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!

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.
