A continuous Function from $R$ to a Banach Space is Borel-Measurable

I think the notation of my book is a bit odd, so I'm having trouble finding any other sources to help me with this proof. The thing I'm trying to prove is that a continuous function, $$f$$, from $$R$$ to a Banach Space, $$B$$, is Borel-Measurable. The definition that my book uses is that a function, $$f$$ is Borel-Measurable if there exists a sequence of functions that converges pointwise to $$f$$, s.t every entry of the sequence is the finite sum of scaled indicator sets of the $$\sigma$$-ring formed by the Borel subsets of $$R$$.

My thoughts so far are that any point, $$b \in B$$ is closed, and so the pre-image of $$b$$ under $$f$$ should also be a closed set, and thus $$f^{-1}[b]$$ is in my $$\sigma$$-field. As such, I could use the indicator function for this set and express $$f$$ as $$f = \sum_{b \in B}b\dot E_{f^{-1}[b]}$$ where $$E_{f^{-1}[b]}$$ is the indicator function for the set $$f^{-1}[b]$$. However, this sum may not be finite. My next thought was to instead of summing over $$b \in B$$, sum over balls of radius $$\frac{1}{n}$$ and then this would converge to my function in the pointwise limit. However, this would assume that my Banach Space is totally bounded. It seems no matter what I think of I'm missing something. Can anyone offer any advice or suggestions?

Thank you!

• Hint: consider $f_n:=\sum_{k=-2^{2n}}^{2^{2n}}~~\frac{k}{2^n}I_{f^{-1}((\frac{k}{2^n},\frac{k+1}{2^n}])}$, where $I_A$ denotes the indicator function of a set $A$. – Eric Yau Nov 1 '18 at 4:26

Fix $$n$$. Then $$f([-n,n])$$ is a compact set. hence it can be covered by a finite number of open balls of radius $$\frac 1 n$$, say $$A_{n,1},A_{n,2},\cdots, A_{n,k}n$$. Let $$B_{n,1}=A_{n,1},B_{n,2}=A_{n,2}\setminus A_{n,1}$$, $$\cdots$$, $$B_{n,k_n}=A_{n,k_n}\setminus \cup_{j=1}^{k_n-1} A_{n,j}$$. Then the sets $$f^{-1}(B_{n,j})$$ are Borel sets in $$\mathbb R$$ (because they are differences of two open sets). For $$|t| \leq n$$ define $$f_n(t)=f(t)$$ if $$f(t) \in B_{n,j}$$. Take $$f_n(t)$$ to be $$0$$ if $$|t| >n$$. I leave it to you to verify that this sequence has the required properties.