Let $C_b(\mathbb{R})$ be the set of real-valued bounded continuous functions on $\mathbb{R}$. The $\sigma-$algebra generated by $C_b(\mathbb{R})$ is the smallest $\sigma-$algebra, say $\sigma(C_b(\mathbb{R}) )$, such that all functions in $C_b(\mathbb{R})$ are measurable with respect to $\sigma(C_b(\mathbb{R}) )$.
Do the following equations hold?
-
$
\sigma(C_b(\mathbb{R}) )= \sigma(\{f^{-1}(A):A\in\mathcal{B}(\mathbb{R}), f \in C_b(\mathbb{R}) \}).
$ -
$
\sigma(C_b(\mathbb{R}) ) = \mathcal{B}(\mathbb{R})
$
Best Answer
You are right on both counts.
More generally, let $\{f_i: X \to X_i\}_{i \in I}$ be a collection of functions that take values in measurable spaces $(X_i, \mathcal{F}_i)$. The smallest $\sigma$-algebra on $X$ such that all $f_i$'s become measurable is precisely $$\sigma\{f_i^{-1}(A_i): i \in I, A_i \in \mathcal{F}_i\}.$$
Your first claim follows from this general fact.
Next, let us show that $$\sigma(C_b(\mathbb{R})) = \mathcal{B}(\mathbb{R}).$$
Clearly the inclusion $\supseteq$ holds, because any closed interval of $\mathbb{R}$ is the Borel-preimage of a function in $C_b(\mathbb{R})$ [If the interval is $[a,b]$, consider the function $f: \mathbb{R}\to \mathbb{R}$ defined by $f(x) = x$ if $a \le x \le b$ and extend it to a continuous bounded function on all of $\mathbb{R}$ such $f$ does not attain the values $[a,b]$ outside the interval $[a,b]$ (draw a picture if you don't see that this exists).] Then recall that closed intervals generate the Borel $\sigma$-algebra on $\mathbb{R}$.
Conversely, clearly all preimages $f^{-1}(A)$ where $A$ is Borel and $f \in C_b(\mathbb{R})$ are again Borel, because continuous functions are measurable. Hence, the inclusion $\subseteq$ also holds.