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Let $f \in C_b(S)$ (set of all bounded and continuous functions) and $\mu$ be a measure on $S$ where $S$ is a complete separable metric space. Then a book (Probability Theory by Borkar) claims that for a given $\epsilon > 0$ there exists $N \geq 1$ and $a_0 < a_1 < a_2<\dots <a_N$ such that

(a) $\|f\| -1 = a_0 < a_1 < a_2<\dots <a_N = \|f\| + 1$

(b) $\mu(\{x | f(x) = a_i\})=0$ for all $i$ and

(c) $a_i - a_{i-1} \leq \epsilon$, $1 \leq i \leq N$ .

I am not understanding how can he claim point (b) for a general $\mu$.

Then he claims $\bigcup_{i=1}^{N} \{x|a_{i-1} \leq f(x) \lt a_i\}=S$.

$\|f\| = \sup_x|f(x)|$

Anonymous
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    Is $\mu(S)$ finite (probably equal to $1$)? The reason this is possible: there can be only countably many numbers $a$ such that the set ${x : f(x)=a}$ has positive measure. Otherwise their (disjoint) union would have infinite measure. –  Dec 27 '14 at 06:08
  • @Behaviour: OK. But, how what is the meaning of (1). It does not cover the whole range of $f$. – Anonymous Dec 27 '14 at 06:14
  • (a) could have a type, maybe the left side should be $-|f|-1$. –  Dec 27 '14 at 06:16
  • @Behaviour: hmm, looks like so. – Anonymous Dec 27 '14 at 06:19
  • @Behaviour: Also, what is the reason of $-1$ and $+1$ in (1) ? – Anonymous Dec 27 '14 at 06:25
  • Just to have a little room around. If you don't add $1$, then the union does not capture the points where $f(x)=|f|$ because of the strict inequality. But most likely it has something to do with the way the proof goes later. –  Dec 27 '14 at 06:30

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I think we established that:

  1. The measure space, being finite, cannot contain an uncountable family of disjoint sets of positive measure. Therefore, there are only countably many values of $a$ that have to be avoided when choosing $a_i$.

  2. The inequality in (a) should probably be $$ -\|f\| -1 = a_0 < a_1 < a_2<\dots <a_N = \|f\| + 1$$

  3. Every value of $f$ falls strictly between $-\|f\| -1$ and $\|f\| + 1$, hence the sets $\bigcup_{i=1}^{N} \{x|a_{i-1} \leq f(x) \lt a_i\} $ cover $S$.

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