$A_1,...,A_n,...$ measurable sets such that: ... Then $ \lambda (\bigcup _{k=1}^{\infty} A_k) \geq 1 $. $A_1,...,A_n,...$  measurable sets such that:
i) $\lambda (A_n) \geq 1/2 $ forall $k$
ii) $\lambda (A_s \cap A_k ) \leq 1/4 $ forall $k \neq s$
Then $ \lambda (\bigcup _{k=1}^{\infty} A_k) \geq 1 $.
Attemp:
We consider the metric Boolean algebra $^\mathcal{A } / _\sim$ defined from : $d(A,B) = \mu (A \triangle B)$ .
We know that $^\mathcal{A } / _\sim$ is complete and when $\mathcal{A}= \mathbb{R} $ then  $^\mathcal{A } / _\sim$ is also seperable.
Also we have that
$d(A_n,A_m) \geq \frac{1}{2}$ for all $n \neq m$.
 A: A quantitative approach to the problem:
Let $B_n = \bigcup_{k=1}^n A_k$, and suppose $\lambda(B_n) \leq 1$ for a given $n$. Let $f_k$ be the indicator function of $A_k$. Then by the Cauchy-Schwarz inequality, we have
$$\left(\int_{B_n} 1 \,d\lambda\right) \left(\int_{B_n} \left(\sum_{k=1}^n f_k\right)^2 \,d\lambda\right) \geq \left( \int_{B_n} \sum_{k=1}^n f_k \,d\lambda\right)^2.$$
The second integral on the left satisfies
$$\int_{B_n} \left(\sum_{k=1}^n f_k\right)^2 \,d\lambda = \sum_{k=1}^n \int_{B_n} f_k^2 \,d\lambda + \sum_{i \neq j} \int_{B_n} f_i f_j \,d\lambda \leq n + n(n-1)(1/4) = \frac{n(n+3)}{4}$$
while the integral on the right has $\int_{B_n} \sum_{k=1}^n f_k \,d\lambda \geq n/2$, and putting these together gives $\lambda(B_n) \cdot \frac{n(n+3)}{4} \geq \frac{n^2}{4}$, which implies $\lambda(B_n) \geq \frac{n}{n+3}$, so $\lambda(B_n) \geq \frac{n}{n+3}$ in any case.
Thus for each $n$, $\lambda(\bigcup_{k=1}^n A_k) \geq \frac{n}{n+3}$, from which it follows that $\lambda(\bigcup_{k=1}^\infty A_k) \geq 1$.
