The Continuity Properties of the Lebesgue Measure

Theorem 1 (The First Continuity Property of the Lebesgue Measure): Let $\{ A_n \}_{n=1}^{\infty}$ be a collection of Lebesgue measurable sets such that $A_n \subseteq A_{n+1}$ for all $n \in \mathbb{N}$. Then $\displaystyle{m \left ( \bigcup_{n=1}^{\infty} A_n \right ) = \lim_{n \to \infty} m(A_n)}$.

• Proof: There are two cases to consider.

• Case 1: Suppose that one of the sets in $\{ A_n \}_{n=1}^{\infty}$ has infinite Lebesgue measure. Then there exists an $n_0 \in \mathbb{N}$ such that $m(A_{n_0}) = \infty$. So by countable subadditivity of the Lebesgue measure we have that:

(1)

\begin{align} \quad m \left ( \bigcup_{n=1}^{\infty} A_n \right ) = \infty \quad (*) \end{align}

• Since $A_n \subseteq A_{n+1}$ for all $n \in \mathbb{N}$ we have that $A_{n_0} \subseteq A_{n_0 + k}$ for all $k \geq 1$. Hence $\infty = m(A_{n_0}) \leq m(A_{n_0 +k})$ for all $k \geq 1$, so:

(2)

\begin{align} \quad \lim_{n \to \infty} m(A_n) = \lim_{k \to \infty} m(A_{n_0 + k}) = \infty \quad (**) \end{align}

• Combining $(*)$ and $(**)$ give us that $\displaystyle{m \left ( \bigcup_{n=1}^{\infty} A_n \right ) = \lim_{n \to \infty} m(A_n)}$.

• Case 2: Suppose that none of the sets in $\{ A_n \}_{n=1}^{\infty}$ has infinite Lebesgue measure. Then $m(A_n) < \infty$ for all $n \in \mathbb{N}$. We can rewrite the union $\displaystyle{\bigcup_{n=1}^{\infty} A_n}$ as follows:

(3)

\begin{align} \quad \bigcup_{n=1}^{\infty} A_n &= A_1 \cup (A_2 \setminus A_1) \cup (A_3 \setminus A_2) \cup ... \\ &= A_1 \cup \left ( \bigcup_{n=2}^{\infty} (A_n \setminus A_{n-1}) \right ) \end{align}

• Since $A_n \subseteq A_{n+1}$ for all $n \in \mathbb{N}$ we see that $\{ A_1, A_n \setminus A_{n-1} : n \in \mathbb{N}, n \geq 2 \}$ is a collection of mutually disjoint sets.

• We take the Lebesgue measure of both sides of the equality above and use The Excision Property of the Lebesgue Measure to get:

(4)

\begin{align} \quad m \left ( \bigcup_{n=1}^{\infty} A_n \right ) = m \left ( A_1 \cup \left ( \bigcup_{n=2}^{\infty} (A_n \setminus A_{n-1}) \right ) \right ) &= m(A_1) + \sum_{n=2}^{\infty} m(A_n \setminus A_{n-1}) \\ &= m(A_1) + \sum_{n=2}^{\infty} [m(A_n) - m(A_{n-1})] \\ &= m(A_1) + \lim_{n \to \infty} \sum_{k=2}^{n} [m(A_k) - m(A_{k-1})] \\ &= m(A_1) + \lim_{n \to \infty} [m(A_n) - m(A_1)] \\ &= \lim_{n \to \infty} m(A_n) \quad \blacksquare \end{align}

Theorem 2 (The Second Continuity Property of the Lebesgue Measure): Let $\{ B_n \}_{n=1}^{\infty}$ be a collection of Lebesgue measurable sets such that $B_n \supseteq B_{n+1}$ for all $n \in \mathbb{N}$. Furthermore, let $m(B_1) < \infty$. Then $\displaystyle{m \left ( \bigcap_{n=1}^{\infty} B_n \right ) = \lim_{n \to \infty} m(B_n)}$.

Note that Theorem 2 is very similar to Theorem 1 in hypotheses however Theorem 2 has the additional condition that $m(B_1) < \infty$.

• Proof: Since $m(B_1) < \infty$ and $B_n \supseteq B_{n+1}$ for all $n \in \mathbb{N}$ we have that $m(B_n) < \infty$ for all $n \in \mathbb{N}$ so $\displaystyle{m \left ( \bigcap_{n=1}^{\infty} B_n \right ) < \infty}$. By the Excision property of the Lebesgue measure we have that:

(5)

\begin{align} \quad m \left ( B_1 \setminus \left ( \bigcap_{n=1}^{\infty} B_n \right ) \right ) = m(B_1) - m \left ( \bigcap_{n=1}^{\infty} B_n \right ) \quad (*) \end{align}

• Now note that:

(6)

\begin{align} \quad B_1 \setminus \left ( \bigcap_{n=1}^{\infty} B_n \right ) = B_1 \cap \left ( \bigcap_{n=1}^{\infty} B_n \right )^c = B_1 \cap \left ( \bigcup_{n=1}^{\infty} B_n^c \right ) = \bigcup_{n=1}^{\infty} (B_1 \cap B_n^c) = \bigcup_{n=1}^{\infty} (B_1 \setminus B_n) \quad (**) \end{align}

• Let $A_n = B_1 \setminus B_n$ for each $n \in \mathbb{N}$. Then $A_n \subseteq A_{n+1}$ for all $n \in \mathbb{N}$ since $B_n \supseteq B_{n+1}$ for all $n \in \mathbb{N}$.

• So by the previous theorem we have that:

(7)

\begin{align} \quad m \left ( B_1 \setminus \bigcap_{n=1}^{\infty} B_n \right ) = m \left ( \bigcup_{n=1}^{\infty} (B_1 \setminus B_n) \right ) = \lim_{n \to \infty} m(B_1 \setminus B_n) = \lim_{n \to \infty} [m(B_1) - m(B_n)] = m(B_1) - \lim_{n \to \infty} m(B_n) \quad (***) \end{align}

• Using the equality at $(**)$ and $(*)$ with $(***)$ gives us:

(8)

\begin{align} \quad m(B_1) - m \left ( \bigcap_{n=1}^{\infty} B_n \right ) = m(B_1) - \lim_{n \to \infty} m(B_n) \quad \Leftrightarrow \quad m \left ( \bigcap_{n=1}^{\infty} B_n \right ) = \lim_{n \to \infty} m(B_n) \quad \blacksquare \end{align}