# Is there an analogue of Lebesgue’s Dominated Convergence Theorem for a net $(f_{\alpha})_{\alpha \in A}$ of measurable functions? [duplicate]

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 $\displaystyle \lim_{\alpha \in A} \int_{\Omega} f_{\alpha} ~ \mathrm{d}{\mu} = \int_{\Omega} f ~ \mathrm{d}{\mu}$?

## marked as duplicate by Alex M., Claude Leibovici integration StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 23 '18 at 9:20

• Is $\{f_{\alpha}\}$ a countable collection? – Alex Schiff Jun 24 '14 at 19:55
• I believe you can use the usual version as long as you know the pointwise convergence is true. – Tunococ Jun 24 '14 at 21:01
• what is the meaning of net? – Guy Fsone Jan 2 '18 at 20:03

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}

• To be more precise, $\displaystyle \lim_{\alpha \to \omega_{1}} f_{\alpha}$ denotes the pointwise limit of the net $(f_{\alpha})_{\alpha < \omega_{1}}$. – Berrick Caleb Fillmore Jun 25 '14 at 0:11

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 $\lim_\alpha \| f-f_\alpha\|_1 = 0$‎.

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?

• 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.) – Alex M. Jan 2 '18 at 19:56
• @AlexM indeed. Thank you for the comment - I'm debating whether it's worth leaving my flawed answer for its educational value. – Clement C. Jan 3 '18 at 16:05