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.


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