On the equality of two sets (a doubt from Probability with Martingales). Let $(S, \Sigma, \mu) $ be $([0,1], \mathcal{B}[0,1], Leb)$. Let $\epsilon(k)$ be a sequence of strictly positive numbers s.t. $\epsilon(k) \downarrow 0$.
Let $V = Q \cap [0,1],$ the set of rationals in $[0,1]$. Since $V$ is a countable union of singletons it has Lebesgue measure $0$. We can include $V$ in an open subset of $S$ of measure at most $4 \epsilon(k)$ as follows:
$$V \subset G_k = \bigcup_{n \in N} [(v_n - \epsilon(k)2^{-n}, v_n + \epsilon(k)2^{-n}) \cap S] =: \bigcup_n I_{n,k} $$
The text then states that $\bigcap_k G_k \ne V$. I am having troubles seeing why this is true. 
My (wrong) reasoning is as follows: I am taking the countable intersection of sets of the form $[(v_n - \epsilon(k)2^{-n}, v_n + \epsilon(k)2^{-n}) \cap S]$ for every $k$, I notice that $\lim_{k \rightarrow \infty} (v_n - \epsilon(k)2^{-n}) = v_n$ and $\lim_{k \rightarrow \infty} (v_n - \epsilon(k)2^{-n}) = v_n$ but $v_n$ is present in every set so the intersection is just the single point $v_n$ and the same for all other sets. This would imply that $\bigcap_k G_k $ is included in $V$ but then $\bigcap_k G_k = V$.
SO why is $\bigcap_k G_k \ne V$? or in other words why is $\bigcap_k G_k $ not included in $V$.
 A: We can prove that the intersection contains irrational numbers:
By construction, each $G_k$ is an open subset of $S$, and $V \subset G_k$, hence $G_k$ is a dense open subset of $S$. Then $U_k := G_k \setminus \{v_k\}$ is also a dense open subset of $S$. By Baire's theorem,
$$\bigcap_{k = 0}^\infty U_k \neq \varnothing,$$
and
$$\bigcap_{k = 0}^\infty U_k = \bigcap_{k = 0}^\infty (G_k \setminus \{v_k\}) = \Biggl(\bigcap_{k = 0}^\infty G_k\Biggr) \setminus \Biggl(\bigcup_{k = 0}^\infty \{v_k\}\Biggr) = \Biggl(\bigcap_{k = 0}^\infty G_k\Biggr) \setminus V$$
then shows $\bigcap_k G_k \not\subset V$. In fact, $S \setminus \bigcap_k U_k = \bigcup_k (S\setminus U_k)$ is a union of countably many nowhere dense subsets of $S$, and such sets are called meagre (or "of the first category"). Meagre sets are in a topological sense small (when the ambient space is not itself meagre), so topologically, most points of $S$ lie in $\bigcap_k U_k$ - although that is a set of measure $0$, so measure-theoretically, most points of $S$ lie outside of $S$; despite the close connection of the Lebesgue measure and the topology of $S$, the two concepts have very different ideas about "largeness" of sets.
For completeness, we can include a proof of Baire's theorem, where we have the free choice whether we use the version for complete metric spaces or the version for locally compact spaces. Let's use the version for locally compact spaces.
Since $U_0$ is dense and $S\neq \varnothing$,, we have $U_0 \neq \varnothing$. Pick an $x_0 \in U_0$. Since $S$ is locally compact, and $U_0$ is a neighbourhood of $x_0$, there is a compact neighbourhood $K_0$ of $x_0$ contained in $U_0$.
Once we have picked points $x_0,x_1,\dotsc, x_{k-1}$ and compact neighbourhoods $K_j$ of $x_j$ satisfying $K_j \subset K_{j-1}$ for $0 < j < k$, we can continue the construction by noting that $U_k$ is a dense open set, and $K_{k-1}$ has nonempty interior (since it is a neighbourhood of $x_{k-1}$), hence $W_k := U_k \cap \overset{\Large\circ}{K}_{k-1} \neq \varnothing$. Now we pick an $x_k \in W_k$, and by local compacteness of $S$, a compact neighbourhood $K_k$ of $x_k$ such that $K_k \subset W_k$.
In that way we obtain a nested sequence $K_0 \supset K_1 \supset K_2 \supset \dotsc$ of nonempty compact sets, and by compactness,
$$\bigcap_{k = 0}^\infty K_k \neq \varnothing.$$
But by construction $K_k \subset U_k$ for all $k$.

So why does your reasoning suggest that $\bigcap_k G_k = V$?
You look at one interval $I_{n,k}$ instead of the whole set $G_k$.
Indeed, as $k \to \infty$, for each fixed $n$, the intervals
$$I_{n,k} = (v_n - \epsilon(k)2^{-n}, v_n + \epsilon(k)2^{-n}) \cap S$$
shrink to $\{v_n\}$ as $k \to \infty$. But if we write $\bigcap_k G_k$ as a union of intersections, we get
$$\bigcap_k \Biggl(\bigcup_n I_{n,k}\Biggr) = \bigcup_{f\colon \mathbb{N}\to{\mathbb{N}}} \Biggl( \bigcap_k I_{f(k),k}\Biggr),$$
a union of uncountably many ($2^{\aleph_0}$) intersections of countably many intervals. It is trivial that
$$\bigcap_k I_{f(k),k}$$
is nonempty for every constant $f\colon \mathbb{N}\to \mathbb{N}$ - those give the $\{v_n\}$ - but we also have
$$\bigcap_k I_{f(k),k} \neq \varnothing$$
for many non-constant functions $f$ - it is harder to find functions $f\colon \mathbb{N} \to \mathbb{N}$ that are not eventually constant for which the intersection is nonempty, but Baire's theorem shows that there are uncountably many such functions.
