A net version of dominated convergence theorem (Wrong)

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 \(f\). ◻

The following is not ture!

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, Peter. 2012. Banach Lattices. Springer Science &Business Media.