# Find two set $E_1, E_2 \in \mathscr{M}$ such that $E_1 \subset E_2, \ \mu(E_1) = \mu(E_2)$ and $\mu(E_2 \setminus E_1) > 0$

I'm learning about Measure Theory and need some help with this problem:

Find two set $E_1, E_2 \in \mathscr{M}$ such that $E_1 \subset E_2, \ \mu(E_1) = \mu(E_2)$ and $\mu(E_2 \setminus E_1) > 0$.

A brief explanation of the symbolism:

$\mathscr{M}$ is the set of measurable subsets of $\mathbb{R}$;

The set function $\mu : \mathscr{M} \rightarrow [0, +\infty]$ is the Lebesgue measure.

I should also mention that we are working in the real line $(\mathbb{R})$.

Since $E_1$ is a subset of $E_2$ I'm having some difficulties understanding how their measures could be equal. How should I think this problem?

• Note that if $\mu E_1 = \mu E_2$ with $E_1 \subset E_2$, then $\mu E_2 = \mu (E_1 \cup E_2\setminus E_1) = \mu E_1 +\mu (E_2\setminus \mu E_1)$ and so the only way for this to hold if is $\mu E_1 = \mu E_2 = \mu (E_2\setminus E_1) = ?$ – sqtrat Mar 1 '16 at 14:02

## 1 Answer

HINT :

The Lebesgue measure can reach infinity.

• Thank you for your helpful tip. Let $E_1 = (1, +\infty)$, $E_2 = (0, +\infty)$. Then $E_2 \setminus E_1 = [0, 1]$ and $\mu(E_2 \setminus E_1) = 1 >0$. Is this a correct solution? – Von Kar Mar 1 '16 at 14:24
• Yes it is correct. – nicomezi Mar 1 '16 at 14:29