In a question paper I got the following three questions.
Let $E \subset \mathbb{R}$ with $0 < \mu(E) < \infty$.
- Show that the function $f : \mathbb{R} \rightarrow \mathbb{R}$ with $f(x) = \mu(E \cap (-\infty, x])$ is continuous.
- Show that there exist a measurable set $F \subset E$ such that $\mu(F) = \frac{1}{3}\mu(E)$.
- Show that there exist a closed set $F \subset E$ such that $\mu(F) = \frac{1}{3}\mu(E)$.
Remark: we are working on the real line $(\mathbb{R})$, and $\mu$ is the Lebesgue measure ($\mathscr{M}$ is the set of measurable subsets of $\mathbb{R}$).
Edit:
Following the explanation of John Ma here's my attempt for question $(2)$.
Let $A_n = E \cap (-\infty, n]$ (increasing sequence of measurable sets)
and $B_n = E \cap (-\infty, -n]$ (decreasing sequence of measurable sets).
It follows that $$\mu(\bigcup_{n = 1}^{+\infty}A_n) = \lim_{n \to +\infty} \mu(A_n) = \lim_{n \to +\infty} f(n) = \mu(E)$$
and
$$\mu(\bigcap_{n = 1}^{+\infty}B_n) = \lim_{n \to +\infty} \mu(B_n) = \lim_{n \to +\infty} f(-n) = \mu(\emptyset) = 0.$$
Since $\lim_{n \to +\infty} f(n) = \mu(E)$, we choose $n_0$, $n_1$ (large enough) such that $$f(n_0) > \frac{9}{10} \mu{(E)} \;\; \text{and} \;\; f(-n_1) < \frac{1}{10} \mu{(E)}.$$
From the Intermediate Value Theorem it follows that $f$ is taking every value between $\frac{1}{10} \mu{(E)}$ and $\frac{9}{10} \mu{(E)}$. Therefore
$$\exists x_0 \in (-n_1, n_0) \;\; \text{such that} \;\; f(x_0) = \mu(E \cap (-\infty, x_0]) = \frac{1}{3}\mu(E).$$
Setting $F = E \cap (-\infty, x_0]$ we conclude that $$F \subset E, F \in \mathscr{M} \;\; \text{and} \;\; \mu(F) = \frac{1}{3}\mu(E).$$