Let $A_1,...,A_n$ sets such that every $x\in[0,1]$ is in at least $r$ of these sets. Then $\exists k:\mu(A_k)\geq r/n$ Let $A_1,...,A_n\subseteq[0,1]$ Borelian sets such that every $x\in[0,1]$ is in at least $r$ of these sets. Prove that there is some $A_k$ such that $\mu(A)\geq \frac{r}{n}$.
Since the Lebesgue measure is regular, we can suppose that every $A_k$ is a (numerable) disjoint union of open intervals. I tried induction, but I ended up in a cumbersome argument. There must be a simpler way.
 A: You can make it as simple as you would like. The assumption tells you in other words:
\begin{align} 
r\leq\sum_{k=1}^n \mathbf{1}_{A_k}
\end{align} 
for all $x\in [0,1]$. Agree with me? Now Integrate both sides, i.e. apply $\int_{[0,1]}(\cdot)\,d\mu$ to get:
\begin{align} \tag{1}
r\leq \sum_{k=1}^n \mu(A_k) 
\end{align} 
Okay now assume that for all $k$ we have $\mu(A_k) <\frac{r} {n} $. This assumption implies:
\begin{align} \tag{2}
 \sum_{k=1}^n \mu(A_k) < n\frac r n =r
\end{align} 
But we had something different in $(1)$. Hence a contradiction. There must be some $k$ such that $\mu(A_k) \geq \frac r n$. 
A: A suggestion/hint, with only modest confidence of correctness:
Let $U$ be $[0, 1] \times \{1, 2, \ldots, r+1\}$, and $U_0 = [0, 1] \times \{1, 2, \ldots, r\} \subset U$. Now define a map 
$$
f_1 : I \to U : x \mapsto \begin{cases} (x, 1) & x \in A_1 \\
(x, r+1) & \text{else}
\end{cases}
$$
and note that $f_1^{-1}(U_0) = A_1$.
Now define $f_2$ similarly, but with a little twist:
$$
f_2 : I \to U : x \mapsto \begin{cases} 
(x, 1) & x \in A_2 \setminus A_1 \\
(x, 2) & x \in A_2 \cap A_1 \\
(x, r+1) & \text{else}
\end{cases}
$$
Informally: $f_2(x)$ has second coordinate $1$ if $A_2$ is the first set in which $x$ appears, but has second coordinate $2$ if $A_2$ is the second set in which $x$ appears. 
Similarly define $f_3, \ldots, f_n$. In general, if $x$ appears in more than $r$ sets, then $f_i(x)$ will be $(x, r+1)$ for all $i$ after the first $r$ sets in which $x$ appears. 
Note that for each $i$, $f_i^{-1}(U_0) \subset A_i$. Also note that fpr $i \ne j$, we have $f_i(x) \ne f_j(x)$ if both are in $U_0$. 
Finally, note that $\mu(f_i(A_i) \cap U_0) \le \mu(A_i)$. 
Now sum that last observation over $i$ to get that 
$$\mu(U_0) \le \sum_i \mu A_i.$$
Now the left side is $r$, because $U_0$ consists of $r$ copies of the unit interval. If all terms on the right side are smaller than $\frac{r}{n}$, then the right side is smaller than $n \cdot \frac{r}{n} = r$, and we get $r < r$, a contradiction. 
Of course, there's a bunch of stuff to show about measurability, etc., and I've been sloppy in ignoring that, but I think that this is the gist of a correct approach. 
