## Definitions for measurable functions

I am currently proceeding on a self-guided study of the book The Elements of Integration and Lebesgue Measure, will post some updates and elaborations of the proofs in the book. Every book is constrained by the number of pages the publisher allows, hence some authors will write rather terse and concise proofs, the worst example of which is simply “Proof: Trivial”. Bartle is a very good author, he does provide details of proofs 90% of the time.

Definition: A function f on X to R is said to be X-measurable (or simply measurable) if for every real number $\alpha$ the set $\{x\in X: f(x)>\alpha\}$ belongs to X.

This definition of measurability is not unique, there are other possible forms which are discussed in the lemma below.

Lemma: The following statements are equivalent for a function f on X to R:

((X,X) is a measurable space where X is a set and X is a $\sigma$-algebra of subsets of X.)

(a) For every $\alpha\in\mathbb{R}$, the set $A_\alpha = \{x\in X: f(x)>\alpha \}$ belongs to X,

(b) For every $\alpha\in\mathbb{R}$, the set $B_\alpha = \{x\in X: f(x)\leq\alpha \}$ belongs to X,

(c) For every $\alpha\in\mathbb{R}$, the set $C_\alpha = \{x\in X: f(x)\geq\alpha \}$ belongs to X,

(d) For every $\alpha\in\mathbb{R}$, the set $D_\alpha = \{x\in X: f(x)<\alpha \}$ belongs to X,

Proof:

Note that $B_\alpha$ and $A_\alpha$ are complements of each other, hence statement (a) is equivalent to statement (b). This is due to one of the properties of $\sigma$-algebra, namely that if A belongs to X, then the complement X\A also belongs to X.

Similarly, statements (c) and (d) are equivalent. We will prove that (a) is equivalent to (c).

Assume (a) holds, we can say that $A_{\alpha-1/n}$ belongs to X for each n.

And since $C_\alpha=\cap_{n=1}^{\infty}{A_{\alpha-1/n}}$, it follows that $C_\alpha\in\mathbf{ X}$. Thus, (a) implies (c).

We also have $A_\alpha =\cup_{n=1}^{\infty} C_{\alpha+1/n}$, and hence if (c) is true, each $C_{\alpha +1/n}$ is in X, and the union of them is also in X (definition for $\sigma$-algebra). It thus follows that (c) implies (a).

What this lemma says is that there is nothing special about the “>” in the definition of measurability. It could very well have been “$\leq$, or even “<” and nothing would change!