Theorem 24.6.5 (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 25.2.2, $\int f_{\alpha} d\mu \le \int f d\mu$ for each $\alpha \in A$.
On the other hand, using Lemma 25.2.3, 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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