Just how continuous is measure

Question

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 $\ell^1(\mathbb N)$.

@Davide: I think $\mathcal L$ must be the $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb R$. — hmakholm left over Monica, Sep 17 '12 at 18:54
@DavideGiraudo As Henning, said it is indeed the Lebesgue measurable subsets of the reals. — JSchlather, Sep 17 '12 at 19:03
You might want to consider looking for a topology on $X = \mathcal{L} / \sim$ where $A \sim B$ when $m((A-B) \cup (B-A)) = 0$ instead. — Mike F, Sep 17 '12 at 22:06
Maybe you are looking for the Frechet-Nikodym metric? See my answer here: http://math.stackexchange.com/questions/44171/the-pseudometric-induced-by-a-measure/44205#44205 — , Sep 17 '12 at 23:38
So you can do that and it gives you a metric on $\mathcal L / \sim$. But pulling the topology generated by it back onto $\mathcal L$ gives you a non-Hausdorff topology. — JSchlather, Sep 18 '12 at 00:05

score 3 · Accepted Answer · answered Oct 21 '12 at 12:16

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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 $\mathcal L$.

answered Oct 21 '12 at 12:16

Sam

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Is it obvious why any topology finer than the topology induced by that inclusion is Hausdorff? – JSchlather Oct 23 '12 at 02:13
@Jacob Schlather: Yes. If $U$ is open in the coarser topology, then it is also open in the finer topology. So if we have disjoint neighbourhoods of two given points in the coarser topology, we also have disjoint neighbourhoods in the finer topology (generally). – Sam Oct 23 '12 at 14:28
Of course, ${0,1}^{\mathbb R}$ is Hausdorff. I guess we can be sure that this topology is not discrete as well since we have the desired convergence. – JSchlather Oct 23 '12 at 14:36
@Jacob Schlather: I'm still wondering whether there might be a more natural topology on $\mathcal L$, though. I mean, this one is just thought up to exactly fit the requirements of your question. So it seems quite artificial... – Sam Oct 23 '12 at 15:54

Michael Greinecker · Answer 2 · 2012-09-17T21:36:42.147

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Identify sets with their indicator functions, endow $\{0,1\}^\mathbb{R}$ with the product topology and take the trace topology with respect to $\mathcal{L}$. This topology is certainly Hausdorff and satisfies the condition for unions. The continuity of $m$ should follow from the fact that for a convergent sequence of sets $\limsup$ and $\liminf$ are equal and their measure coincides.

It might be that $m$ is only continuous for finite measure spaces.

edited Sep 17 '12 at 21:36

answered Sep 17 '12 at 21:02

Michael Greinecker

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$\newcommand{\R}{\mathbb R}$I looked at this topology. If you'll notice any open set in ${0,1}^\R$ misses at most finitely many $x \in \R$ in particular any open set in the topology given by ${0,1}^\R$ contains sets of arbitrary measure. So every open set has a measurable set of infinite measure, thereby $m$ cannot be continuous. This same argument shows that if you looked at lebesgue measurable subsets of $[0,1]$ with the topology given by ${0,1}^{[0,1]}$ that every open set contains a set of arbitrary measure so $m$ is still not continuous. – JSchlather Sep 17 '12 at 22:11
@JacobSchlather You are right. The topology is merely sequentially continuous for finite measure spaces, which is essentially the dominated convergence theorem. – Michael Greinecker Sep 18 '12 at 09:39

Just how continuous is measure

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