A monotone convergence theorem for nets in Reed and Simon, Vol. I

Question

Reed and Simon state the following monotone convergence theorem for nets (Theorem IV.15, Vol. I).

Let $\mu$ be a regular Borel measure on a compact Hausdorff space $X$. Let $\{f_\alpha\}_{\alpha\in I}$ be an increasing net of continuous functions. Then, $f=\lim_\alpha f_\alpha \in L^1(X,d\mu)$ if and only if $\sup_\alpha \| f_\alpha \|_1 < \infty$ and in that case $\lim_\alpha \|f-f_\alpha\|_1 = 0 $.

Unfortunately, they do not give a proof. Can anyone provide one, or give a reference to one? In the notes to the chapter containing the theorem, Reed and Simon cite Bourbaki's Integration, but I've found that text difficult to navigate. I do not have access to the other text they cite (Nachbin).

This is not quite your classical monotone convergence theorem, as it requires each $f_\alpha$ to be continuous. This is necessary in order to rule out common counter-examples (see, for instance, this question or these answers) to the statement of the dominated convergence theorem for nets instead of sequences.

I could try adapting a usual proof for the MCT. However, I don't quite see the role that continuity of the $f_\alpha$'s play in the result, so I'm not sure how to make use of it.

Also, I'm not actually certain that the hypotheses of the theorem rule out the example here. I suspect the monotonicity requirement would get in the way of choosing the net of functions required in the example, but I don't have a proof of that either.

Reference

Reed, Michael; Simon, Barry, Methods of modern mathematical physics. I: Functional analysis. Rev. and enl. ed, New York etc.: Academic Press, A Subsidiary of Harcourt Brace Jovanovich, Publishers, XV, 400 p. $ 24.00 (1980). ZBL0459.46001.

Answer 1

Here is a proof with the added assumption that for each $\alpha \in I$, $f_\alpha \geq 0$ (but without the assumption that $X$ is compact). In this case, all of the integrals involved are automatically well-defined as elements of $[0,\infty]$ and the result boils down to showing that $$\int f d\mu = \sup_{\alpha \in I} \int f_\alpha d\mu$$ since it is easy to check that the right hand side is the same as $\lim_\alpha \int f_\alpha d \mu$. In fact, it is clear that $\int f d\mu \geq \sup_\alpha \int f_\alpha d\mu$ so we only have the other inequality to show.

If $U_j^n = f^{-1}((j2^{-n}, \infty))$ then $$\varphi_n = 2^{-n} \sum_{j=1}^{2^{2n}} \chi_{U_j^n}$$ is a sequence of simple functions increasing to $f$ (this is one of the standard constructions of such a sequence). The key point is that since $f = \sup_{\alpha} f_\alpha$ and the $f_\alpha$ are continuous, $f$ is lower semicontinuous and so each $U_j^n$ is open. This will allow us to leverage regularity of $\mu$.

Fix $\varepsilon > 0$. For $n$ sufficiently large, by the M.C.T. for sequences, we have that $\int \varphi_n d \mu > \int f d\mu - \varepsilon$. Fix also an $n$ large enough such that this inequality holds.

By regularity of the measure $\mu$, we can find compact sets $K_j \subset U_j^n$ such that if $\Phi = 2^{-n} \sum_{j=1}^{2^{2n}} \chi_{K_j}$ then $\int \Phi d\mu > \int f d\mu - \varepsilon$ also. Then, since $f > \varphi_n \geq \Phi$, for each $x \in X$ there is an $\alpha_x$ such that $f_{\alpha_x}(x) > \Phi(x)$. The point of this construction is that since $X$ is Hausdorff, each $K_j$ is closed so that $\Phi$ is upper-semicontinuous. Hence $f_{\alpha_x} - \Phi$ is lower semicontinuous and so $$O_x = \{y: f_{\alpha_x}(y) - \Phi(y) > 0\}$$ is open. Now $\{O_x: x \in X\}$ is an open cover for the compact set $\bigcup_{j=1}^{2^{2n}} K_j$ and hence there are $x_1, \dots, x_m$ such that $\{O_{x_i}: i = 1, \dots, m\}$ is also an open cover.

We can pick an $\alpha_0 \in I$ such that $\alpha_0 \geq \alpha_{x_i}$ for each $i$. This guarantees that $f_{\alpha_0} - \Phi \geq 0$ everywhere. Hence $$\sup_\alpha \int f_\alpha d \mu \geq \int f_{\alpha_0} d \mu \geq \int \Phi d \mu \geq \int f d \mu - \varepsilon.$$

$\varepsilon > 0$ was arbitrary so the result follows.

---

In the comments, I gave an example that shows that with only the assumptions given in the question the result "$f \in L^1(\mu)$ if and only if $\sup_\alpha \|f_\alpha\|_1 < \infty$" is incorrect.

