TITLE: Comparing two $\sigma$-algebras on $B(\ell^1)$
QUESTION [9 upvotes]: Let us consider  $B(\ell^1)$, bounded linear operators on $\ell^1$. We recall the weak operator topology, denoted by $w$, on  $B(\ell^1)$ is determined as follow 
$$w-\lim T_i=T \Longleftrightarrow \lim\langle (T_i-T)x,f\rangle=0$$
for every $x\in \ell^1 , f\in\ell^{\infty}$. 
Let $M$ be the Borel sigma algebra coming from the weak operator topology on $B(\ell^1)$ and let $M_0$ be the $\sigma$-algebra generated by the $w$-basic neighborhoods in $B(\ell^1)$. 
Q. Do we have $M=M_0$?
Let us give some information concerning this problem which may be helpful. It is proved by  Ju Myung Kim that the weak operator topology on the unit ball of $B(X)$ is relatively second countable if $X^*$ is separable. By an interesting argument in enter link description here, Matthew Daws showed the converse is also valid. So we have that: 
$X^*$ is separable if and only if  the weak operator topology on the unit ball of $B(X)$ is relatively second countable.
It is clear that, if $X^*$ is separable then $M_0=M$.  I feel the converse is not true and  $B(\ell^1)$ may serve a counterexample, however I have no clear proof.  
With taking account to the comments  mentioned in this problem, another approach  of  this question is: 
Is $(B(\ell^1),w)$  hereditary lindelof?

REPLY [2 votes]: Combination of the following two facts implies that $(B(\ell^1),w)$ is hereditary Lindelof.
1- $(B(\ell^1),sot)$ is hereditary Lindelof. 
2- (Kuratowski and Sierpinski [1921]) A regular space  $X$ is hereditary Lindelof if and only if  every uncountable set $A\subseteq X$ has a condensation point which is contained in $A$