It is well-known that if there is a function $f: \Omega \subset \mathbb R^n \rightarrow X$ with $\Omega$ open and $X$ is a Hilbert space, then continuity of $f$ implies also Bochner measurability of $f$.


I was wondering whether this is also true if $\Omega$ is an open subset of a Hilbert space. 


Does it hold if we additionally assume $f$ to be linear?



Bochner measurable; continuous operator
Can you specify which $\sigma$-algebra and measure you have on $\Omega$? Bochner measurability is defined as being the limit a.e. of measurable finitely valued functions. By Pettis' Theorem this is the same as being weakly measurable and almost separably valued.


Now take a non-separable $\Omega\subset X$ and let $f$ be the identity. Then $f$ is not separably valued, but it might be almost separably valued, depending on the measure.


On the other hand, if $\Omega$ is separable, then every continuous function is separably valued. The weak measurability then depends on the $\sigma$-algebra on $\Omega$. 


