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Show that $\sigma(C(\Bbb R^n,\Bbb R))=\mathcal L_0(\Bbb R^n,\beta_n,\overline{\Bbb R})$.

Im stuck in this exercise. Some previous theoretical context:

Here $\mathcal L_0(\Bbb R^n,\beta_n,\overline{\Bbb R})$ is the set of functions $\Bbb R^n\to\overline{\Bbb R}$ measurables by $\beta_n$, that is the restriction of the Lebesgue measure to the Borel $\sigma$-algebra in $\Bbb R^n$, named as $\mathcal B(n)$.

The operator $\sigma(X)$ represent the induced Baire space by $X$, that is: the minimal Baire space that contains $X$.

My definition of Baire space, as I had seen in wikipedia, is not the standard: a Baire space $B\subset\overline{\Bbb R}^X$ (where $X$ is a Banach space) holds these 3 conditions:

  1. If $f+g\in\overline{\Bbb R}^X$ for some pair $f,g\in B$ then $f+g\in B$
  2. If $f\in B$ then $\alpha f\in B$ for $\alpha\in\Bbb R$
  3. If $(f_j)$ is a sequence in $B$ then $\sup_j f_j\in B$

Also in the following $\mathcal S(\Bbb R^n,\beta_n,\Bbb R)$ is the vector space of $\beta_n$-simple measurable functions. If $f\in\mathcal S(\Bbb R^n,\beta_n,\Bbb R)$ then it have the form $f=\sum_{k=1}^m c_k\cdot\chi_{A_k}$ for some $c_k\in\Bbb R$ and borelian sets of finite measure $A_k\subset\Bbb R^n$.


My way to try to solve this exercise is this:

I know that $\mathcal L_0(\Bbb R^n,\beta_n,\overline{\Bbb R})$ is a Baire space (assuming that $0\cdot f=0$ for any $f\in\mathcal L_0(\Bbb R^n,\beta_n,\overline{\Bbb R})$). And because $C(\Bbb R^n,\Bbb R)\subset\mathcal L_0(\Bbb R^n,\beta_n,\overline{\Bbb R})$ then we find that $\sigma(C(\Bbb R^n,\Bbb R))\subset\mathcal L_0(\Bbb R^n,\beta_n,\overline{\Bbb R})$.

Now my idea to show the other inclusion is that: if $\mathcal S(\Bbb R^n,\beta_n,\Bbb R)\subset\sigma(C(\Bbb R^n,\Bbb R))$ then $\sigma(\mathcal S(\Bbb R^n,\beta_n,\Bbb R))=\mathcal L_0(\Bbb R^n,\beta_n,\overline{\Bbb R})\subset\sigma(C(\Bbb R^n,\Bbb R))$ by a previous result.

To show that $\mathcal S(\Bbb R^n,\beta_n,\Bbb R)\subset\sigma(C(\Bbb R^n,\Bbb R))$ by the properties of Baire spaces listed above it will be enough to show that there is a sequence of continuous functions $(f_j)$ such that $(f_j)\uparrow \chi_A$ for any chosen borelian $A\subset\Bbb R^n$ with finite Lebesgue measure.

However Im not sure how to approach this last task or if this line of action will be fruitful. I need some help to finish the exercise, maybe using a different way to prove it.


UPDATE: I think that finally I find a correct proof, can someone check it?

Note that a Borel set is defined by countable set operations of union and complementation over the topology of $\Bbb R^n$. Also note that:

  1. For every open ball $O:=\Bbb B(x,\epsilon)$ (with $x\in\Bbb Q^n$ and $\epsilon\in\Bbb Q_{>0}$) there are sequences of continuous functions $(f_j)\uparrow \chi_O$ and $(g_j)\downarrow\chi_O$.

  2. Every open set in $\Bbb R^n$ is a countable union of open balls as in the point above, because these balls are a countable basis of the topology of $\Bbb R^n$.

  3. If there is a sequence of continuous functions $(f_{j,k})_j\uparrow \chi_{A_k}$ for each $k\in\Bbb N$ then $(\tilde f_n)\uparrow\chi_{\bigcup_k A_k}$ is also a sequence of continuous functions, for $\tilde f_n:=\max_{j,k\le n}f_{j,k}$. Similarly $(\tilde g_n)\downarrow\chi_{\bigcup_k A_k}$ for $\tilde g_n:=\min_{j,k\le n}g_{j,k}$ if there are sequences $(g_{j,k})_j\downarrow\chi_{A_k}$.

  4. If there is a sequence of continuous functions $(f_j)\uparrow \chi_A$ then there is other sequence of continuous functions $(\tilde f_j)\downarrow\chi_{A^\complement}$, for $\tilde f_j:=1-f_j$. Similarly if there is some sequence $(g_j)\downarrow\chi_A$ then $(\tilde g_j)\uparrow\chi_{A^\complement}$ for $\tilde g_j:=1-g_j$.

The four previous points ensures that for any borelian set $B$ there are monotone sequences of continuous functions such that $(f_j)\uparrow\chi_B$ and $(g_j)\downarrow\chi_B$ because any borelian set can be defined by countable operations of union and complementation of a countable basis of the topology of $\Bbb R^n$.$\Box$

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Coming back to this question some time later I finally solved it. Note that

$$\chi_{\bigcup_j A_j}=\sup_j\chi_{A_j}\qquad\text{and}\qquad \chi_{A^\complement}=1-\chi_A\tag1$$

Now let $(a,b)\subset\Bbb R$ and define a sequence of continuous functions $(f_j)$ such that $\sup_j f_j=\chi_{(a,b)}$ (any positive $j$-th root of an arc on $(a,b)$ with positive values but less or equal to 1 and zero in $\Bbb R\setminus\{(a,b)\}$ would work).

This procedure can be generalized easily to open intervals of the kind $I(a,b):=\prod_{k=1}^n(a_k,b_k)$ on $\Bbb R^n$ (for $a:=(a_1,\ldots, a_n)$ and $b$ defined equivalently), and consequently $\chi_{I(a,b)}\in\sigma(C(\Bbb R^n,\Bbb R))$.

(In the previous notation if there is some $a_k\ge b_k$ then we set $I(a,b):=\emptyset$, and $\chi_\emptyset=0$.)

By a previous result I know that the collection of open intervals

$$\Bbb J(n):=\left\{\prod_{k=1}^n(a_k,b_k)\in\Bbb R^n: a_k,b_k\in\Bbb Q\right\}\tag2$$

generates the Borel $\sigma$-algebra under the operations of countable union and complementation. Then $(1)$ and $(2)$ together imply that the indicator function of any borelian set in $\Bbb R^n$ belongs to $\sigma(C(\Bbb R^n,\Bbb R))$, then using the definition of $\beta_n$-simple function and the first two properties of Baire spaces listed in the question then we finally find that

$$\mathcal S(\Bbb R^n,\beta_n,\Bbb R)\subset\sigma(C(\Bbb R^n,\Bbb R))$$

and because the operator $\sigma$ is idempotent and by a previous result I know that $\sigma(\mathcal S(\Bbb R^n,\beta_n,\Bbb R))=\mathcal L_0(\Bbb R^n,\beta_n,\overline{\Bbb R})$ then we are done.

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