# Example Problems Regarding the Lebesgue Measure

So far we have defined the Lebesgue measure to be the set function $m : \mathcal M \to [0, \infty) \cup \{ \infty \}$ defined for all Lebesgue measurable sets $E$ by $m(E) = m^*(E)$.

Furthermore we have proven that:

**1. If $A$ and $B$ are Lebesgue measurable sets with $A \subseteq B$ and if $m(A) < \infty$ then:**

**2. If $(E_n)_{n=1}^{\infty}$ is a sequence of mutually disjoint Lebesgue measurable sets then:**

**3. If $\{ A_n \}_{n=1}^{\infty}$ is a collection of Lebesgue measurable sets such that $A_n \subseteq A_{n+1}$ for all $n \in \mathbb{N}$ then:**

**4. If $\{ B_n \}_{n=1}^{\infty}$ is a collection of Lebesgue measurable sets such that $B_{n} \supseteq B_{n+1}$ for all $n \in \mathbb{N}$ and $m(B_1) < \infty$ then:**

We will now look at some example problems regarding the Lebesgue measure.

## Example 1

**Let $E_1$ and $E_2$ be Lebesgue measurable sets. Prove that $m(E_1 \cup E_2) + m(E_1 \cap E_2) = m(E_1) + m(E_2)$.**

We first note that since $E_1$ and $E_2$ are Lebesgue measurable sets we have also that $E_1 \cup E_2$ and $E_1 \cap E_2$ are Lebesgue measurable sets.

There are two cases to consider:

**Case 1:** Suppose that at least one of $E_1$ or $E_2$ has infinite measure. Then the equality clearly holds.

**Case 2:** Suppose that both $E_1$ and $E_2$ have finite measure, that is, $m(E_1) < \infty$ and $m(E_2) < \infty$.

Note the following equality of sets:

(5)The three sets on the right are disjoint. Therefore:

(6)Note that $E_1 \cap E_2 \subseteq E_1, E_2$ and $m(E_1), m(E_2) < \infty$. So by the excision property of Lebesgue measurable sets we have that:

(7)Therefore $m(E_1 \cup E_2) + m(E_1 \cap E_2) = m(E_1) + m(E_2)$.

## Example 2

**For the continuity of the Lebesgue measure, we saw that if $\{ B_n \}_{n=1}^{\infty}$ is a collection of Lebesgue measurable sets with $m(B_1) < \infty$ and such that $B_n \supseteq B_{n+1}$ for all $n \in \mathbb{N}$ then $\displaystyle{m \left (\bigcap_{n=1}^{\infty} B_n \right ) = \lim_{n \to \infty} m(B_n)}$. If $m(B_1) = \infty$ show that the conclusion of this theorem does not always hold.**

For each $n \in \mathbb{N}$ let:

(8)Then clearly $B_n \supseteq B_{n+1}$ for all $n \in \mathbb{N}$. Furthermore $B_1 = [1, \infty)$ and $m(B_1) = \infty$. Now:

(9)However:

(10)So clearly we see that:

(11)