Elementary Lebesgue measure problem

Define $X:=\sum_{k=1}^n 1_{E_k}$, the number of $E_k$ that occur. Observe that $1_{\{X=n\}}\ge X-(n-1)$. Consequently, $$ \Bbb P(\cap_{k=1}^n E_k)=\Bbb P(X=n)=\Bbb E[1_{\{X=n\}}]\ge\Bbb E[X-(n-1)]=\Bbb E[X]-(n-1)>0. $$

The same answer as John Dawkins, without the probabilistic formalism.

Set $A_i:=[0,1]\setminus E_i$. Then $$\mu\left(\bigcup_{i=1}^n A_i \right)\leq \sum_{i=1}^n \mu(A_i)=\sum_{i=1}^n(1-\mu(E_i))=n-\sum_{i=1}^n \mu(E_i)<1.$$

So $[0,1]\setminus\bigcup_{i=1}^n A_i\neq\emptyset$, which is what you want.

Partial answer, for the cases $n=3$ and $n=4$:

$n=3$: Assume $E_1\cap E_2 \cap E_3=\phi$. Let $F_2=E_2\setminus E_3\cup E_3\setminus E_2$ and $F_3=E_2\cap E_3$. We can easily see:

• $\mu(E_2)+\mu(E_3)=\mu(E_2\setminus E_3\cup E_2\cap E_3)+\mu(E_3\setminus E_2\cup E_2\cap E_3)=\mu(F_2)+2\mu(F_3)$
• $E_1 \cap F_3=\phi\Rightarrow \mu(E_1)+\mu(F_3)\leq 1$
• $F_2 \cap F_3=\phi\Rightarrow \mu(F_2)+\mu(F_3)\leq 1$

Using this we deduce \begin{eqnarray} \mu(E_1)+\mu(E_2)+\mu(E_3) & = & \mu(E_1)+\mu(F_2)+2\mu(F_3)\\ & = & \mu(E_1)+\mu(F_3)+\mu(F_2)+\mu(F_3)\\ & \leq & 2 \end{eqnarray}

which is a contradiction.

$n=4$: Assume $E_1\cap E_2 \cap E_3\cap E_4=\phi$. Let \begin{eqnarray} F_2 & = & (E_2\setminus (E_3\cup E_4))\cup (E_3\setminus (E_2\cup E_4))\cup (E_4\setminus (E_2\cup E_3))\\ F_3 & = & ((E_2\cap E_3)\setminus E_4)\cup ((E_2\cap E_4)\setminus E_3)\cup ((E_3\cap E_4)\setminus E_2)\\ F_4 & = & E_2\cap E_3\cap E_4 \end{eqnarray} We can see that:

• $\mu(E_2)+\mu(E_3)+\mu(E_4)=\mu(F_2)+2\mu(F_3)+3\mu(F_4)$
• $E_1 \cap F_4=\phi\Rightarrow \mu(E_1)+\mu(F_3)\leq 1$
• $F_3 \cap F_4=\phi\Rightarrow \mu(F_3)+\mu(F_4)\leq 1$
• $F_2 \cap F_3 \cap F_4=\phi\Rightarrow \mu(F_2)+\mu(F_3)+\mu(F_4)\leq 1$

Using this we deduce \begin{eqnarray} \mu(E_1)+\mu(E_2)+\mu(E_3)+\mu(E_4) & = & \mu(E_1)+\mu(F_2)+2\mu(F_3)+3\mu(F_4)\\ & = & (\mu(E_1)+\mu(F_4))+(\mu(F_2)+\mu(F_3)+\mu(F_4))+(\mu(F_3)+\mu(F_4))\\ & \leq & 3 \end{eqnarray}

which is a contradiction.