Measurable Image

Theorem
Let $$\mathfrak {M}$$ be the set of measurable sets of $$\mathbb{R}$$. For any extended real-valued function $$f:\mathbb{R}\to\mathbb{R}\cup\left\{{-\infty,+\infty}\right\} \ $$ whose domain is measurable, the following statements are equivalent:


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

These statements imply

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

Proof
Let the domain of $$f \ $$ be $$D \ $$. We have $$1 \Longrightarrow 4$$, since $$\left\{{x:f(x) \leq \alpha}\right\} = D - \left\{{x:f(x)>\alpha}\right\}$$, and the difference of two measurable sets is measurable, as a known property of measurable sets. Similarly, $$4 \Longrightarrow 1$$ and $$2 \iff 3$$.

We also have $$1 \Longrightarrow 2$$, since $$\left\{{x:f(x) \geq \alpha }\right\} = \bigcap_{n=1}^\infty \left\{{x:f(x)>\alpha-\tfrac{1}{n}}\right\}$$, and the intersection of a sequence of measurable sets is measurable. Similarly, $$2 \Longrightarrow 1$$, since $$\left\{{x:f(x) > \alpha}\right\} = \bigcup_{n=1}^\infty \left\{{x:f(x) \geq \alpha + \tfrac{1}{n}}\right\}$$. This shows that $$1 \iff 2 \iff 3 \iff 4$$.

For the fifth statement, we have $$\left\{{x:f(x)=\alpha}\right\} = \left\{{x:f(x) \geq \alpha}\right\} \cap \left\{{x:f(x) \leq \alpha}\right\}$$, and so 3 and 4 imply 5 for $$\alpha \in \mathbb{R}$$. Since $$\left\{{x:f(x)= +\infty }\right\} = \bigcap_{n=1}^\infty \left\{{x:f(x) \geq n }\right\}$$, 2 implies 5 for $$\alpha=+\infty$$, and similarly 4 implies 5 for $$\alpha = - \infty$$.