A continuous path between shapes Let $A$ and $B$ be two geometric shapes in the plane (two measureable subsets of $\mathbb{R}^2$) such that $A\subseteq B$.
Define a $path$ from $A$ to $B$ as a function $f$ from $[0,1]$ to subsets of $\mathbb{R}^2$ such that:


*

*$f(0)=A$.

*$f(1)=B$.

*$f$ is monotonically increasing, i.e. for every $t'>t$, $f('t)\supseteq f(t)$.

*$Area(f(t))$ is a continuous function of $t$.


Intuitively, a path describes how the shape $A$ grows continuously until it becomes $B$. It makes sense that such a function exists. But how can I prove it?
Alternatively, if a path does not always exist, what are the conditions on $A$ and $B$ that guarantee that it does?
 A: Let's assume $B$ is bounded.
If $A$ is nonempty, pick a point in $A$, call it $0$. Define $C_r$ ($r \geq 0$) to be the closed disc of radius $r$ around $0$. Since $B$ is bounded there exists $c \in \mathbb{R}$ such that $C_c \supset B$.  
If $A$ is empty but $B$ is not then pick $0 \in B$, and define $C_r$ ($r \geq 0$) to be the open disc of radius $r$ around $0$; again since $B$ is bounded there exists $c \in \mathbb{R}$ such that $C_c \supset B$.
(If $A$ and $B$ are both empty the result is trivial).
Either way, define $f(t) = (A \cup C_{ct}) \cap B$ for $0 \leq t \leq 1$.
Clearly $f(0) = A$, $f(1) = B$, and $f$ is monotonically increasing. 
Finally, that $Area(f(t))$ is continuous can be seen from the fact that $Area(C_{ct})$ is continuous and $Area(f(t+h)) - Area(f(t)) \leq Area(C_{c(t+h)}) - Area(C_{ct})$.
I think that if $B$ is not bounded then defining $f(t) =  (A \cup C_{\frac{t}{1-t}}) \cap B$ for $0 \leq t < 1$ and $f(1) = B$ works, but I don't know enough about measure theory to completely convince myself that $Area(f(t))$ is continuous at $1$ then.
