Let $0=t_0\le t_1\le...\le t_{n}=1$ and suppose that $\Delta=\max_{i\in\{1,...,n\}}(t_{i}-t_{i-1})\to 0$ as $n\to \infty$. Let $E_{i}=[t_{i-1}, t_i)$ and for each $i=1,...,n$ fix $\xi_{i}\in E_i$. Let us now define
$$f_{n}(t)=\sum_{i=1}^{n}f(\xi_{i})\chi_{E_{i}}(t)$$
for all $t\in [0,1]$ and $n\in \mathbb{N}$. Fix $t\in [0,1]$, $\varepsilon>0$ and observe that
$$\|f(t)-f_{n}(t)\| \le \sum_{i=1}^{n}\|f(t)-f(\xi_{i})\|\chi_{E_{i}}(t)=\|f(t)-f(\xi_{i(t)})\|<\varepsilon$$
for sufficiently large $n$ which is a consequance of uniform continuity of $f$ on compact set $[0,1]$. This imples that
$$\lim_{n\to\infty}\|f(t)-f_{n}(t)\|=0$$
for all $t\in [0,1]$.
Let $X$ be a Banach space and $(\Omega,\Sigma,\mu)$ be a measure space. By $L_0(\Omega,\mu, X)$ we denote the linear space of $X$-valued measurable functions.
1) This is wrong counterexample, because anotherr definiton of measurability should ne used. If $(\Omega,\Sigma,\mu)$ is not complete the result is not true. Indeed the measure is not complete we have $E\in\Sigma$ and $F\subset E$ such that $F\notin \Sigma$ and $\mu(E)=0$. Let $f_n(\omega)=0$ for all $\omega\in\Omega$ and $n\in\mathbb{N}$, then $f_n\to\chi_F$ a.e, $\{f_n:n\in\mathbb{N}\}\subset L_0(\Omega,\mu,\mathbb{C})$ but $\chi_F\notin L_0(\Omega,\mu,\mathbb{C})$.
2) If $(\Omega,\Sigma,\mu)$ is complete we use Pettis measurability theorem. By assumption we have $E\in\Sigma$ such that $f(\omega)=\lim\limits_{n\to\infty} f_n(\omega)$ for all $\omega\in E$ and $\mu(\Omega\setminus E)=0$. Take arbitrary $x^*\in X^*$, then for all $\omega\in E$ we have $x^*(f(\omega))=\lim\limits_{n\to\infty} x^*(f_n(\omega))$. Since $f_n\in L_0(\Omega,\mu,X)$, then by Pettis measrability theorem $x^*\circ f_n$ is scalar valued measurable function. Since $x^*\circ f$ is limit on $E$ of measurable scalar valued functions $x^*\circ f_n$, then $x^*\circ f$ is scalar valued measurable on $E$ function. Since $\mu(\Omega\setminus E)=0$ and measure is complete then $x^*\circ f$ is scalar valued measurable on $\Omega$. Since $x^*$ is arbitrary $f$ is weakly $\mu$-measurable.
Since $f_n\in L_0(\Omega,\mu,X)$, then by Pettis measurability theorem $f_n$ is separably valued, i.e. there exist countable $S_n\subset X$, $E_n\in \Sigma$ such that $f_n(E_n)\subset \operatorname{cl}(S_n)$ and $\mu(\Omega\setminus E_n)=0$. Define $E_0=(\bigcap E_n)\cap E\in \Sigma$, then $\mu(\Omega\setminus E_0)=0$. Consider countable set $S=\bigcup_{n\in\mathbb{N}} S_n$, then
$f_n(E_0)\subset \operatorname{cl}(S_n)\subset \operatorname{cl}(S)$ and $\bigcup_{n\in\mathbb{N}} f_n(E_0)\subset \operatorname{cl}(S)$. Recall $f(\omega)=\lim\limits_{n\to\infty} f_n(\omega)$ for all $\omega\in E$. Since $E_0\subset E$ we conclude that each point in $f(E_0)$ is limit of some sequence in $\bigcup_{n\in\mathbb{N}} f_n(E_0)$. In other words $f(E_0)\subset \operatorname{cl}(\bigcup_{n\in\mathbb{N}} f_n(E_0))$. Now we have $f(E_0)\subset \operatorname{cl}(\bigcup_{n\in\mathbb{N}}f_n(E_0))\subset$$\subset \operatorname{cl}(\operatorname{cl}(S))=\operatorname{cl}(S)$. Since $S$ is countable, $f(E_0)$ is separable. Since $\mu(\Omega\setminus E_0)=0$, then $f$ is separably valued.
Since $f$ is weakly measurable and separably valued by Pettis measurability theorem it is $\mu$ measurable, i.e. $f\in L_0(\Omega,\mu,X)$.
Best Answer
Actually, there is a small "issue" with the definition of strong measurability for non-complete measure spaces.
To see it take the Banach space to be $\Bbb R$. Let $(\Omega, \Sigma, \mu)$ to be a measure space that is not complete.
Then using the definition of strong measurability as "$f$ strongly measurable means that there exists a sequence of simple functions $(f_n)$ converging almost everwhere to $f$" will result in having non-measurable functions that will be "strong measurable".
However, given any such strong measurable function $h$ that is non-measurable, there is a measurable function $k$ such that $k=h$ a.e. (and of course, $k$ is also strong measurable).
Remark In fact, strong measurability is not defined for individual functions $f$ but for the class $[f]$ (class of functions $h$ such that $h=f$ a.e.).
It is similar to what happens with $L^p$. We say that a function is in $L^p$, but to be precise the elements of $L^p$ are equivalence classes of functions by relation $=$ a.e.. For a function $f$, we may write $f \in L^p$, to actually mean $[f] \in L^p$.