Let $\mathcal{B} = \mathcal{B}_1\oplus\ldots\oplus \mathcal{B}_n$ be a direct sum of Banach spaces $\mathcal{B}_i$ each with norm $\|\cdot\|_{\mathcal{B}_i}$. The Banach space $\mathcal{B}$ has many equivalent norms. For instance, letting $v = (v_1,\ldots,v_n)\in\mathcal{B}, $
$$\|v\|_\infty = \max_{i}\|v_i\|_{\mathcal{B}_i}$$
and
$$\|v\|_1 = \|v_1\|_{\mathcal{B}_i} + \ldots + \|v_n\|_{\mathcal{B}_n}.$$
Let $(\Omega,\Sigma,\mathbb{P})$ be a probability space. My question is to do with the Bochner integrability of a measurable function $f:\Omega\rightarrow \mathcal{B}$ of the form $f(\omega) = (f^{(1)}(\omega),\ldots,f^{(n)}(\omega))$.
Claim: A function $f:\Omega\rightarrow\mathcal{B}$ is Bochner integrable if and only if $f^{(i)}:\Omega\rightarrow\mathcal{B}_i$ is Bochner integrable for each $i$.
Proof: Suppose first that each $f^{(i)}$ is Bochner integrable and so let $s^{(i)}_k$ be the corresponding sequences of simple functions. Then, using $\|\cdot\|_1$ we have
$$\lim_{k\rightarrow\infty} \left[\sum_{i=1}^n\int_\Omega\|f^{(i)}-s^{(i)}_k\|_{\mathcal{B}_i}\,\mathrm{d}\mathbb{P}\right] = \lim_{k\rightarrow\infty} \left[\int_\Omega\|f – s_k\|_1\,\mathrm{d}\mathbb{P}\right] = 0,$$
where $s_k = (s^{(1)}_k,\ldots,s^{(n)}_k)$. Thus $f$ is Bochner integrable.
Suppose now that $f$ is Bochner integrable and let $s_k$ be the corresponding sequence of simple functions. Then, now using $\|\cdot\|_\infty$, since for each $i$, $0\leq \|f^{(i)}-s_k^{(i)}\|_{\mathcal{B}_i}\leq \|f-s_k\|_\infty$ and by the squeeze theorem,
$$ \lim_{k\rightarrow\infty} \left[\int_\Omega\|f – s_k\|_\infty\,\mathrm{d}\mathbb{P}\right] = 0 \implies \lim_{k\rightarrow\infty} \left[\int_\Omega\|f^{(i)} – s^{(i)}_k\|_{\mathcal{B}_i}\,\mathrm{d}\mathbb{P}\right] = 0.$$
Thus each $f_i$ is Bochner integrable.
Question: Is this proof valid? In particular, am I allowed to freely interchange the choice of norm for $\mathcal{B}$ since they are equivalent?
EDIT: Is the claim also true for a countably infinite product of Banach spaces $\mathcal{B} \subseteq \prod_{i\in \mathbb{N}}\mathcal{B}_i$, where $\mathcal{B}$ consists of elements $v$ such that $\|v\|_\infty$ is finite? Is the claim also true where $\mathcal{B}$ consists of elements $v$ such that $\|v\|_1$ is finite?
Best Answer
Yes, you can use either norm in the proof. Note that $\|v\|_{\infty} \leq \|v\|_1\leq n \|v\|_{\infty}$ and $n$ is fixed.