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 


$$\lim\_{n\rightarrow\infty}\int\_S ||f(s) - \phi\_n(s)||\_{K}\,d\mu(s) = 0?$$



Bochner integrability within a subspace
The Bochner integral is a red herring here. Let $S = \mathbb{N}$ and let $\mu$ be counting measure. Then a "strongly measurable function" is just a sequence, and it is "Bochner integrable" iff it is absolutely summable (in $H,K$ respectively). 


If the $H$ and $K$ norms are not equivalent (which is necessarily the case when $K$ is not closed) then we can find a sequence $y\_n \in K$ with $\|y\_n\|\_H \le 2^{-n}$ but $\|y\_n\|\_K \ge 2^n$. Set $x\_1 = y\_1$, $x\_{n+1} = y\_{n+1} - y\_n$, and think of the function $f(n) = x\_n$. Note that $\|x\_n\|\_H \le 2^{-(n-1)}$ and $\|x\_n\|\_K \ge 2^{n-1}$. 


Now $x\_n$ is absolutely summable in $H$-norm and $\sum\_n x\_k = 0$ because the sum telescopes. That is to say that $f$ is Bochner integrable in $H$ and the value of the integral $\int\_S f\,d\mu$ is 0 which is in $K$.


On the other hand, $\sum\_n x\_n$ diverges in $K$, which is to say that the Bochner integral in $K$ does not exist.


