Is there an analogue of Lebesgue’s Dominated Convergence Theorem for a net $ (f_{\alpha})_{\alpha \in A} $ of measurable functions? Is there an analogue of Lebesgue’s Dominated Convergence Theorem for a net $ (f_{\alpha})_{\alpha \in A} $ of measurable functions defined on a measure space $ (\Omega,\Sigma,\mu) $, where the index set $ A $ (assumed to be directed) is not necessarily $ \mathbb{N} $? That is, if


$ (f_{\alpha})_{\alpha \in A} $ converges pointwise almost everywhere to a measurable $ f $,
there exists a measurable $ g $ such that $ |f_{\alpha}| \leq |g| $ almost everywhere for each $ \alpha \in A $, and
$ \displaystyle \int_{\Omega} |g| ~ \mathrm{d}{\mu} \leq \infty $,


then does $ \displaystyle \lim_{\alpha \in A} \int_{\Omega} f_{\alpha} ~ \mathrm{d}{\mu} = \int_{\Omega} f ~ \mathrm{d}{\mu} $?
Here is an interesting case study involving a directed and well-ordered index set.

Assume that the Continuum Hypothesis holds, i.e., $ 2^{\aleph_{0}} = \aleph_{1} $. As the cardinality of the closed interval $ [0,1] $ is $ 2^{\aleph_{0}} $, we can find a bijection $ \phi: \omega_{1} \to [0,1] $. Define a net $ (f_{\alpha}: [0,1] \to [0,1])_{\alpha &lt; \omega_{1}} $ of Lebesgue-measurable functions by
$$
\forall \alpha &lt; \omega_{1}, ~ \forall x \in [0,1]: \quad
{f_{\alpha}}(x) \stackrel{\text{def}}{=}
\begin{cases}
1 &amp; \text{if $ x = \phi(\beta) $ for some $ \beta \leq_{\mathbf{On}} \alpha $}; \\
0 &amp; \text{elsewhere}.
\end{cases}
$$
Consider an ordinal $ \alpha &lt; \omega_{1} $. As $ \alpha $ is countable, it follows that $ f_{\alpha} $ assumes the value $ 1 $ for at most countably many arguments and the value $ 0 $ everywhere else. Hence, $ f_{\alpha} $ is Lebesgue measurable and is zero almost everywhere. We thus have
$$
\forall \alpha &lt; \omega_{1}: \quad
\int_{[0,1]} f_{\alpha} ~ \mathrm{d}{\mu} = 0.
$$
However, we have $ \displaystyle \lim_{\alpha \to \omega_{1}} f_{\alpha} = 1_{[0,1]} $, which yields
\begin{align}
       \lim_{\alpha \to \omega_{1}} \int_{[0,1]} f_{\alpha} ~ \mathrm{d}{\mu}
&amp; =    \lim_{\alpha \to \omega_{1}} 0 \\
&amp; =    0 \\
&amp; \neq 1 \\
&amp; =    \int_{[0,1]} 1_{[0,1]} ~ \mathrm{d}{\mu} \\
&amp; =    \int_{[0,1]}
       \left( \lim_{\alpha \to \omega_{1}} f_{\alpha} \right)
       \mathrm{d}{\mu}.
\end{align}
Most likely wrong, see comment below (leaving it as the mistake may be instructive)

The same theorem does apply, from what I recall. Take any sequence $(a_n)$ such that $a_n \xrightarrow[n\to\infty]{} a$, and set $g_n\stackrel{\rm def}{=} f_{a_n}$. Can't you apply your version of Lebesgue's Dominated Convergence Theorem to $(g_n)_{n\in\mathbb{N}}$, which implies what you want as $(a_n)_{n\in\mathbb{N}}$ is arbitrary?