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Let $\{0,1\}^{\mathbb{N}}$ be the set of infinite 0-1 sequences, let $(f_n)$ be a sequence of functions such that $f_n:2^{\{1,\ldots,n\}} \to \mathbb{R}_+$, and let $$ F(s) = \lim_{n\to\infty} f_n(s_1,\ldots,s_n). $$ for $s\in\{0,1\}^{\mathbb{N}}$, assuming that it always exists.

What does it mean the statement "$F(s)$ is continuous"?

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    Well, obviously $F:\{0, 1\}^{\mathbb{N}}\to\mathbb{R}$. So I assume the question is about $F$ being continous as a function between two topological spaces. But the real question is: what is the topology on $\{0, 1\}^{\mathbb{N}}$? Is it compact-open? Or something else? Then you'll have your answer. – 2017-02-23
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    On $\{0,1\}^{\Bbb N}$ you have the product topology if you give $\{0,1\}$ the discrete topology. This topology is actually the same as the cofinite topology, meaning a set $U$ is open in it if and only if $\{0,1\}^{\Bbb N}-U$ is finite or $U$ is the empty set. In this topology the only continuous maps to $\Bbb R$ (actually to any Hausdorff space) are the constant maps. – 2017-02-23
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    @s.harp can you please point to me to a reference saying what you said? That is, "In this topology the only continuous maps to $\mathbb{R}$ (actually to any Hausdorff space) are the constant maps". I'd like to understand it better. My (little) background is "Introduction to topology" by Mendelson. – 2017-03-11
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    If $X$ is a Hausdorff space and let $f:\{0,1\}^{\Bbb N}\to X$ be continuous. If $x\neq y$ are two points in $X$, let $U_x,U_y$ be disjoint open neighbourhoods of $x$ and $y$. $f^{-1}(U_x)$ and $f^{-1}(U_y)$ are open from continuity and disjoint because $U_x,U_y$ are. However you can verify that in the topology described above (cocountable) two open sets can be disjoint if and only if one is the empty set, this implies either $x$ or $y$ are not in the image of $f$. I can't really recommend any specific book, I learned pointset topology from Bourbaki's book General Topology. – 2017-03-11

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