Just how continuous is measure

It's a classical theorem of real analysis that Lebesgue measure is "continuous" that is for an ascending chain of subsets $A_k$ we have

$$\lim_{k\rightarrow \infty} m(A_k)=m\left(\bigcup_{k=1}^\infty A_k\right)$$

and we have an analogous condition for descending chains of finite measure. Of course we can also talk about an arbitrary measure space having a continuous measure as well.

I'm curious if we can actually make Lebesgue measure continuous in a meaningful way. What I mean by this is if we can find a Hausdorff topology on $\mathcal L$, the Lebesgue measurable subsets of $\mathbb R$, so that $m: \mathcal L \rightarrow [0,\infty]$ is continuous and for $A_n \in \mathcal L$ we have that $U_n=\bigcup_{k=1}^n A_k$ converges in the topology to the union of the $A_k$. Ideally I'd like to see an example or such a topology or if it's impossible a proof that it isn't possible.

A couple of things to note. It's not hard to get any two of the three conditions. For instance we could put the discrete topology to get Hausdorff and continuous or we could put the pullback topology of $[0,\infty]$ under $m$ and get continuous and convergence, but not Hausdorff. If it's too difficult a question in Lebesgue measure setting, I'd also be interested in the analagous question for $\ell^1(\mathbb N)$.

Solution 1:

Here is a cheap way of getting a suitable topology:

Give $\mathcal L$ the initial topology with respect to the two maps $j: \mathcal L \to \{0,1\}^{\mathbb R}$, where $j(A) = \chi_A$ is the characteristic function of $A$, and $m: \mathcal L \to [0,\infty]$. Then $m$ is continuous by definition and $\mathcal L$ is Hausdorff, because it's topology is finer than the topology induced by inclusion $j(\mathcal L)\subset \{0,1\}^{\mathbb R}$.

Now setting $U_k = \bigcup_{i\le k} A_i$ for $A_i\in \mathcal L$, we furthermore have that $$U_k \to U_\infty \iff j(U_k)\to j(U_\infty)\text{ and }m(U_k)\to m(U_\infty)$$ by definition of the initial topology. Hence this topology also satisfies that $U_k \to U_\infty$ in $\mathcal L$.