# continuity from above and below

- Property of a measure space (see Measures and Probability Measures)
- Continuity from above:
- if \(A_1 \supseteq A_2 \supseteq \cdots\) and at least one \(A_i\) has finite measure
- then \(\mu \left( \cap_{i=1}^{\infty}A_i \right) = \lim_{i\rightarrow\infty} \mu(A_i)\)

- Continuity from below:
- if \(A_1 \subseteq A_2 \subseteq \cdots\)
- then \(\mu \left( \cup_{i=1}^{\infty}A_i \right) = \lim_{i\rightarrow\infty} \mu(A_i)\)

- What does this have to do with continuity?
- you can read this as \(\mu (\lim_{i\rightarrow\infty} A_i) = \lim_{i\rightarrow\infty}\mu(A_i)\) (see this answer)
- how do we give meaning to \(\lim_{i\rightarrow\infty} A_i\)? see linked answer also wikipedia article on set theoretic limit

- Why is the cumulative distribution function not left continuous? See this answer

## 1. proof of continuity from below

- taken from this answer
- Statement:
- Let \(E_1 \subseteq E_2 \subseteq \cdots\)
- Then \(\mu \left( \bigcup_{i=1}^{\infty} E_i \right) = \lim_{i\rightarrow \infty} \mu(E_i)\)

- proof:
- First, construct \(A_1 = E_1\), \(A_2 = E_2 \setminus E_1\), and \(A_{i} = E_{i} \setminus \cup_{i=1}^{i-1}E_i\) for \(i \geq 2\)
- Note that all \(A_i\) are disjoint. This is what lets us use countable additivity (see below).
- You can think of each \(A_i\) as another layer of onion that we are building up (see useful graphics in this answer). Note then that \(\cup_{i=1}^{k}A_i = \cup_{i=1}^{k}E_i\). In particular \(\cup_{i=1}^{\infty}E_i = \cup_{i=1}^{\infty}A_i\), so \(\mu(\cup_{i=1}^{\infty} A_i) = \mu(\cup_{i=1}^{\infty} E_i)\).
- We get \(\mu(\cup_{i=1}^{\infty} A_i) = \sum_{i=1}^{\infty} \mu(A_i)\) from countable additivity (see Measures and Probability Measures)
- this is the key point that allows us to move the limit "outside" the measure \(\mu\)

- Then, \(\sum_{i=1}^{\infty} \mu(A_i) = \lim_{k\rightarrow\infty} \sum_{i=1}^{k} \mu(A_i)\)
- And, \(\lim_{k\rightarrow\infty}\sum_{i=1}^{k} \mu(A_i) = \lim_{k\rightarrow\infty}\mu(\cup_{i=1}^{k}A_i) = \lim_{k\rightarrow\infty}\mu(E_k)\) where we used the fact that \(\cup_{i=1}^{k}E_i = \cup_{i=1}^{k}A_i\).