Continuity from below and above

Question

In Folland's Real analysis, two of properties of measures are stated as follows:

Let $(X,\mathcal{M}, \mu)$ be a measure space.

Continuity from below: If $\{E_j\}_1^{\infty} \subset \mathcal{M}$ and $E_1 \subset E_2 \subset \cdots$, then $\mu(\bigcup_1^{\infty} E_j) = \lim_{j \to \infty} \mu(E_j)$

Similarly,

Continuity from above: If $\{E_j\}_1^{\infty} \subset \mathcal{M}$, $E_1 \supset E_2 \supset \cdots$ and $\mu(E_1) < \infty$. Then $\mu(\bigcap_1^{\infty} E_j) = \lim_{j \to \infty} \mu(E_j)$

I do not understand the point of these statements. How they are related to continuity for example?

Answer 1

It's reasonable for a continuous increasing function $f$ with real values, defined on the real line, to have $f(t_n)\uparrow f(t)$ when $t_n\uparrow t$ and $f(t_n)\downarrow f(t)$ when $t_n\downarrow t$. As a measure can take possibly infinite values, we have to be careful in this context, but that's the idea.

The set functions $\mu\colon\cal M\to \overline{\Bbb R}_+$ is increasing in the sense that if $A\subset B$ then $\mu(A)\subset \mu(B)$. If $A_n\uparrow A$, then $\mu(A_n)\uparrow \mu(A)$.

We have to assumed that at least a set in $\{B_n\}$ is of finite measure in the case of decreasing sequence, to avoid counter-examples likes $X=\Bbb R$ with Lebesgue measure and $B_n=[n,+\infty)$ (like if we had a function defined on the real line with values in $\overline{\Bbb R}_+$ and such that $f(x)=0$ if $x\leqslant 0$, $f(x)=+\infty$ otherwise).

Answer 2

How they are related to continuity for example?

As stated in the comments continuity of a function $\mu$ can be interpreted as a property that allows for the calculation $$\mu(\lim_{j \to \infty} E_j) = \lim_{j \to \infty}\mu(E_j).\tag{1}$$

$\mu$ is a measure, i.e. it is a mapping into a field of numbers, therefore the right-hand side of (1) makes sense (if the sequence converges).

Convergence for a sequence $(E_j)_{j \in \mathbb{N}}$ of sets

We want to give a meaning to the expression $\lim_{j \to \infty} E_j$. One way, would be to define a topology on a suitable collection of sets that includes the sets $(E_j)$ and check, whether convergence in this topological space implies (1).

A more pragmatical approach is the following

Definition (limes superior, limes inferior and limit).

Let $(E_j)_{j \in \mathbb{N}}$ a sequence of subsets of $X$. Then we call $$ \limsup_{j \to \infty} E_j := \bigcap_{j = 1}^\infty \; \bigcup_{j = k}^\infty A_j $$ and $$ \liminf_{j \to \infty} E_j := \bigcup_{j = 1}^\infty \; \bigcap_{j = k}^\infty A_j $$ the limes superior and limes inferior of $(E_j)$, respectively.

If $\limsup_{j \to \infty} E_j = \liminf_{j \to \infty} E_j$, we say that $(E_j)$ converges and define the limit $$ \lim_{j \to \infty} E_j := \limsup_{j \to \infty} E_j = \liminf_{j \to \infty} E_j. $$

Definition and Lemma (increasing and decreasing sequences). Let $(E_j)_{j \in \mathbb{N}}$ a sequence of subsets of $X$.

If $E_j \subset E_{j + 1}$ for all $j \in \mathbb{N}$ we call the sequence $(E_j)$ increasing. Every increasing sequence converges with limit $$ \lim_{j \to \infty} E_j = \bigcup_{j =1}^\infty E_j. $$

If $E_j \supset E_{j + 1}$ for all $j \in \mathbb{N}$ we call the sequence $(E_j)$ decreasing. Every decreasing sequence converges with limit $$ \lim_{j \to \infty} E_j = \bigcap_{j =1}^\infty E_j. $$

Interpretation of (1) in the sense of set-convergence

Using the notion of continuity, OP's reference may be read as follows:

Let $(X,\mathcal{M}, \mu)$ be a measure space.

Continuity from below: If $\{E_j\}_1^{\infty} \subset \mathcal{M}$ and $E_1 \subset E_2 \subset \cdots$, then $\mu(\lim_{j \to \infty} E_j) = \lim_{j \to \infty} \mu(E_j)$

Similarly,

Continuity from above: If $\{E_j\}_1^{\infty} \subset \mathcal{M}$, $E_1 \supset E_2 \supset \cdots$ and $\mu(E_1) < \infty$. Then $\mu(\lim_{j \to \infty} E_j) = \lim_{j \to \infty} \mu(E_j)$