Prove that $\mathscr{B}(\mathbb{R}^d)$ is a subset of $\mathscr{F}$, a $\sigma$-field

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 $$-\infty we have a continuous function $$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$$
• Thank you! Why is it important that $f(J)$ does not contain 0? – John Simpleton Oct 9 '18 at 15:55
• Note the equality $f^{-1}(R-\{0\})=J$ – S.D. Oct 9 '18 at 15:57
• Construction of $f_i$ is followed form graph of $f_i$ : join $(a_i,0)\in R^2$ with $(\frac{a_i+b_i}{2},1)\in R^2$ by line segment and join $(\frac{a_i+b_i}{2},1)\in R^2$ with $(b_i,0)\in R^2$ by another line segment and finally consider the two ray $(-\infty,a_i)$, $(b_i,\infty)$ as parts of graph. – S.D. Oct 9 '18 at 16:19