# Measurable Image

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## Theorem

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

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

- $(1): \quad \forall \alpha \in \R: \set {x: \map f x > \alpha} \in \mathfrak M$
- $(2): \quad \forall \alpha \in \R: \set {x: \map f x \ge \alpha} \in \mathfrak M$
- $(3): \quad \forall \alpha \in \R: \set {x: \map f x < \alpha} \in \mathfrak M$
- $(4): \quad \forall \alpha \in \R: \set {x: \map f x \le \alpha} \in \mathfrak M$

These statements imply:

- $(5): \quad \forall \alpha \in \R \cup \set {-\infty, +\infty}: \set {x: \map f x = \alpha} \in \mathfrak M$

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## Proof

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

We have that Measurable Sets form Algebra of Sets.

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

So:

- $\set {x: \map f x \le \alpha} = D - \set {x: \map f x > \alpha}$

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:

- $\ds \set {x: \map f x \ge \alpha} = \bigcap_{n \mathop = 1}^\infty \set {x: \map f x > \alpha -\dfrac 1 n}$

because if:

- $x \in \set {x: \map f x \ge \alpha}$

that is:

- $\map f x \ge \alpha$

and since:

- $\forall n \in \N: n > 0: \alpha > \alpha - \dfrac 1 n$

then:

- $\forall n \in \N: n > 0: \map f x > \alpha - \dfrac 1 n$

That is:

- $\forall n \in \N: n > 0: x \in \set {x: \map f x > \alpha - \dfrac 1 n}$

Hence:

- $\ds x \in \bigcap_{n \mathop = 1}^\infty \set {x: \map f x > \alpha - \dfrac 1 n}$

Conversely, suppose:

- $\ds x \in \bigcap_{n \mathop = 1}^\infty \set {x: \map f x > \alpha - \dfrac 1 n}$

that is:

- $\forall n \in \N: n > 0: x \in \set {x: \map f x > \alpha - \dfrac 1 n}$

Claim $\map f x \ge \alpha $.

Otherwise $\map f x < \alpha$, say for example $\map f x = \alpha - \size \epsilon$.

Choose $N = \ceiling {\dfrac 1 {\size \epsilon} } + 1 \in \N$.

Therefore:

- $N > \ceiling {\dfrac 1 {\size \epsilon} } \ge \dfrac 1 {\size \epsilon}$

and so:

- $\alpha - \dfrac 1 N > \alpha - \size \epsilon$

By hypothesis:

- $\forall N \in \N: \map f x > \alpha - \dfrac 1 N$

and therefore by the just previous:

- $\map f x > \alpha - \size \epsilon$

But we had $\map f x = \alpha - \size \epsilon$, a contradiction.

Therefore:

- $\map f x \ge \alpha $

that is: $x \in \set {x: \map f x \ge \alpha}$ which was to be shown.

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

Similarly:

- $\ds \set {x: \map f x > \alpha} = \bigcup_{n \mathop = 1}^\infty \set {x: \map f x \ge \alpha + \dfrac 1 n}$

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

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

For the fifth statement, we have:

- $\set {x: \map f x = \alpha} = \set {x: \map f x \ge \alpha} \cap \set {x: \map f x \le \alpha}$

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

Since:

- $\ds \set {x: \map f x = +\infty} = \bigcap_{n \mathop = 1}^\infty \set {x: \map f x \ge n}$

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

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

$\blacksquare$