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Let $B_{\infty}(\Omega)$ be the space of bounded measurable functions on the measurable space $\Omega$. For a given Banach space $X$, let us denote $B_{\infty}(\Omega,X)$ by the set of all bounded measurable functions $f:\Omega\to X$ (meaning $f^{-1}(O)$ is measurable for every open $O$ in $X$).

Is the space $B_{\infty}(\Omega,X)$ just the invective tensor product of $X$ and $B_{\infty}(\Omega)$?

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For $\Omega=\mathbb N$ with the disrete $\sigma$-algebra measurability is no condition and $$\ell^\infty(\mathbb N)\hat\otimes_\varepsilon X \cong C(\beta\mathbb N)\hat\otimes_\varepsilon X\cong C(\beta\mathbb N,X)$$ is (via restriction to $\mathbb N$ isomorphic to) the space of pre-compact sequences in $X$.

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  • $\begingroup$ Thanks. Seems that $B_{\infty}\hat\otimes_\varepsilon X$ is always embedded into $B_{\infty}(\Omega,X)$. Right? $\endgroup$
    – ABB
    Commented Jun 25, 2018 at 10:11
  • $\begingroup$ Yes, with respect to uniform convergence this is a closed subspace. $\endgroup$
    – Jochen Wengenroth
    Commented Jun 25, 2018 at 11:24

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