Let $\mathscr{F}$ be a $\sigma$-field on $\mathbb{R}^d$ such that every continuous function $f:\mathbb{R}^d\to\mathbb{R}$ that vanishes outside a bounded interval is $\mathscr{F}$-measurable. Prove that $\mathscr{B}(\mathbb{R}^d)\subseteq\mathscr{F}.$
I interpereted "vanishes outside a bounded interval" as the following:
Let $g$ be a continuous function such that $g:\mathbb{R}^d\to\mathbb{R}$ and $f=g1_{A}$ where $A\subseteq\mathbb{R}^d$ bounded. Then since $f$ is measurable, take any open bounded set $(a,b)\in\mathbb{R}$ and we have $f^{-1}(a,b)\in\mathscr{F}.$ Intuitively, since $\mathscr{B}(\mathbb{R}^d)$ is a $\sigma$-field every collection $\mathscr{A}\subseteq\mathscr{B}(\mathbb{R}^d)$ consists of subsets of $\mathbb{R}^d$.
Let $\mathscr{A}=\{A_1,A_2,A_3...\}$ where each $A_i\subseteq\mathbb{R}^d$. Is it true that $f(\mathscr{A})=\{(a_1,b_1),(a_2,b_2),...\}$ where each $(a_i,b_i)\subseteq\mathbb{R}$? If it's the case that $f$ maps each $A_i\in\mathbb{R}^d$ to a $(a_i,b_i)\in\mathbb{R}$, then isn't it clear that $\mathscr{A}\in\mathscr{F}$ from the argument in the above paragraph? What am I missing here?
Note that $\mathscr B(R^d)$ is generated by the set $S:=\{ \prod _{i=1} ^ d I_i : I_i\ are\ bounded\ open\ intervals\ of\ R\} $ i.e. $\mathscr B(R^d)$ is the smallest $\sigma-algebra$ contining $S$. Now for each element $J$ of $S$ you can construct a continuous map $f:R^d\rightarrow R$ which vanishes outside $J$ and $f(J)$ does not contains $0$.Notcice we can do when $d=1$ i.e. for $-\infty<a_i<b_i<\infty$ we have a continuous function $f_i$ such that $0 \notin f_i((a_i,b_i))$ and $f_i(R-(a_i,b_i))=\{0\}$ for $i=1,2,..,d$. So whenever $ J=\prod _{i=1} ^ d (a_i,b_i)$ your $f$ will be $\prod _{i=1} ^ d f_i$.
Then by hypothesis $f$ is measurable,so that $f^{-1}(R-\{0\})=J$ is in $\mathscr F$ , each element of $S$ is in $\mathscr F$ but $\mathscr B(R^d)$ is the smallest $\sigma-$algebra containing $S$, hence $\mathscr B(R^d) \subseteq \mathscr F$