any metric space $M$). Otherwise it has a similar problem: $d(A,\overline{A})=0$.

In that situation assume that $x,y\in[0,1]$. Then I leave as an exercise that
$$d(f(x),f(y))=\lVert (x,x)-(y,y)\rVert=\sqrt{2}\cdot|x-y|$$
and so $f$ is not only continuous but also Lipschitz.

Note that there is also a different way to look at the problem. Instead of defining a metric on $\mathcal{P}(X)$ you can actually treat your $f$ as a multivalued mapping. In that scenario there are various different definitions of hemicontinuity (upper and lower) that you can apply. They do correspond to topological continuity if so called Vietoris topology (upper and lower) is considered on whole $\mathcal{P}(X)\backslash\{\emptyset\}$, but in general that topoogy is not metrizable.
