an interesting topology question about open sets Suppose we are in $\mathbb{R}^n$ and say $\mathcal{B}$ is the collection of all open sets of $\mathbb{R}^n$ : all the open balls. we know $\mathcal{B}$ is a basis for $\mathbb{R}^n$. Now, put
$$ T : \mathcal{B} \to \mathbb{R} $$
with $T(B) = \operatorname{measure}(B) $ 
is $T$ a continuous map? An open map? Why is it important such  a function? or is it not important at all?
 A: Your original question does not make sense unless you specify a topology on the set of all open sets.
I am not aware of any such topology, but there is something related to your idea that is more standard: the Hausdorff distance, which measures the difference of two bounded subsets of $\mathbb R^n$.
(It also works in greater generality, but that's not important here.)
To make the Hausdorff distance into a metric, you should consider only compact subsets.
If you consider all bounded subsets, you still get a topology.
Consider the OP's mapping $T$ on the collection of compact sets.
It is not continuous with respect to the Hausdorff distance:
The cube $Q=[0,1]^n$ maps to one but the finite (and compact and zero measure) sets $A_k=(2^{-k}\mathbb Z)\cap Q$ approach $Q$ in Hausdorff distance as $k\to\infty$.
That is, $d_H(Q,A_k)\to0$ as $k\to\infty$ but $T(A_k)=0$ for all $k$ whereas $T(Q)=1$.
The same construction works for bounded open sets if you replace the finite sets $A_k$ with unions of finitely many small balls (centers at $A_k$ if you wish) whose radius goes to zero fast enough as $k\to\infty$.
The mapping $T$ is not continuous with respect to the Hausdorff distance, no matter how you interpret it.
But this, of course, can change if you change the topology.
