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We say that the non-empty set $S$ with partial order $\leq$ is directed set if for any $s,t\in S$ we have a $u\in S$ such that $s\leq u, t\leq u$. A net is a function from directed $S$ into the any space $X$.

In the topological space $X$ we say that a net $\{s_\alpha\}\subset X$ converges to $s\in X$ ( where $s_\alpha=f(\alpha)$ for some directed set $S$) if for any open set $V\subset X$ with $s\in V$ we have some $\alpha_0$ such that for any $\alpha_0\leq\alpha$, $s_\alpha\in V$ and we write $s_\alpha\to s$.

When $S$ is countable we name $\{s_\alpha\}$ a sequence. On the other side it was proven that for any positive measure $\mu$ on $X$ and any sequence $\{h_n\}$ of $\mu$-measurable complex function if $h(x)=\lim_{n\to\infty} h_n(x)$ and if there is $g\in L^1(x,\mu)$ such that for any $n\in N$ and any $x\in X$ $$|h_n(x)|\leq g(x)$$ then we'll have $h\in L^1(x,\mu)$ and $$\lim_{n\to\infty}\int_X h_nd\mu=\int_Xh d\mu$$ Now the question is this:

Is the above theorem true for nets which are not sequence?

Martin Sleziak
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Hamid Shafie Asl
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    I think that the word "uniform" has nothing to do with your question, you may want to remove it from the title. Nevertheless, I would like to add to your question a second one, related: if $X$ is compact and $\lim \limits {\delta \in \Delta} f\delta = f$ uniformly (as a net, not as a sequence, with $\Delta$ a directed set), is it true that $\lim \limits {\delta \in \Delta} \int \limits _X f\delta \Bbb d \mu = \int \limits _X f \Bbb d \mu$? – Alex M. Jun 18 '15 at 13:06
  • Thanks! Also your question is good one. – Hamid Shafie Asl Jun 18 '15 at 13:26
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    Related: A net version of dominated convergence? There is a counterexample with nets of functions on $[0,1]$. – Martin Sleziak Jun 18 '15 at 14:13
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    @AlexM. It definitely works for continuous functions, since uniform convergence is simply convergence in $C(X)$ endowed with sup-norm and $f\mapsto \int_X f d\mu$ is a continuous function from $C(X)$ to $\mathbb R$. – Martin Sleziak Jun 18 '15 at 15:53

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