Measurable Image

Theorem
Let $$\mathfrak {M}$$ be the set of measurable sets of $$\R$$.

For any extended real-valued function $$f: \R \to \R \cup \left\{{-\infty, +\infty}\right\} \ $$ whose domain is measurable, the following statements are equivalent:


 * $$(1): \qquad \forall \alpha \in \R, \left\{{x: f \left({x}\right) > \alpha}\right\} \in \mathfrak {M}$$
 * $$(2): \qquad \forall \alpha \in \R, \left\{{x: f \left({x}\right) \ge \alpha}\right\} \in \mathfrak {M}$$
 * $$(3): \qquad \forall \alpha \in \R, \left\{{x: f \left({x}\right) < \alpha}\right\} \in \mathfrak {M}$$
 * $$(4): \qquad \forall \alpha \in \R, \left\{{x: f \left({x}\right) \le \alpha}\right\} \in \mathfrak {M}$$

These statements imply:


 * $$(5): \qquad \forall \alpha \in \R \cup \left\{{-\infty, +\infty}\right\}: \left\{{x:f \left({x}\right) = \alpha}\right\} \in \mathfrak {M}$$

Proof
Let the domain of $$f \ $$ be $$D \ $$.

We have that Measurable Sets are an Algebra of Sets.

First we note that, from Properties of Algebras of Sets, the difference of two measurable sets is measurable.

So:
 * $$\left\{{x: f \left({x}\right) \le \alpha}\right\} = D - \left\{{x: f \left({x}\right) > \alpha}\right\}$$

and so $$(1) \iff (4)$$.

Similarly, $$(2) \iff (3)$$.

Next we note that, also from Properties of Algebras of Sets, the intersection of a sequence of measurable sets is measurable.

Now:
 * $$\left\{{x: f \left({x}\right) \ge \alpha}\right\} = \bigcap_{n=1}^\infty \left\{{x: f \left({x}\right) > \alpha -\tfrac{1}{n}}\right\}$$

because if:
 * $$x \in \left\{{x: f \left({x}\right) \ge \alpha}\right\}$$

that is:
 * $$f \left({x}\right) \ge \alpha$$

and since:
 * $$\forall n \in \N: n > 0: \alpha > \alpha - \frac{1}{n}$$

then:
 * $$\forall n \in \N: n > 0: f \left({x}\right) > \alpha - \frac{1}{n}$$

That is:
 * $$\forall n \in \N: n > 0: x \in \left\{{x: f \left({x}\right) > \alpha -\tfrac{1}{n}}\right\}$$

Hence:
 * $$x \in \bigcap_{n=1}^\infty \left\{{x: f \left({x}\right) > \alpha -\tfrac{1}{n}}\right\}$$

Conversely, suppose:
 * $$x \in \bigcap_{n=1}^\infty \left\{{x: f \left({x}\right) > \alpha -\tfrac{1}{n}}\right\}$$

that is:
 * $$\forall n \in \N: n > 0: x \in \left\{{x: f \left({x}\right) > \alpha -\tfrac{1}{n}}\right\}$$

Claim $$f \left({x}\right) \ge \alpha $$.

Otherwise $$f \left({x}\right) < \alpha$$, say for example $$f \left({x}\right) = \alpha - |\epsilon|$$.

Choose $$N = \left \lceil \frac{1}{|\epsilon|} \right \rceil + 1 \in \N$$.

Therefore:
 * $$N > \left \lceil \frac{1}{|\epsilon|} \right \rceil \ge \frac{1}{|\epsilon|}$$

and so:
 * $$\alpha - \frac{1}{N} > \alpha - |\epsilon|$$

By hypothesis:
 * $$\forall N \in \N: f \left({x}\right) > \alpha - \frac{1}{N}$$

and therefore by the just previous:
 * $$f \left({x}\right) > \alpha - |\epsilon|$$

But we had $$f \left({x}\right) = \alpha - |\epsilon|$$, a contradiction.

Therefore:
 * $$f \left({x}\right) \ge \alpha $$

that is: $$x \in \left\{{x: f \left({x}\right) \ge \alpha}\right\}$$ which was to be shown.

So $$(1) \implies (2)$$.

Similarly:
 * $$\left\{{x: f \left({x}\right) > \alpha}\right\} = \bigcup_{n=1}^\infty \left\{{x: f \left({x}\right) \ge \alpha + \tfrac{1}{n}}\right\}$$.

and so $$(2) \implies (1)$$.

This shows that $$(1) \iff (2) \iff (3) \iff (4)$$.

For the fifth statement, we have:
 * $$\left\{{x: f \left({x}\right) = \alpha}\right\} = \left\{{x: f \left({x}\right) \ge \alpha}\right\} \cap \left\{{x: f \left({x}\right) \le \alpha}\right\}$$

and so $$(3) \and (4) \implies (5)$$ for $$\alpha \in \R$$.

Since:
 * $$\left\{{x: f \left({x}\right) = +\infty }\right\} = \bigcap_{n=1}^\infty \left\{{x: f \left({x}\right) \ge n }\right\}$$

we have that $$(2) \implies (5)$$ for $$\alpha = +\infty$$.

Similarly $$(4) \implies (5)$$ for $$\alpha = - \infty$$.