Lemma 1. Suppose $E$ is a Riesz space, $F$ is an order ideal of $E$
and $(f_i)_{i\in I}$ is a net in $F$. Then $f_i$ converges (in order) to
an element $f$ in $F$ iff and only if $f_i$ converges to $f$ in $E$ and
$|f_i|\leq \bar f$ for some $\bar f\in F$ when $i$ is sufficiently
large.
Proof. Suppose $f$ is the order limit of $(f_i)$ in $F$, then we can
find a decreasing net $(g_j)_{j\in J}$ in $F$ with $\inf_{j\in J} g_j=0$
such that for any $j\in J$ there exists some $i_0\in I$ satisfying
$|f_i-f|\leq g_j$ for all $i\geq i_0.$ Pick $g\in E$ that $g\leq g_j$
for all $j\in J.$ Since $F$ is an ideal, $g$ is an element of $F$ and
thus less than $0$. Therefore, the infimum of $(g_j)_{j\in J}$ in $E$ is
also $0$. As a result, $f_i$ converges to $f$ in $E$ too. The proof for
the
only if part ends by taking $\bar f=|f|+g_{i_0}$.
Conversely, we suppose $f_i$ converges to $f$ in $E$ and
$|f_i|\leq \bar f$ for some $\bar f\in F$ when $i$ is sufficiently
large. Pick a decreasing net $(g_j)_{j\in J}$ with in $E$ with
$\inf_{j\in J} g_j=0$ such that for any $j\in J$ there exists some
$i_0\in I$ satisfying $|f_i-f|\leq g_j$ for all $i\geq i_0.$ We may
assume $g_j\leq |f|+\bar f,$ otherwise, replace $g_j$ with
$g_j\wedge(|f|+\bar f)$. It follows from $F$ is an ideal of $E$ that
$g_j\in F$. Since $F$ is a subset of $E$, any $g\in F$ satisfying
$g\leq g_j$ for all $j\in J$ satisfies $g\leq 0$. Consequently, the
infimum of $(g_j)_{j\in J}$ in $F$ is also $0$ and hence $f_i$ converges
to $f$ in $F$. ◻
The following theorem is an abstract statement of dominated convergence
theorem in Banach lattice.
Theorem 2. Suppose $E$ is a Riesz space, $F$ is an order ideal of
$E$ equipped with an order continuous norm. Let $(f_i)$ be a net order
convergent to some $f$ in $E$ and $|f_i|\leq \bar f$ for some
$\bar f\in F$. Then $f_i$ is norm convergent to $f$ in $F$.
Proof. By the above lemma, $f_i$ is order convergent to $f$ in $F$.
The order continuity of the norm on $F$ yields that $f_i$ is norm
convergent to $f$. ◻
Now, we can give a net version of the canonical Dominated Convergence
Theorem.
Corollary 3. Let $(\Omega,\mathcal{A},\mu)$ be a measure space,
$(f_i)$ a net of ${\mathcal {A}}$-measurable functions. Assume the net
$(f_i)$ converges $\mu$-almost everywhere to an
$\mathcal {A}$-measurable function $f$, and is dominated by a
$g \in L^1(\Omega,\mathcal{A},\mu)$ i.e., for each $i$ we have:
$|f_i| \leq g$, $\mu$-almost everywhere. Then f is in
$L^1(\Omega,\mathcal{A},\mu)$ and
$$\lim_{i} \int_\Omega |f_i-f|\,d\mu = 0$$ which also implies
$$\lim_i\int_\Omega f_i\,d\mu = \int_\Omega f\,d\mu.$$
Proof. Let $E$ be the linear space of all $\mathcal{A}$-measurable
functions. It is an Riesz space, with respect to the order defined by
$f\leq g$ if $f(x)\leq g(x)$ $\mu$-a.e., in which a net $(f_i)$ is order
convergent if and only if $(f_i(x))$ is convergent $\mu$-a.e.. By
Theorem 2.4.2, (Meyer-Nieberg, 2012), the canonical norm on
$L^1(\Omega,\mathcal{A},\mu)$ is order continuous. Applying
Theorem 2 with
$F=L^1(\Omega,\mathcal{A},\mu)$, we finishes the proof. ◻
Meyer-Nieberg, P. (2012). Banach lattices. Springer Science &Business Media.