Measurable Image

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[edit] 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:

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

[edit] 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.

So:

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

and 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 $\alpha = - \infty$ .

$\blacksquare$

This article or a section of it, namely: We need to show (or provide a link to) the results: $\left\{{x: f \left({x}\right) \ge \alpha}\right\} = \bigcap_{n=1}^\infty \left\{{x: f \left({x}\right) > \alpha -\tfrac{1}{n}}\right\}$ and $\left\{{x: f \left({x}\right) > \alpha}\right\} = \bigcup_{n=1}^\infty \left\{{x: f \left({x}\right) \ge \alpha + \tfrac{1}{n}}\right\}$ . Should be straightforward (reminds me of something I did in my first assignment on my topology course some years ago). needs explaining. You can help ProofWiki by explaining it. If this has already been done, please remove {{Explain}} from the code.

Measurable Image

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[edit] Theorem

[edit] Proof

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