What's the relationship between continuity property of Lebesgue measure and continuity on a metric space?

Question

This is a topic from Lebesgue measure in $\textit {Carothers' Real Analysis}$:

I know how to prove Theorem 16.23. However, I can not figure out why he names this property as continuity? Besides what's the relationship between continuity here and continuity on a metric space?(I mean it is a little bit weird to say continuous Lebesgue measure)

Some definitions on the book:

1. Continuity properties on a metric space:

2. Lebesgue outer measure:

3. Capital $M$ is introduced here:

4. Lebesgue measure:

Answer 1

If the underlying space has finite measure, then continuity of measure actually corresponds directly to continuity in a pseudometric space. Specifically, the pseudometric is $d(A,B)=\| \chi_A - \chi_B \|_{L^1}=m(A \Delta B)$, where $\Delta$ denotes the symmetric difference. After quotienting out by the relation $d(A,B)=0$, Royden and Fitzpatrick call this the Nikodym metric space.

Answer 2

Not clearly. Continuity property of Lebesgue measure is just a nature term defined that measure of a limit is equal to limit of the measure. And with this, we can take limit in parentheses out (like continuity in a metric space that $f$ is continuous <=> $limit_{n->+∞} f(x_n) = f(limit_{n->+∞} x_n$) whenever $limit_{n->+∞} x_n = x$).

Answer 3

To expand a little on your own answer to your own question, if you have a probability space $(X, \mathcal{M}, m)$ where $\mathcal{M}$ is complete, then we can set up a pseudo metric by letting $d(A, B) = m(A\cup B) - m(A\cap B).$ Then, if you have a sequence of sets $\{A_n\}_{n\in\mathbb N}$ we say that $\lim_{n\to\infty} A_n$ exists iff $$\liminf_{n \rightarrow \infty} A_n = \bigcup_{n \ge 1} \bigcap_{j \geq n} A_j = \bigcap_{n \ge 1} \bigcup_{j \geq n} A_j = \limsup_{n \rightarrow \infty} A_n$$ where the first and third equalities are definitions.

We then want $m(\lim_{n\to\infty} A_n) = \lim_{n\to\infty} m(A_n).$ The results you quote are exactly what we need to get that.

In fact, if you factor $\mathcal M$ by the equivalence relation $A \sim B$ iff $d(A, B)=0$ then $M$ is a complete metric space. There's even a pretty interesting group action on $M/\sim$ where $A\oplus B = (A\cup B) - (A\cap B)$ which preserves the metric. The isometries that fix the origin, $\emptyset,$ are invertible measure preserving transformations $T:X\to X$. You even get a weird, quite geometric, result that given $A, B, C$, there is a unique $D$ such that $d(A, D) + d(D, B) = d(A, B)$, $d(B, D) + d(D, C) = d(B, C)$ and $d(A, D) + d(D, C) = d(A, C).$