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In Real Analysis written by Royden, the definition of measurable function is as follows.

An extended real-valued function f defined on E is said to be Lebesgue measurable, or simply measurable, provided its domain E is measurable and it satisfies one of the four statements of Proposition1.

Proposition 1 Let the function f have a measurable domain E. Then the following statements are equivalent:

(1) For each real number c, the set $\{x \in E | f(x) \gt c\}$ is measurable
(2) For each real number c, the set $\{x \in E | f(x) \ge c\}$ is measurable
(3) For each real number c, the set $\{x \in E | f(x) \lt c\}$ is measurable
(4) For each real number c, the set $\{x \in E | f(x) \le c\}$ is measurable

But I don't understand why mathematicians define LMS function as above. In my thought, a condition that a domain of a function is a measurable set is enough to define that a function is measurable.
What a property is added if the four statements is included in the definition?

alryosha
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    Eventually you'll want to find a sequence of functions in the form $f_k=\sum_{j=1}^{n_k} a_{k,j} 1_{E_{k,j}}$ such that all $E_{k,j}$ are measurable and, for all $i\ne j$, $E_{k,j}\cap E_{k,i}=\emptyset$; and you'll want that $f_k\to f$. You'll want to define $\int f=\lim_{k\to\infty} \sum_{j=1}^{n_k} a_{k,j}\mu(E_{k,j})$. It turns out that such functions $f$ that are limits like that are like the author says and not like you say. –  Apr 10 '19 at 13:26

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