About measurability of operators I'm triyng without success, to find some examples of functions that:
$\bullet$Are WOT-measurable, but not SOT-measurable. 
$\bullet$Are SOT-measurable, but not $||\cdot||$-measurable.

I give the definitions I'm dealing with. We have:
$X=\cal{L}$($E_1$,$E_2$)
$f:\Omega\to X$$\;,\;\;$$(X,\Sigma,\mu)$ measure space.
$\bullet\; ||\cdot||_{X}$-measurable: $\exists s_n:\Omega\to X$, simple, and $\exists A\in\Sigma,\;\mu(A)=0$, with $||s_n(w)-f(w)||_{X}\xrightarrow[n\to\infty]{}0\;\;\forall w\notin A.$
$\bullet\;$ SOT-measurable: $\forall e_1\in E_1$, the function $w\to f(w)(e_1)\;\;$ ($\Omega\to E_2$) is $||\cdot||_{E_2}$-measurable.
$\bullet\;$ WOT-measurable: $\forall e_1\in E_1$, and $\forall e_2^{}*\in E_2^{*}$, the function $w\to <f(w)(e_1),e_2^{*}>\;\;$ ($\Omega\to \mathbb{K}$) is measurable (in the usual sense).
Thanks a lot for any help finding those examples.
 A: If $E_2$ is separable, then WOT-measurable is the same as SOT-measurable. Indeed, if $f:\Omega\to \mathcal L(E_1,E_2)$ is WOT-measurable then, for each $e_1\in E_1$, the map $f_{e_1}:\omega\mapsto f(\omega)e_1$ is weakly measurable from $\Omega$ into $E_2$. Since $E_2$ is separable, this implies that $f_{e_1}$ is strongly measurable for every $e_1\in E_1$ (this is ``Pettis' measurability theorem"). In other words, $f$ is SOT-measurable.
To show that SOT-measurable does not imply $\Vert\,\cdot\,\Vert$-measurable, note first that if $f$ is $\Vert\,\cdot\,\Vert$-measurable, then its range is ``essentially separable", i.e. there exists a set $B$ with $\mu(\Omega\setminus B)=0$ such that $f(B)$ is separable. So it is enough to produce an SOT-measurable map $f$ whose range is not essentially separable.
Here is one example. Take $E_1=\mathcal C([0,1])$, the space of all continuous functions on $[0,1]$ and $E_2=\mathbb R$. So $\mathcal L(E_1,E_2)=M([0,1])$, the space of all real Borel measures on $[0,1]$. Define $f:[0,1]\to M([0,1])$ by $f(\omega)=\delta_\omega$, the Dirac point mass at $\omega$. (The measure $\mu$ on $[0,1]$ is Lebesgue measure). Note that if $\Vert f(\omega)-f(\omega')\Vert=2$ whenever $\omega\neq \omega'$. It follows that for any uncountable set $B\subset [0,1]$, the set $f(B)$ cannot be separable; hence the range of $f$ is not essentially separable. On the other hand, $f$ is SOT-measurable because for any $e_1\in E_1=\mathcal C([0,1])$, we have $f(\omega)e_1=e_1(\omega)$ and hence the map $\omega\mapsto f(\omega)e_1$ is even continuous.
Here is another example. Take $E_1=E_2=L^2([0,1]):=H$ and define $f:[0,1]\to \mathcal L(H)$ as follows: for any $\omega\in [0,1]$, the operator $f(\omega)$ is the multiplication operator by the function $\mathbf 1_{[0,\omega]}$. Then $f$ is SOT-measurable, but $\Vert f(\omega)-f(\omega')\Vert=1$ whenever $\omega\neq\omega'$, so the range of $f$ is not essentially separable.
Finally, here is an example of a WOT-measurable but not SOT-measurable. Take $E_1=\mathbb R$ and $E_2=\ell^2([0,1])$, i.e. a nonseparable Hilbert space having an orthonormal basis $(e_\omega)_{\omega\in [0,1]}$ of cardinality $\mathfrak c$. Then $\mathcal L(E_1,E_2)=E_2=\ell^2([0,1])$. Consider the map $f:[0,1]\to \ell^2([0,1])$ defined by $f(\omega)=e_\omega$. This map is not SOT-measurable because its range is not essentially separable (and SOT-measurable just means strongly measurable in the present case). On the other hand, $f$ is WOT-measurable because for any $u\in \ell^2([0,1])$, we have $\langle u,f(\omega)\rangle=0$ except for a countable set of $\omega$'s.
