# Measure Theory/Basic Structures And Definitions/Measurable Functions

This section defines measurable functions, which will be used in the development of integration.

Let $f:E \to \mathbb{R}$ be a function from the measurable domain $E$. We say that $f$ is measurable if the preimage of every measurable set in $\mathbb{R}$ is measurable. What is interesting with this definition is it's strong relation to the defintion of continuity between topological spaces, which is, the preimage of every open set is open. A further investigation of this topic is left as an exercise. When proving measurability, there are another set of tools which are useful.

## Equivalent Definitions

Proposition:

Let $f: E \rightarrow [-\infty, \infty]$ be an extended real-valued function defined on a measurable domain E. Fix some $\alpha \in \mathbb{R}$. If $f$ is measurable, the following sets are equivalent, and measurable:

$(i) \{ x \in E : f(x) < \alpha \}$
$(ii) \{ x \in E : f(x) > \alpha \}$
$(iii) \{ x \in E : f(x) \leq \alpha \}$
$(iv) \{ x \in E : f(x) \geq \alpha \}$
We are using such sets Proof: It should be clear that (i) and (iv) are equivalent by complements of algebras, as with (ii) and (iii). What remains to be shown is the equivalence of (i) and (iii). We do this by establishing the following identities:
$\{ x \in E : f(x) \leq \alpha \} = \cap_{n=1}^{\infty} \{ x \in E : f(x) < \alpha - \frac{1}{n} \}$
$\{ x \in E : f(x) < \alpha \} = \cup_{n=1}^{\infty} \{ x \in E : f(x) \leq \alpha + \frac{1}{n} \}$
Finally, because the countable intersection or union of measurable sets is measurable, the resulting set is measurable.

Examples: First, we will give a couple examples of measurable functions. Let $f,g: E \rightarrow [-\infty, \infty]$ be extended real-valued mappings from the measurable domain $E$.

Exercises Let $f$ be a measurable function, and $g$ be continuous, where $f: E \rightarrow F$ and $g: F \rightarrow G$