# Continuity implies $\mu$-strongly measurability?

In view of the definitions below (That can be found in Infinite Dimensional Analysis: A Hitchhiker's Guide, of ALIPRANTIS and BORDER):

Definition 1. Suppose $$\Omega$$ is a set equipped with an álgebra $$\mathcal{A}$$. Also, let $$X$$ be a vector space. As in the real case, a function $$\varphi: \Omega \longrightarrow X$$ thah assumes only a finite number of values, say $$x_1,x_2,\cdots, x_n$$ is called X-simple function if $$A_i=\varphi^{-1}(\{x_i\}) \in \mathcal{A}$$, for each $$i=1,\cdots,n.$$

Definition 2. Let $$(\Omega, \Sigma, \mu)$$ be a measure space, and let $$f:\Omega \longrightarrow X$$ be a vector function. We say that $$f$$ is strongly $$\mathbf{\mu}$$-measureble if there exists a sequence $$\{\varphi_n\}_{n\in\mathbb{N}}$$ of X-simple functions such thath $$\displaystyle \lim_{n \rightarrow \infty} ||f(\omega)-\varphi_n(\omega)||=0$$ for almost all $$\omega \in \Omega$$.

If $$f$$ is continuous then $$f$$ is $$\mu$$-strongly measurability?

I could not prove (or give a example) because I could not even relate all these concepts. Is that true?

To discuss the continuity of a function $$f:\Omega\to X$$, one needs a topology on $$\Omega$$ and I will assume all open sets are in $$\mathcal{A}$$. If $$X$$ is a separable Banach space, every continuous function $$f:\Omega\to X$$ will be strongly $$\mu$$-measurable. Just pick a countable dense set $$D=\{d_1,d_2,\ldots\}$$ and $$f_n$$ have value $$d_m$$ on $$f^{-1}(B_{1/n}(d_m))$$ for $$m\leq n$$ and value $$0$$ everywhere else. Then $$\langle f_n\rangle$$ converges pointwise to $$f$$.
The other direction is a bit more complicated. If there is a sequence $$\langle\phi_n\rangle$$ of simple functions convergin almost everywhere to $$f$$, $$f$$ must take almost everywhere values in the closure of the countable set $$\bigcup_n \phi_n(\Omega)$$. So a strongly $$\mu$$-measurable function must have almost all values in a separable subspace. From the argument above, this is also sufficient.
If $$X$$ is a nonseparable Banach space, we can take $$\Omega$$ to be $$X$$ endowed with its Borel $$\sigma$$-algebra and $$\mu$$-counting measure. Then almost everywhere means everywhere. The identity function has range a nonseparable Banach space and is therefore not strongly continuous.
If $$X$$ is a nonseparable Banach space, it is essentially a question that cannot be decided on the usual assumptions of set theory whether any Borel probability measure on $$X$$ is supported on separable subspace (the appendix of the first edition of Billinglsey's book Convergence Of Probability Measures has the details.) If it is not, the identity can again give a counterexample.