## Proposition 7.3.11: Equivalent Measurable Sets

Let

*f*be a real-valued function with a measurable set as domain. Then the following are quivalent:

- the set
{ f(x) a }is measurable for all real numbersa- the set
{ f(x) < a }is measurable for all real numbersa- the set
{ f(x) a }is measurable for all real numbersa- the set
{ f(x) > a }is measurable for all real numbersa

The proof is not too bad. Remember, we do not need to prove that any of these
sets are measurable per se; we need to prove that *if* one of them is measurable
then the others are, too.

So, assume that (1) is true. Then (4) must be true since it is the complement of (1), and we know that complements of measurable sets are measurable. This clearly works the other way around as well, so that (1) is equivalent to (4).

Again by considering complements we can see that (2) is equivelent to (3).

Finally we can explore the fact that unions and intersections of measurable sets are measurable together with:

{ f(x) < a } =_{n > 0}{ f(x) a - 1/n }

to see that (1) implies (2). Similarly because:

{ f(x) a } =_{n > 0}{ f(x) < a + 1/n }

we see that (2) implies (1). But then our proof is finished.