## 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.