I first specify the setting and then formulate the question precisely. (A very long post follows.)

**Definitions 1.** For $E$ a (real Hausdorff) locally convex space, say that $E$ is *suitable* iff there is a Banach(able) space $E_0$ from which $E$ is obtained by possibly weakening the topology so that the set of bounded sets does not enlarge. Thus $E$ and $E_0$ have the same underlying (abstract) vector space, and the topology of $E$ is (possibly) weaker. Since $E$ and $E_0$ have the same bounded sets, the topology of $E_0$ can be recovered from that of $E$. Call any norm $\nu$ determining the topology of $E_0$ *admissible* for $E$. We may also say that $E$ is a *weakening* of $E_0$.

For example, if $E$ is Banach, then $E$, $E_\sigma$ and $E_\sigma^{\kern.4mm\prime}$ are suitable since $E_\sigma$ is a weakening of $E$ and $E_\sigma^{\kern.4mm\prime}$ is one of $E_\beta^{\kern.4mm\prime}$. Note that $E_\sigma^{\kern.4mm\prime}$ may be weaker than $(E_\beta^{\kern.4mm\prime})_\sigma$. This holds e.g. when $E=c_0(\mathbb N_0)$. Let $E'$ denote the *set* of continuous linear functionals $u:E\to\mathbb R$.

For simplicity, let $\mu$ be the Lebesgue measure restricted to the standard Borel $\sigma\,$−algebra of the unit interval $I=[0,1]$. Considering functions $x:I\to E$ where $E$ is suitable, say that $x$ is *scalarly measurable* iff $u\circ x:I\to\mathbb R$ is measurable for all $u\in E'$. Say that $x$ is *simple* iff its range is finite and $x^{-1}[\{\xi\}]\in{\rm dom\kern.8mm}\mu$ holds for all $\xi$ in $E$. Say that $x$ is *simply measurable* iff there are a sequence $\langle\kern.8mm\sigma_i:i\in\mathbb N_0\kern.4mm\rangle$ of simple functions and $N\in\mu^{-1}[\{0\}]$ such that $\langle\kern.6mm\sigma_i(t):i\in\mathbb N_0\kern.6mm\rangle\to x(t)$ in the topology of $E$ for every fixed $t\in I\setminus N$. Write $x\approx z$ iff $x,z:I\to E$ are scalarly measurable such that $u(x(t))=u(z(t))$ holds for a.e. $t\in I$ for every fixed $u\in E'$. Say that $x$ is *worth Bochner* in $E$ iff there is a simply measurable $z$ with $x\approx z$.

For functions $x:I\to E$ and $\varphi:I\to[\,0\,,+\infty\,]$, say that $\varphi$ *measurably dominates* $x$ iff $\varphi$ is (Borel) measurable with $\mu(\varphi^{-1}[\{+\infty\}])=0$ and $\nu(x(t))\le\varphi(t)$ for all $t\in I$ and some admissible norm $\nu$ for $E$. Say that $x$ is *measurably dominated* iff some measurably dominating $\varphi$ exists. Now the question is

Q.Do there exist a (necessarily nonseparable) reflexive Banach space $E$ and a measurably dominated scalarly measurable $x:I\to E$ such that $x$ isnotworth Bochner in $E$ ?

**Remark 2.** (on the motivation behind Q) If $E$ is a separable Banach space, by Pettis' theorem we have $L_s^p(\mu,E)=L^p(\mu,E)$ where the former space is explained in Constructions 3 below and the latter space is the classical Bochner space. For $1\le p<+\infty$, for example from R. E. Edwards' book *Functional Analysis*, Theorems 8.18.2 and 8.20.3 on pages 588 and 606, respectively, we know that $(*)$ the strong dual of $L_s^p(\mu,E)$ is naturally represented by $L_s^{p^*}(\mu,E_\sigma^{\kern.4mm\prime})$ when $E$ is separable. Here $p^*$ denotes the "conjugate" exponent, i.e. $p^*=(1-p^{-1})^{-1}$ for $p\not=1$, and $1^*=+\infty$. Likewise, if $E$ is any reflexive Banach space, the dual of $L^p(\mu,E)$ is similarly represented by $L^{p^*}(\mu,E_\beta^{\kern.4mm\prime})$. If the answer to Q is "no", we can then state $(*)$ in a more "unified" manner by saying that it holds whenever $E$ is a separable or reflexive Banach space. Thus the scalarwise vector valued Lebesgue space $L_s^p$ would be more "natural" in connection with duality.

