Definition 24.6.2 (Ky Fan Metric).label Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and
then:
- (1)
$\alpha$ is a metric on $L^{0}(X; Y)$.
- (2)
$\alpha$ induces the uniform structure of convergence in measure on $L^{0}(X; Y)$.
The mapping $\alpha$ is the Ky Fan metric on $L^{0}(X; Y)$.
Proof. (1): Let $f, g, h \in \mathcal{L}^{0}(X; Y)$, then
- (M)
If $\alpha(f, g) = 0$, then by continuity from above,
\[\mu\bracs{d(f, g) > 0}= \limv{n}\mu\bracs{d(f, g) > 1/n}\le \limv{n}\frac{1}{n}= 0\]so $f = g$ almost everywhere.
- (PM3)
For each $\eps > 0$,
\[\bracs{d(f, h) > \eps}\subset \bracs{d(f, g) > \eps/2}\cup \bracs{d(g, h) > \eps/2}\]so $\alpha(f, h) \le \alpha(f, g) + \alpha(g, h)$.
so $\alpha$ is a metric on $\mathcal{L}^{0}(X; Y)$, modulo almost everywhere equality.
(2): Let $f, g \in \mathcal{L}^{0}(X; Y)$. For any $\eps, \delta > 0$, if $\alpha(f, g) < \eps \wedge \delta$, then there exists $r \in (0, \eps \wedge \delta]$ such that $\mu\bracs{d(f, g) > r}\le r$. Thus
On the other hand, if $\mu\bracs{d(f, g) > \eps}\le \eps$, then $d(f, g) \le \eps$. Therefore $\alpha$ induces the uniform structure of convergence in measure.$\square$
Post a Comment