Consider the probability space $([0,1],\mathcal{B}([0,1]),\lambda)$, where $\lambda(\cdot)$ is the Lebesgue measure and let $\{F_\gamma\}_\Gamma$ be an uncountable collection of closed sets in $[0,1]$ such that $(\forall \Gamma'\subset\Gamma)(|\Gamma'|<\aleph_0\Rightarrow\lambda(\bigcap_{\gamma\in\Gamma'}F_\gamma)\geq\frac{1}{2})$. Show that $\lambda(\bigcap_{\gamma\in\Gamma}F_\gamma)\geq\frac{1}{2}$.
If $\Gamma$ was countable this would be obvious but dealing with the the uncountable case is where I am not sure how to proceed.
Since the $F_\gamma$ are closed any intersection of $F_\gamma$'s are also closed and so Borel measurable and hence they are also Lebesgue measurable so $\lambda(\bigcap_{\gamma\in\Gamma} F_\gamma)$ is well defined.
For any $\Gamma'\subset\Gamma$ with $|\Gamma'|=\aleph_0$, wlog we can say $\Gamma'=\{\gamma_n\}_{n\in\mathbb{N}}$, so we trivially have the decreasing sequence $A_n= (\bigcap_{i = \gamma_i}^{\gamma_n} F_{\gamma_i})$ and so by continuity of probability measures we have $\lambda(\bigcap_{\gamma\in\Gamma'} F_\gamma)=\lim_n\lambda(A_n)\ge\frac{1}{2}$.
Also just to note as $F_\gamma$'s (and any intersections of them) are closed in $[0,1]$ they are also compact in $[0,1]$.
My only idea was to consider the outer-measure $\lambda^*$ with closed(hence compact) sets of $[0,1]$ as the generating set of the Borel-algebra of $[0,1]$. Then I tried to show that for any cover (of compact sets $\bigcup_n C_n$) of $\bigcap_{\gamma\in\Gamma}F_\gamma$ we can find a countable $\Gamma'$ such that $\bigcap_{\gamma\in\Gamma'}F_\gamma\subset \bigcup_nC_n$ which will then give the required result, but I've gotten nowhere with this and don't think it is correct.
So any help or feedback with this problem is greatly appreciated. Thanks in advance.
Take $U_{\gamma}=[0,1]\setminus F_{\gamma}$. What we want to show is equivalent to $\lambda(\bigcup_{\Gamma}U_{\gamma})\leq \frac{1}{2}$. Take an open set $U'$ so that $\lambda(U'\setminus([0,1]\setminus \bigcup\limits_{\Gamma}{U_{\gamma}}))\leq \varepsilon$ and $[0,1]\setminus \bigcup\limits_{\Gamma}{U_{\gamma}}\subseteq U'$. Then $\{U' \cap [0,1] \} \cup \{U_{\gamma}\}_{\Gamma}$ is an open covering of $[0,1]$, by compactness we can extract a finite family $\Gamma'\subseteq \Gamma$ so that $$\bigcup_{\Gamma}U_{\gamma} \subseteq \bigcup_{\Gamma'}U_{\gamma} \cup (U'\cap \bigcup_{\Gamma}U_{\gamma} )$$ Then $$\lambda(\bigcup_{\Gamma}U_{\gamma})\leq \lambda(\bigcup_{\Gamma'}U_{\gamma}) + \lambda(U'\setminus([0,1]\setminus \bigcup\limits_{\Gamma}{U_{\gamma}}))$$ But $$\lambda(U'\cap \bigcup_{\Gamma}U_{\gamma})\leq \lambda(U'\setminus([0,1]\setminus \bigcup\limits_{\Gamma}{U_{\gamma}}))\leq \varepsilon$$ And by our hypothesis $$\lambda(\bigcup_{\Gamma'}U_{\gamma})=\lambda([0,1]\setminus \bigcap_{\Gamma}F_{\gamma})\leq \frac{1}{2}$$ So $$\lambda(\bigcup_{\Gamma}U_{\gamma})\leq \frac{1}{2}+\varepsilon$$ By taking $\varepsilon \to 0$ the result follows.