Let $H$ be a separable infinite dimensional Hilbert space. Denote by $\mathcal{B}(H)$ the space of bounded operators on $H$, and $\mathcal{K}(H)$ the ideal of compact operators. When endowed with the strong (or weak) operator topology, is $\mathcal{K}(H)$ Borel in $\mathcal{B}(H)$?


Remarks: 


* It is well known that, unlike in the norm topology, $\mathcal{K}(H)$ is not closed; the identity operator is a strong limit of finite rank projections.
* $\mathcal{K}(H)$ is analytic: $\mathcal{K}(H)$ is Polish in the norm topology, and the inclusion map $\mathcal{K}(H)\to\mathcal{B}(H)$ is norm-weak continuous.



Is the ideal of compact operators strongly Borel?
Yes.


This can be deduced from the argument given for Corollary 3.2 in G. A. Edgar, [Measurability in a Banach space](http://dx.doi.org/10.1512/iumj.1977.26.26053), Indiana Univ. Math. J. 26 (1977), 663-677, [MR542944](http://www.ams.org/mathscinet-getitem?mr=542944).


The proof (attributed to Talagrand) is easy, so I reproduce it: 


Choose a countable norm-dense subset $\lbrace d\_k : k \in \mathbb{N}\rbrace \subseteq \mathcal{K}(H)$ and let $B$ be the unit ball in $\mathcal{B}(H)$. Then $B$ is compact in the weak operator topology and
$$
\begin{align\*}
\mathcal{K}(H) & = \bigcap\_{n \in \mathbb{N}} \left(\mathcal{K}(H) + \tfrac{1}{n}B\right) \supseteq \bigcap\_{n\in\mathbb{N}}\bigcup\_{k \in \mathbb{N}}\left(d\_k + \tfrac{1}{n}B\right) \supseteq \mathcal{K}(H)
\end{align\*}
$$
shows that $\mathcal{K}(H)$ is a $K\_{\sigma\delta}$ in $\mathcal{B}(H)$ with the weak operator topology.


Therefore $\mathcal{K}(H)$ is Borel with respect to all [the usual topologies](http://en.wikipedia.org/wiki/Topologies_on_the_set_of_operators_on_a_Hilbert_space) considered on $\mathcal{B}(H)$.


