26.8 Local Convergence in Measure
Definition 26.8.1 (Locally In Measure*).label Let $(X, \cm, \cf, \mu)$ be a scaffolded measure space and $(Y, d)$ be a separable metric space. For each $\eps, \delta > 0$ and $A \in \cf$, let
then
forms a fundamental system of entourages for a uniformity.
The uniformity defined by $\fB$ is the uniform structure of local convergence in measure, and $\mathcal{L}_{\cf}^{0}(X; Y)$ denotes $\mathcal{L}^{0}(X; Y)$ equipped with this uniformity.
Proof. It is sufficient to check the conditions of Proposition 6.1.8:
- (FB1)
For each $\eps, \eps', \delta, \delta' > 0$ and $A, A' \in \cm$ with $\mu(A), \mu(A') < \infty$,
\[U(A \cup A', \delta \wedge \delta', \eps \wedge \eps') \subset U(A, \delta, \eps) \cap U(A', \delta', \eps')\] - (UB3)
For each $\eps, \delta > 0$, $A \in \cf$, and $f, g, h \in \mathcal{L}^{0}(X; Y)$,
\[\bracs{d(f, h) > \delta}\subset \bracs{d(f, g) > \delta}\cup \bracs{d(g, h) > \delta}\]so $U(A, \delta/2, \eps/2) \circ U(A, \delta/2, \eps/2) \subset U(A, \delta, \eps)$.
$\square$
Proposition 26.8.2.label Let $(X, \cm, \cf, \mu)$ be a scaffolded measure space, $(Y, d)$ be a separable metric space, and $\fF$ be a filter of $(\cm, \cb_{Y})$-measurable functions, then $\fF$ is Cauchy in measure if and only if:
- (L)
$\fF$ is definition:locally-in-measure.
- (T)
For each $\eps, \delta > 0$, there exists $F \in \fF$ and $A \in \cf$ such that
\[\sup_{f, g \in F}\mu(A^{c} \cap \bracs{d(f, g) > \delta}) < \eps\]
Proof. (L) + (T) $\Rightarrow$ (In Measure): Let $\eps, \delta > 0$. By (T) then there exists $F_{1} \in \fF$ and $A \in \cf$ such that
By (L), there exists $F_{2} \in \fF$ with $F_{2} \subset F_{1}$ such that
Therefore
$\square$
Theorem 26.8.3.label Let $(X, \cm, \cf, \mu)$ be a scaffolded localisable measure space and $(Y, d)$ be a Polish space, then $\mathcal{L}^{0}_{\cf}(X; Y)$ is complete.
Proof. Let $\fF \subset \mathcal{L}^{0}_{\cf}(X; Y)$ be a Cauchy filter. By Theorem 26.7.4, for each $A \in \cf$, there exists an almost everywhere unique $f_{A} \in \mathcal{L}^{0}(A; Y)$ such that $\fF$ converges to $f_{A}$ when restricted to $A$. Thus for any $A, B \in \cf$, $f_{A \cup B}|_{A \cap B}= f_{A}|_{A \cap B}= f_{B}|_{B \cap A}$ almost everywhere. By the gluing lemma for measurable functions, there exists $f \in \mathcal{L}^{0}(X; Y)$ such that $f|_{A} = f_{A}$ for all $A \in \cf$. Thus $\fF \to f$ locally in measure, and $\mathcal{L}^{0}(X; Y)$ is complete.$\square$
Theorem 26.8.4 (Monotone Convergence Theorem (in Measure)).label Let $(X, \cm, \mu)$ be a semifinite measure space, $\net{f}\subset \mathcal{L}^{+}(X, \cm)$, and $f \in \mathcal{L}^{+}(X, \cm)$ such that
- (a)
For each $x \in X$, $f_{\alpha}(x) \upto f(x)$.
- (b)
$f_{\alpha} \to f$ locally in measure.
then
Proof. By Definition 27.2.2, $\int f_{\alpha} d\mu \le \int f d\mu$ for each $\alpha \in A$.
On the other hand, using Lemma 27.2.3, it is sufficient to show that
for any $\phi \in \Sigma^{+}(X, \cm)$ satisfying:
- (i)
There exists $\delta > 0$ such that $\phi + \delta \le f$ on $\bracs{\phi > 0}$.
- (ii)
$\phi \in L^{1}(X, \cm)$.
To this end, let $\eps > 0$. Since $\mu\bracs{\phi > 0}< \infty$, by (b), there exists $\alpha \in A$ such that
In which case, by linearity,
As the above holds for all $\eps > 0$, $\int \phi d\mu \le \sup_{\alpha \in A}\int f_{\alpha} d\mu$. Therefore
$\square$
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