A net version of dominated convergence theorem

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.

Leave a Reply