Show that $\exists $ indices $i_1<i_2<i_3<i_4$ such that $F_{i_1}\cap F_{i_2}\cap F_{i_3}\cap F_{i_4}\neq \emptyset$

Question

Let $(X,\mathcal F,\mu )$ be a measure space such that $\mu (X)=1$ and suppose that $F_1,F_2,\ldots ,F_7$ are measurable sets.$\mu (F_i)\ge \frac{1}{2}$. Show that $\exists $ indices $i_1<i_2<i_3<i_4$ such that $F_{i_1}\cap F_{i_2}\cap F_{i_3}\cap F_{i_4}\neq \emptyset$

Let forall indices $i_1<i_2<i_3<i_4$ we have that $F_{i_1}\cap F_{i_2}\cap F_{i_3}\cap F_{i_4}= \emptyset$

Then $\mu (F_{i_1}\cup F_{i_2}\cup F_{i_3}\cup F_{i_4}\cup F_{i_5}\cup F_{i_6}\cup F_{i_7})\le \frac{7}{2}$

How to arrive at a contradiction from here?

Please help.

Answer 1

Let us define $$ F(x) = \sum_{j=1}^7 1_{F_j}(x). $$ Assume to the contrary that there is no $i_1<i_2<i_3<i_4$ such that $\cap_{k=1}^4 F_{i_k} \neq \varnothing$. Since $F(x)$ counts the number of $j$ for which $x\in F_j$, it follows that $F(x) \le 3$ for all $x$. But this leads to a contradiction: $$\begin{eqnarray} \frac{7}{2}\le \sum_{j=1}^7 \mu(F_j)&=&\sum_{j=1}^7\int_X 1_{F_j}d\mu\\&=&\int_X \left(\sum_{j=1}^7 1_{F_j}\right)d\mu \\&=&\int_X F(x)d\mu\\&\le& \int_X 3d\mu=3\mu(X) =3. \end{eqnarray}$$ Hence there must a quadruple $i_1<i_2<i_3<i_4$ such that $\cap_{k=1}^4 F_{i_k} \neq \varnothing$.

Answer 2

The statement is wrong: take $X=\{1,\ldots,7\}$ with the uniform probability measure and $F_i=\{i\}$.