How does one go about finding an appropriate $c$ for the statement below?
Let $E$ be a measurable subset of $[0,1]$. Show that there exists some $c \in [0,1]$ such that $m(E \cap [0,c]) = \frac{m(E)}{2017}.$
I am currently going over former analysis qualifying exam questions (from 2017 as one may quickly guess) and stumbled across this one. If $E$ has measure zero then the result follows immediately so lets assume that $E$ does not have measure zero. Then it is clear that $\frac{m(E)}{2017} \leq c \leq 1$ by the monotonicity of measure. I am not exactly sure how to narrow in on a value of $c$ that works and so any hints, observations or solutions are welcome.