However, in the above, I don't use the assumption that $X$ is compact. With this assumption, one can show that with the assumptions given in the question the result "$f \in L^1(X, d\mu)$ if and only if for every $\alpha_0 \in I$, $\sup_{\alpha \geq \alpha_0} \|f_\alpha\|_1 < \infty$" is true.

The key point is that this tells us that for each $\alpha_0$, $\min_{x \in X} f_{\alpha_0}(x) > - \infty$ and that $\mu(X) < \infty$ by regularity of $X$. The result I give then follows by applying the result already proved in this question with functions of the form $f_\alpha - \min_{x \in X} f_{\alpha_0}(x)$.

Answer 2

Despite the beautiful answer given by Rhys Steele, I feel we can get a little bit further. Infact in the proposed answer, convergence of the integral $ \int f d \mu$ is assumed. It seems to me however that such an assumption is not necessary. Below I give a modified (but strongly inspired) version of the argument proposed by Rhys, that does not assume convergence of such integral.

Since $f_\alpha \leq f$ it is clear that

\begin{equation*} \sup_{\alpha \in I} \int f d \mu \leq \int f d \mu \end{equation*}

It remains then only to show the reverse inequality.

If $U_j^n = f^{-1}((j2^{-n}, +\infty)), j,n \in \mathbb{N}$ then

\begin{equation*} \varphi_n = \sum_{j = 1}^{2^{2n}} \chi_{U_j^n} \end{equation*}

is a sequence of step functions increasing to $f$ and strictly smaller.

Since $f = \sup_{\alpha \in I} f_\alpha$, $f$ is lower semicontinuous and so each $U_j^n$ is open.

By the monotone convergence theorem $\int \varphi_n d \mu \to \int f d \mu$.

Since $\mu$ is radon, for each $n$, and opens $U_j^n$, we find increasing compact sets $K_j^{n,m} \subseteq U_j^n, m \in \mathbb{N}$ such that $\mu(U_j^n) = \sup_{m} \mu(K_j^{n,m})$, let then $\phi_{n,m} = \sum_{j = 1}^{2^{2n}} \chi_{K_j^{n,m}}$, so that $\phi_{n,m} \leq \varphi_n < f$, and $\lim_{m \to +\infty} \int \phi_{n,m} d \mu = \int \varphi_n d \mu$. Hence for every neighborhood $N$ of $\int f d \mu$, and $n$ large enough, $\int \varphi_n d \mu \in N$, so that $N$ is a neighborhood for $\int \varphi_n d \mu$ also, and we can take $m$ large enough so that $\int \phi_{n,m} \in N$. Chosing in such a way an $m(n)$ for every $n$, we get a sequence $\phi_n := \phi_{n, m(n)}$ such that $\int \phi_n d \mu \to \int f d \mu$, $\phi_n < f$.

Since $X$ is hausdorff, every compact is closed, so that $\phi_{n,m}$ is upper semicontinuous for all $n, m \in \mathbb{N}$, hence same is true for $\phi_n$, so that $f_\alpha - \phi_n$ is lower semicontinuous for all $\alpha$ . Note also that $\phi_n$ has compact support.

Now, fix $n$, by the definition of $f$, and since $f > \phi_n$, for every $x \in X$ there is $\alpha_x \in I$ such that $f_{\alpha_x}(x) > \phi_n(x)$. Since $f_{\alpha_x} - \phi_n$ is lower semicontinuous, the set $O_{\alpha_x} := \{x : f_{\alpha_x}(x) - \phi_n(x) > 0\}$ is open, so that $\bigcup_{\alpha_x} O_{\alpha_x}$ is an open cover for $\text{supp}(\phi_n)$, and so there are finitely many indices $\alpha_1 ... \alpha_k \in I$ such that $\text{supp}(\phi_n) \subseteq O_{\alpha_1} \cup ... \cup O_{\alpha_k}$. Take $\alpha \geq \alpha_i \; i \in \{1,...k\}$, then $f_\alpha \geq f_{\alpha_i} \; i \in \{1,...k\}$, and $f_\alpha \geq \phi_n$.

We have thus showed that for every $n \in \mathbb{N}$, we can find $\alpha_n \in I$ such that $f_{\alpha_n} \geq \phi_n$, and thus

\begin{equation*} \int f_{\alpha_n} d \mu \geq \int \phi_n d \mu \end{equation*} so that, taking the supremum on the left, and the limit on the right, gives

\begin{equation*} \sup_{\alpha \in I} \int f d \mu \geq \int f d \mu \end{equation*}

Note that the hausdorff condition is used only to conclude that compact sets are closed, so one can exchage the two hypothesis.