Let $(H,||\cdot||_H)$ be a Banach space and $K$ a (not necessarily closed) subspace. Suppose that $K$ is a Banach space under another norm $||\cdot||_K$, which satisfies

$$||x||_H\leq ||x||_K$$

for all $x\in K$. Let $(S,\mu)$ be a measure space and $f:S\rightarrow K$ a strongly measurable function in the sense of Bochner, with respect to both $||\cdot||_K$ and $||\cdot||_H$. That is, there exist sequences of simple functions $\phi_n$ and $\psi_n$ on $S$ taking values in $K$ and $H$ respectively such that

$$\lim_{n\rightarrow\infty} ||f(s) - \phi_n(s)||_{K} = \lim_{n\rightarrow\infty} ||f(s) - \psi_n(s)||_{H} = 0$$

for $\mu$-almost all $s\in S$.

Suppose now that $f$ satisfies

$$\int_S ||f(s)||_H\,d\mu(s) < \infty,$$

so that, by a criterion of Bochner, $f$ is integrable as a function with values in $H$. Denote $$h := \int_S f(s) d\mu(s).$$

**Question:** If $h\in K$, then does it follow that $f$ is Bochner-integrable in $K$? That is, does there exist a sequence of simple functions $\phi_n$ such that

$$\lim_{n\rightarrow\infty}\int_S ||f(s) - \phi_n(s)||_{K}\,d\mu(s) = 0?$$