1

I'm learning about measure theory, specifically measurable sets, and need help with the following exercises:

$(1)$ Find the measure of the set $E_1 = \mathbb{Z} \cup \mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{N})$;

$(2)$ Find the measure of the set $E_2 = \bigcap_{n=1}^{\infty}\big[1 - {1 \over 2^n}, 3 + {1 \over 3^n}\big]$;

$(3)$ Let $\{a_n\}$ be an increasing sequence of positive terms. If $E_n = (-a_n, a_n]$, find $m(\bigcup_n E_n)$;

$(4)$ (True or False) Let $A \subset \mathbb{R}$. Then $m(A) = 0$ iff all the subsets of $A$ are measurable.

Note : $m$ is the Lebesgue measure.

Since I'm having difficulties for question $(4)$ I am going to share my thoughts on $(1)$, $(2)$ and $(3)$.

$(1)$ Since $\mathbb{N} \subsetneq \mathbb{Q}$, then $(\mathbb{R} \setminus \mathbb{N}) \cup \mathbb{Q} = \mathbb{R}$. But $\mathbb{R} \cup \mathbb{Z}$ is just $\mathbb{R}$. Therefore, since the Lebesgue measure can reach infinity,

$$m(E_1) = m(\mathbb{R}) = \infty.$$

$(2)$ Calculating the first two terms: for $n = 1$ we get the interval $[{1 \over 2}, {10 \over 3}]$. For $n = 2$ we have the interval $[{3 \over 4}, {28 \over 9}]$. So as $n$ increases the interval increases from the left and decreases from the right. As $n$ tends to infinity we get the interval $[1, 3]$. Therefore,

$$m(E_2) = m\big(\bigcap_{n=1}^{\infty}\big[1 - {1 \over 2^n}, 3 + {1 \over 3^n}\big]\big) = m([1, 3]) = 2.$$

$(3)$ Since $E_n = (-a_n, a_n]$ where $\{a_n\}$ is an increasing sequence of positive terms, then $\bigcup_n E_n = \mathbb{R}$. It follows that

$$m(\bigcup_n E_n) = m(\mathbb{R}) = \infty.$$

Is my work correct for $(1)$, $(2)$ and $(3)$? How do I prove or disprove $(4)$?

Community
  • 1
glpsx
  • 2,122

2 Answers2

3

Your work is correct for $1$ and $2$. For $3$, how about the sequence: $a_n = 1 - \frac1n$? It is increasing, though:

$$\bigcup_n E_n = (-1,1)$$

We have to separate two cases. We know that either $(a_n)$ converges or goes to $+ \infty$. If it converges to $a$, $0 < a < \infty$, then we have:

$$\bigcup_n E_n= (-a,a)$$

Otherwise, the union is $(-\infty, \infty)$.

For $4$, recall that the Lebesgue measure is complete, which means that every subset of a null set is measurable. If $m(A) = 0$, then $A$ is a null set and every subset of $A$ is measurable. Conversely, it is known that every set of positive Lebesgue measure contains a nonmeasurable subset (for a proof, see e.g. this)

Community
  • 1
  • Your example for $(3)$ is extremely helpful, so is the link you mentionned for $(4)$. Thanks! – glpsx Jun 12 '16 at 14:14
0

(1) and (2) are correct and nice.

(3) is almost good as well. However, you implicitly assumed $a_n\to\infty$.

For (4), this is true. One direction follows from the definition of Lebesgue measure, and the other one is a consequence of the fact that there exists a non-Lebesgue-measurable set in $[0,1]$.

Berci
  • 90,745