Given a map from $\big([0,1], \mathcal{B}[0,1], m\big)$ to a Banach space $(X, \|\cdot \|)$. There are strong measurable functions (they are the point wise a.e. limit of simple functions) and weak measurable functions (for each $u^* \in X^*$, we have $t\mapsto \langle u^*, f(t)\rangle$ is a measurable as a function from $[0,1] \rightarrow \mathbb{R}$).

I have two questions:

1. Why don't we use the normal measurable definition here? That is a function is measurable if the pre-image of all Borel sets $U\in \mathcal{B}(X)$ are in $\mathcal{B}[0,1]$.
2. Now suppose $X= L^1(\Omega, \mathcal{F}, \mu)$, given a measurable $f:[0,1]\rightarrow L^1$ (either strong, weak or the normal definition), can we say that the function $f(t,\omega)$ is measurable as a function from $[0,1] \times \Omega \rightarrow \mathbb{R}$ with respect to the product sigma on the domain?

Here, assume elements of $L^1$ are just measurable functions, not equivalent classes, so that $f(t,x)$ is well defined.

Thank you for your time.

Given a map from $\big([0,1], \mathcal{B}[0,1], m\big)$ to a Banach space $(X, \|\cdot \|)$. There are strong measurable functions (they are the point wise a.e. limit of simple functions) and weak measurable functions (for each $u^* \in X^*$, we have $t\mapsto \langle u^*, f(t)\rangle$ is a measurable as a function from $[0,1] \rightarrow \mathbb{R}$).

I have two questions:

1. Why don't we use the normal measurable definition here? That is a function is measurable if the pre-image of all Borel sets $U\in \mathcal{B}(X)$ are in $\mathcal{B}[0,1]$.
2. Now suppose $X= L^1(\Omega, \mathcal{F}, \mu)$, given a measurable $f:[0,1]\rightarrow L^1$ (either strong, weak or the normal definition), can we say that the function $f(t,\omega)$ is measurable as a function from $[0,1] \times \Omega \rightarrow \mathbb{R}$ with respect to the product sigma algebra on the domain?

Here, assume elements of $L^1$ are just measurable functions, not equivalent classes, so that $f(t,x)$ is well defined.

Thank you for your time.