**Constructions 3.** For $1\le p\le+\infty$ and $E$ suitable, construct the space $F=L_s^p(\mu,E)$ as follows. In the case $p\not=+\infty$, let $S$ be the set of all scalarly measurable $x:I\to E$ such that there is an integrable function $\varphi:I\to[\,0\,,+\infty\,]$ with $\|x(t)\|^{\,p}\le\varphi(t)$ for all $t\in I$, where $\xi\mapsto\|\xi\|$ is any fixed admissible norm for $E$. For $p=+\infty$, the obvious modification is made. The set $S$ is a vector subspace in $E^{\,I}$, and we let $X$ denote this abstract vector space whose underlying set is $S$. A seminorm $\nu$ for $X$ is obtained by taking $\nu(x)=\inf\big\{\,\big(\int_I\varphi{\kern1.1mm\rm d\kern.7mm}\mu)^{\,p^{-1}}\kern-.5mm:\varphi\in{\rm D\,}(x)\,\}$ where ${\rm D\,}(x)$ is the set all those dominating integrable $\varphi$. (obvious modification for $p=+\infty$) This seminorm determines a vector topology $\mathscr T$ for $X$, and we then put $F=(X,\mathscr T)/N_0$ where $N_0$ is the set of all $x\in S$ with $u(x(t))=0$ for a.e. $t\in I$ for every fixed $u\in E'$.

Under our above assumptions, one can prove that $F$ is Banachable with a compatible (i.e. topology determining) norm given by $\vec x\mapsto\inf\,\{\,\nu(x):x\in\vec x\,\}$. The space $L^p(\mu,E)$ is constructed similarly except that we take as elements of $S$ the simply measurable $x:I\to E$ instead of the scalarly measurable ones.

**Remark 4.** Assuming the continuum hypothesis, there are weakenings $E$ of nonseparable nonreflexive Banach spaces with $L_s^p(\mu,E)$ not naturally linearly homeomorphic to $L^p(\mu,E)$. That is, the linear map $L^p(\mu,E)\to L_s^p(\mu,E)$ determined by $\vec x\mapsto\vec z$, when $\vec x\subseteq\vec z$, is not surjective. Indeed, using the result of this answer by Michael Renardy, for $E$ the vector subspace of $\ell^{\,+\infty}(I)$ formed by the elements $\xi$ with $\xi(s)\not=0$ only for countably many $s\in I$, and equipped with the weak topology against $\ell^{\,1}(I)$, one obtains a scalarly measurable $x:I\to E$ which is not worth Bochner in $E$. Then $x$ belongs to a vector of $L_s^p(\mu,E)$ which does not contain any vector of $L^p(\mu,E)$.

**Remark 5.** If $E$ is a separable Banach space, then one has $L_s^{p^*}(\mu,E_\sigma^{\kern.4mm\prime})=L^{p^*}(\mu,E_\sigma^{\kern.4mm\prime})$. Thus even if the answer to Q is "yes", one can unify $(*)$ by taking there in both places the space $L^p$ instead of $L_s^p$. However, I do not know whether generally the space $L^p(\mu,E_1)$ is Banach when $E_1$ is a weakening of a Banach space. This untouched question of mine has its motivation in obtaining a counterexample to this.

**Remark 6.** The space $L^p(\mu,E)$ of Constructions 3 is precisely the classical Bochner space (considered *normable* instead *normed*) when $E$ is Banachable. To see this, one only needs to prove that $x\in N_0$ implies $x(t)=0_E$ for a.e. $t\in I$.

**Remark 7.** By this answer of Ramiro de la Vega, if we assume ZFC + MA$\,(\aleph_1)\,$, then $L_s^p(\mu,E_\sigma^{\kern.4mm\prime})\cong L^p(\mu,E_\sigma^{\kern.4mm\prime})$ for $E=c_0(I)$ although $E$ is not separable. To see this, it suffices to show that for every scalarly measurable $x:I\to E_\sigma^{\kern.4mm\prime}$ there is a countable set $J_0\subseteq I$ such that only for $s\in J_0$ we may have $(s)$ that $x(t)({\rm e\kern.5mm}s)=0$ does not hold for a.e. $t\in I$, noting that then by Pettis' theorem it follows that $x$ is worth Bochner. Here ${\rm e\kern.5mm}s$ is the "unit vector" at $s$. Indeed, if there were an uncountable $J_0$ with $(s)$ for all $s\in J_0$, by regularity of the Lebesgue measure, for every $s\in J_0$ we could find a compact set $K_s$ of positive measure such that $x(t)({\rm e\kern.5mm}s)\not=0$ for all $t\in K_s$. Then there would exist an uncountable $J_1\subseteq J_0$ and $t\in\bigcap\,\{\,K_s:s\in J_1\,\}$. However, this contradicts the fact that $x(t)$ is represented by an element of $\ell^{\kern.4mm 1}(I)$ and hence $x(t)({\rm e\kern.5mm}s)\not=0$ can hold only for $s$ in some countable set.

**Remark 8.** Still assuming ZFC + MA$\,(\aleph_1)\,$, the argument of Remark 7 similarly also gives $L_s^p(\mu,E)\cong L^p(\mu,E)$ when with $1<q<+\infty$ we take $E=\ell^{\kern.4mm q}(I)$ which is reflexive. Hence then the answer to the question of the title is "yes" and to Q it is "no" *provided* that $E$ is restricted to be one of these spaces.