TITLE: strong measurability question
QUESTION [9 upvotes]: Let $X$ be a separable Banach space and $\mathcal L$ the collection of bounded linear operators on $X$. The strong operator topology has the sub-basis $\{B_{x,y,\epsilon}\colon x,y\in X,\epsilon>0\}$, where $B_{x,y,\epsilon}=\{T\colon \|Tx-y\|<\epsilon\}$. The Borel $\sigma$-algebra generated by this topology is called the strongly measurable $\sigma$-algebra on $\mathcal L$.
Now the question (which has arisen in a study of the multiplicative ergodic theorem on Banach spaces):

Let $K=\{T\in\mathcal L\colon \text{ker}(T)\ne\{0\}\}$. Is this set strongly measurable?

Thanks for any information.

REPLY [6 votes]: This is true at least when $X$ is a separable and reflexive. Take a dense sequence $(x_n)_n$ in the unit sphere of $X$. Then, for any $T \in B(X)$, one has $\ker T \neq 0$ iff $\exists m$ $\forall k$ $\exists n$ such that $\| x_m - x_n \| < 1/2$ and $\| Tx_n \| < 1/k$. Indeed, since the closed unit ball of $X$ is weakly compact and any $T \in B(X)$ is weak-weak continuous, if the latter condition holds, then any weak limit point $x$ of the subsequence $(x_{n(k)})_k$ satisfies $\| x \| \geq \| x_m \| - \| x_m - x \| \geq 1/2$ and $Tx = 0$. This proves the "if" direction. The "only if" direction is easy.