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Consider collections $ \mathcal{V} $ of functions from $ \mathbb{R} $ to $\mathbb{R} $ satisfying the following conditions:

(a) $ \mathcal{V} $ is a vector space

(b) $ \mathcal{V} $ contains the continuous functions

(c) If $ (f_{n})_{n} $ is an increasing sequence of nonnegative functions in $ \mathcal{V} $ and if $ \lim_{n \rightarrow \infty} f_{n}(t) $ exists and is finite for all $ t \in \mathbb{R} $, then $ \lim_{n\rightarrow \infty} f_{n}(t) \in \mathcal{V} $ .

Show that the collection $ \mathcal{V}_{0} $ consisting of the Borel-measurable functions is the smallest such collection of functions. (Hint: define $\mathcal{A}= \{ A \subseteq \mathbb{R}: \chi_{a} \in \mathcal{V} \}$. Show that $ \mathcal{A} $ contains the interval $ (-\infty,a) $, and then contains the Borel sets).

$\chi_{A} :$ characteristic function of $A$.

$\chi_{A}(x)=1$ if $x \in A $ , $\chi_{A}(x)=0$ if $x \notin A, $

I need to find or prove the existence of a sequence of positive continuous functions, increasing to converge to $ \chi_ {A} $, then I could use the basic theorem of measure theory to put all the functions borel measurable in the set. Any help is appreciated, I try with step functions but dont result.

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If you are familiar with the construction of the Borel sets (i.e. $\Sigma^0_\alpha,\Pi^0_\alpha,\Delta^0_\alpha$) then you should prove that $A\in\Sigma^0_\alpha$ if and only if $\chi_A$ is in the $\alpha$-th iteration of limits starting from continuous functions. – Asaf Karagila May 13 '12 at 22:25

It is not true that every $\chi_A$, where $A$ is Borel, can be written as a limit of continuous functions. (Not even $\chi_\mathbb{Q}$ can be written this way.)

Consider the set $\mathcal{A}$ defined in the hint. As Norbert's answer suggests, you can show $(-\infty, a) \in \mathcal{A}$ for each $a \in \mathbb{R}$. Now use Dynkin's $\pi$-$\lambda$ theorem. Show that $\mathcal{A}$ is a Dynkin system (or $\lambda$-system). The collection $\mathcal{P} = \{(-\infty, a) : a \in \mathbb{R}\}$ is a $\pi$-system contained in $\mathcal{A}$, so Dynkin's theorem says $\mathcal{A}$ contains $\sigma(\mathcal{P})$, which is exactly the Borel $\sigma$-algebra.

Once this is done, you have $\chi_A \in \mathcal{V}$ for every Borel set $A$. Then $\mathcal{V}$ contains all the simple functions (finite linear combinations of characteristic functions), and any Borel measurable function is a pointwise limit of simple functions.

What you are proving here is essentially a special case of the functional monotone class theorem.

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I think use $\mathcal{A}$ is a $\sigma$-algebra,soon how $c=(-\infty,a)\in \mathcal{A} $ and c generated the open sets of $\mathbb{R}$, so the borel-sets is in $\mathcal{A}$, but $\mathcal{A}$ is a $\sigma$-algebra? – kEoz May 13 '12 at 17:03
No, in general I think $\mathcal{A}$ is not a $\sigma$-algebra itself. (I don't know a counterexample in this case, but if $\mathbb{R}$ is replaced by another set there are certainly examples, see here.) – Nate Eldredge May 13 '12 at 18:03
ok, thx for your help, i think in other idea, because i dont know about Dynkin system and dont is in my course of measure theory – kEoz May 13 '12 at 18:13

Using the hint it is enough to prove that $\chi_{(-\infty,a)}$ can be pointwise limit of continuous increasing functions. Indeed, consider functions $$ f_n(x)= \begin{cases} 1 & \text{ if }\quad x<a-1/n\\ n(a-x) & \text{ if }\quad a-1/n\leq x<a\\ 0 & \text{ if }\quad x\geq a \end{cases} $$

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thx now i try put the borel-sets in $\mathcal{A}$, – kEoz May 13 '12 at 1:19
some idea for the borel-sets? – kEoz May 13 '12 at 4:04

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