Let $(X,\mathfrak{M})$ be a measurable space and $E$ be a Banach space and $f:(X,\mathfrak{M})\rightarrow E$ be a function.


Are Borel-measurable condition on $f$ and Bochner-measurable condition on $f$ compatible?

If these are not compatible, then is it true that if $f$ is Bochner integrable then $f$ is Borel measurable?

It is written in wikipedia that Bochner-integral is an extension of Lebesgue integral, but I am curious in what sense it is an extension.

If this is true, is there any textbook treating these contents? There is nothing on these in standard textbooks..(such as that by Joseph)


1 Answer 1


The standard reference is probably still the book Vector Measures by Diestel and Uhl.

If $f$ is the pointwise limit of Borel measurable functions, then $f$ is Borel measurable. To see this, note that it suffices to show that the preimage of an open set is measurable and if $\langle f_n\rangle$ is a sequence of Borel measurable functions converging pointwise to $f$ and $O$ is open, then $$f^{-1}(O)=\{\omega:f(\omega)\in O\}=\{\omega:\lim_n{ f_n(\omega)\in O}\}$$ $$=\{\omega:f_n(\omega)\in O\text{ for }n\text{ large enough}\}=\bigcup_{n=1}^\infty\bigcap_{m=n}^\infty f_m^{-1}(O).$$ In particular, Bochner measurable functions are measurable. In a separable Banach space, every Borel measurable function will be the pointwise limit of simple functions. Just pick a countable dense set $D=\{d_1,d_2,\ldots\}$ and for $f$ Borel measurable let $f_n$ 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 $f$. If one is willing to allow for simple functions with countably many values, as is sometimes done, one can make the convergence uniform.

If $f$ is the pointwise limit of simple functions, then the range of $f$ is in the closure of the set of values of the simple functions, so a Bochner measurable function must have values in a separable subspace. If $X$ is a nonseparable Banach space then the identity from $X$ endowed with the Borel $\sigma$-algebra to $X$ is Borel measurable, but not Bochner measurable.

Many of these concepts also come with almost-everywhere provisos instead of the pointwise version, but the difference is inessential if one works with complete measure spaces.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.