Measurable Image

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Theorem

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

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

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

These statements imply:

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


Proof

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


We have that Measurable Sets form 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:

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

because if:

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

that is:

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

and since:

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

then:

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

That is:

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

Hence:

$\displaystyle x \in \bigcap_{n \mathop = 1}^\infty \left\{{x: f \left({x}\right) > \alpha - \dfrac 1 n}\right\}$


Conversely, suppose:

$\displaystyle x \in \bigcap_{n \mathop = 1}^\infty \left\{{x: f \left({x}\right) > \alpha - \dfrac 1 n}\right\}$

that is:

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

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

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

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

Therefore:

$N > \left\lceil{\dfrac 1 {\left\vert{\epsilon}\right\vert} }\right\rceil \ge \dfrac 1 {\left\vert{\epsilon}\right\vert}$

and so:

$\alpha - \dfrac 1 N > \alpha - \left\vert{\epsilon}\right\vert$

By hypothesis:

$\forall N \in \N: f \left({x}\right) > \alpha - \dfrac 1 N$

and therefore by the just previous:

$f \left({x}\right) > \alpha - \left\vert{\epsilon}\right\vert$

But we had $f \left({x}\right) = \alpha - \left\vert{\epsilon}\right\vert$, 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:

$\displaystyle \left\{{x: f \left({x}\right) > \alpha}\right\} = \bigcup_{n \mathop = 1}^\infty \left\{{x: f \left({x}\right) \ge \alpha + \dfrac 1 n}\right\}$

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


Since:

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

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

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

$\blacksquare$