Let $\mathcal C_b(\mathbb R)$ the set of bounded and continuous functions. For any compact interval $I\subset \mathbb R$ of non zero length, define $$\varphi_I:\mathcal C_b(\mathbb R)\to \mathbb R$$ by $$\varphi_I f=\sup_{I}f.$$
1) Let $\mathcal G$ the $\sigma -$field of subset of $\mathcal C_b(\mathbb R)$ generated by the applications of the form $\varphi_I$. What does it mean. Show it's equivalent to a simplet $\sigma -$field.
2) Let $g\in \mathcal C_b(\mathbb R)$. Define $$F:(\mathbb R,\mathcal B)\to (\mathcal C_b(\mathbb R), \mathcal G),$$ by $$F(x)(y)=\sup_{z\in [y-\frac{|x|}{2},y+\frac{|x|}{2}]}g'z).$$ I recall that $\mathcal B$ is the $\sigma -$field of the Borel sets. Prove that $F$ is measurable.
I'm sorry but I even have problem to understand what to do. I followed a course of measure, but here it looks very weird. Even I can't answer on what is $\mathcal G$ because I don't know which tribute is on $\mathbb R$. If it's the Borel set, I would say that $$\mathcal G=\sigma \{\varphi_I^{-1}(B)\mid B\in \mathcal B, I\subset \mathbb R\text{ a non zero interval}\}.$$ But still, I can I guess that we use $\mathcal B$ instead of $\mathcal M$ (the $\sigma -$ field of Lebesgue mesurable function) ?
In all this post $I$ is supposed to be a compact nonzero interval.
Q1) $\mathcal G$ is the smallest algebra s.t. $\varphi_I$ are measurable for all $I$. Notice that $\varphi_If$ are measurable $\iff$ $$\{f\mid \varphi_I f<\alpha \}\in \mathcal G,$$ which is equivalent to prove that $\varphi_I(B)\in \mathcal G$ for all Borel set $B$. So indeed $$\mathcal G=\sigma \{\varphi_I^{-1}(B)\mid B\in \mathcal B, I\subset \mathbb R\text{ nonzero interval}\}.$$ In fact, if we set $$\mathcal G'=\{\delta_x:\mathcal C_b(\mathbb R)\longrightarrow \mathbb R\mid x\in\mathbb R, \delta_x f=f(x)\},$$ we'll have $\mathcal G'=\mathcal G$. To prove this fact, you just have to prove that $\delta_x$ are $\mathcal G$ measurable for all $x$ and $\varphi_I$ are $\mathcal G'$ measurable for all nonzero interval $I$.
Q2) Since $\mathcal G$ is generated by sets of the form $\{f\in \mathcal C_b(\mathbb R)\mid f(y)<\alpha \}$, it's enough to prove that $F^{-1}(\{f\mid f(y)<\alpha \})$ is Borel for all $\alpha $ and all $y\in\mathbb R$.
$$F^{-1}(\{f\mid f(x)<\alpha \})=\{x\mid F(x)(y)<\alpha \}=\left\{x\mid \sup_{[y-|x|/2,y+|x|/2]}g<\alpha \right\}.$$
But $x\longmapsto \sup_{[y-|x|/2,y+|x|/2]}g$ is continuous and bounded, and thus $$\left\{x\mid \sup_{[y-|x|/2,y+|x|/2]}g<\alpha \right\}$$ is an open set. That prove that $F^{-1}(\{f\mid f(x)<\alpha \})$ is open set and thus a Borel set.