I have doubts as to how the dominated convergence theorem is applied in contexts where you have nets and not sequences. For context, my question is related to an answer to this question https://math.stackexchange.com/questions/217702/when-can-we-interchange-the-derivative-with-an-expectation#:~:text=2%20Answers,18%2C%202020%20at%2017:50. The answer is:
Lemma. Let $X\in\mathcal{X}$ be a random variable $g\colon \mathbb{R}\times \mathcal{X} \to \mathbb{R}$ a function such that $g(t, X)$ is integrable for all $t$ and $g$ is continuously differentiable w.r.t. $t$. Assume that there is a random variable $Z$ such that $|\frac{\partial}{\partial t} g(t, X)| \leq Z$ a.s. for all $t$ and $\mathbb{E}(Z) < \infty$. Then $$\frac{\partial}{\partial t} \mathbb{E}\bigl(g(t, X)\bigr) = \mathbb{E}\bigl(\frac{\partial}{\partial t} g(t, X)\bigr).$$
Proof. We have $$\begin{align*} \frac{\partial}{\partial t} \mathbb{E}\bigl(g(t, X)\bigr) &= \lim_{h\to 0} \frac1h \Bigl( \mathbb{E}\bigl(g(t+h, X)\bigr) - \mathbb{E}\bigl(g(t, X)\bigr) \Bigr) \\ &= \lim_{h\to 0} \mathbb{E}\Bigl( \frac{g(t+h, X) - g(t, X)}{h} \Bigr) \\ &= \lim_{h\to 0} \mathbb{E}\Bigl( \frac{\partial}{\partial t} g(\tau(h), X) \Bigr), \end{align*}$$ where $\tau(h) \in (t, t+h)$ exists by the [mean value theorem][1]. By assumption we have $$\Bigl| \frac{\partial}{\partial t} g(\tau(h), X) \Bigr| \leq Z$$ and thus we can use the [dominated convergence theorem][2] to conclude $$\begin{equation*} \frac{\partial}{\partial t} \mathbb{E}\bigl(g(t, X)\bigr) = \mathbb{E}\Bigl( \lim_{h\to 0} \frac{\partial}{\partial t} g(\tau(h), X) \Bigr) = \mathbb{E}\Bigl( \frac{\partial}{\partial t} g(t, X) \Bigr). \end{equation*}$$ This completes the proof.
My question: when using the Dominated convergence theorem, it is necessary that we are working with sequences, but here we are using a net or real numbers $\left(\mathbb{E}\left(\frac{\partial}{\partial t}g(\tau(h),X)\right)\right)_h$. I thought the solution was to just say, consider any subnet, and then take any subsequence $\left(\mathbb{E}\left(\frac{\partial}{\partial t}g(\tau(h(n)),X)\right)\right)_n$ of that subnet. Use the dominated convergence theorem on this subsequence to show that that the subsequence converges to $\mathbb{E}\left(\frac{\partial}{\partial t}g(t,X)\right)$, and then that forces the original net to converge to the same limit (a net converges to a limit if every subnet has a further subnet that converges to that limit). However, after considering this problem Counter example to dominated convergence for nets, I have doubts as to whether this reasoning is correct. Since some nets do not have subsequences, how can we know that every subnet of $\left(\mathbb{E}\left(\frac{\partial}{\partial t}g(\tau(h),X)\right)\right)_h$ will have a subsequence?
EDIT:
The issue is that for a subnet of $\left(\mathbb{E}\left(\frac{\partial}{\partial t}g(\tau(h),X)\right)\right)_h$, we don't generally know what indexing set that subnet is using, and so we can't suppose that subnet will have a subsequence (because the indexing set of the subnet might not have a countable cofinal subset). I think perhaps the solution is that since the net $\left(\mathbb{E}\left(\frac{\partial}{\partial t}g(\tau(h),X)\right)\right)_h$ is a net of real values (and $\mathbb{R}$ is first countable), we can replace the statement,
"The net $\left(\mathbb{E}\left(\frac{\partial}{\partial t}g(\tau(h),X)\right)\right)_h$ converges to a limit if and only if every subnet has a further subnet that converges to that limit"
with
"The net $\left(\mathbb{E}\left(\frac{\partial}{\partial t}g(\tau(h),X)\right)\right)_h$ converges to a limit if and only if every subsequence has a further subsequence that converges to that limit"
Then, since the indexing set $[0,\infty,\geq_{order})$, where $t\geq{order} s$ if and only if $t<s$, clearly has countable cofinal subsets, it is not an issue finding subnets of $\left(\mathbb{E}\left(\frac{\partial}{\partial t}g(\tau(h),X)\right)\right)_h$.
Does this seem correct?
