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This question already has an answer here:

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} $?

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marked as duplicate by Alex M., Claude Leibovici integration May 23 '18 at 9:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Is $\{f_{\alpha}\}$ a countable collection? $\endgroup$ – Alex Schiff Jun 24 '14 at 19:55
  • $\begingroup$ I believe you can use the usual version as long as you know the pointwise convergence is true. $\endgroup$ – Tunococ Jun 24 '14 at 21:01
  • $\begingroup$ what is the meaning of net? $\endgroup$ – Guy Fsone Jan 2 '18 at 20:03
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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 < \omega_{1}} $ of Lebesgue-measurable functions by $$ \forall \alpha < \omega_{1}, ~ \forall x \in [0,1]: \quad {f_{\alpha}}(x) \stackrel{\text{def}}{=} \begin{cases} 1 & \text{if $ x = \phi(\beta) $ for some $ \beta \leq_{\mathbf{On}} \alpha $}; \\ 0 & \text{elsewhere}. \end{cases} $$ Consider an ordinal $ \alpha < \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 $$ \forall \alpha < \omega_{1}: \quad \int_{[0,1]} f_{\alpha} ~ \mathrm{d}{\mu} = 0. $$ 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} & = \lim_{\alpha \to \omega_{1}} 0 \\ & = 0 \\ & \neq 1 \\ & = \int_{[0,1]} 1_{[0,1]} ~ \mathrm{d}{\mu} \\ & = \int_{[0,1]} \left( \lim_{\alpha \to \omega_{1}} f_{\alpha} \right) \mathrm{d}{\mu}. \end{align}

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  • $\begingroup$ To be more precise, $ \displaystyle \lim_{\alpha \to \omega_{1}} f_{\alpha} $ denotes the pointwise limit of the net $ (f_{\alpha})_{\alpha < \omega_{1}} $. $\endgroup$ – Berrick Caleb Fillmore Jun 25 '14 at 0:11
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Here is a nice theorem that might interest you ("‎Methods of Modern Mathematical Physics‎: ‎Functional analysis‎", ‎Volume 1‎, ‎Michael Reed‎, ‎Barry Simon)‎:

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

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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?

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    $\begingroup$ First of all, $a$ does not exist, in general. Second, if a property is true for the subsequences of a net, it is not necessarily true for the net itself. (Nets are really ugly, subnets even more so.) $\endgroup$ – Alex M. Jan 2 '18 at 19:56
  • $\begingroup$ @AlexM indeed. Thank you for the comment - I'm debating whether it's worth leaving my flawed answer for its educational value. $\endgroup$ – Clement C. Jan 3 '18 at 16:05

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