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Measure theory: proof of the "Standardproof" given theorem.

Let $(X, \mathcal E)$ be a measurable space.

Let $W \subseteq \mathcal M(\mathcal E)$ (set of measurable $\mathcal E$-$\mathcal B(\mathbb R)$-functions) and suppose $W$ satisfy:

(i) $1_A \in W$ for all $A \in \mathcal E$.

(ii) $W$ is a subspace of the vectorspace of reel functions defined on $X$.

(iii) If $(f_n)$ is a increasing sequence of functions of $W$ ($f_n \le f_{n+1}$) s.t $\sup_{n \in \mathbb N} f_n(x) < \infty$, $x \in X$, then $\lim_{n \rightarrow \infty} f_n = \sup_{n \in \mathbb N} f_n(x) \in W$.

I want to prove $W =\mathcal M(\mathcal E)$.


I am given the following theorem:

Let $f \in \mathcal M(\mathcal E)$. Then there exist a sequence $(s_n)$ of simple functions s.t

(i) $f(x) = \lim_{n \rightarrow \infty} s_n(x)$ for all $x \in X$.

(ii) $|s_n(x)| \le |f(x)|$ for all $n \in \mathbb N$ and $x \in X$.

(iii) If $f \ge 0$ then $(s_n)$ can be chosen s.t $0 \le s_n \le s_{n+1}$.


I see that $s_n \in W$ for all $n \in \mathbb N$ by considering the standard representation of $s_n$. However I cannot make the sequence $(s_n)$ increasing even by considering a subsequence of $(s_n)$ ?

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  • $\begingroup$ Point (iii) says that $(s_n)_n$ is increasing. You just have to reduce to the case $f \geq 0$ (how?). $\endgroup$ – PhoemueX Sep 18 '14 at 8:19
  • $\begingroup$ I have considered this, but I've no idea. I have already proved the theorem in case $f \in \mathcal M(\mathcal E)^+$. But in this case $f$ might be negative for certain $x$'s and then $f \ge 0$ is not true. $\endgroup$ – Shuzheng Sep 18 '14 at 8:23
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Simply decompose $f \in \mathcal{M}(\mathcal{E})$ as $f = f_+ - f_-$ with $f_+ = \max \{0, f\}$ and $f_- = - \min\{0, f\}$.

Then $f_+, f_- \in \mathcal{M}(\mathcal{E})^+$ (why?). By the part of the proof you have already done, $f_+, f_- \in W$.

Why does this help you?

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  • $\begingroup$ Ahh, yes - and then I use the fact that $\mathcal M(\mathcal E)$ is a vector space to conclude that $f \in \mathcal M(\mathcal E)$. $\endgroup$ – Shuzheng Sep 18 '14 at 8:31
  • $\begingroup$ @NicolasLykkeIversen: Yes, exactly. $\endgroup$ – PhoemueX Sep 18 '14 at 8:35

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