Let $H$ be a separable, infinite-dimensional Hilbert space and let $\mathbb{B}(H)$ be the algebra of bounded operators on $H$. The norm topolology on $\mathbb{B}(H)$ is stricly finer, hence the identity map from the strong to the norm topology is not continuous.

However, it is Borel measurable. To see this, let $v_1, v_2, \dots$ be a countable dense subset of the unit sphere of $H$. Then $$\{A \in \mathbb{B}(H) \mid \|A\|<1\} = \bigcap_{n \in \mathbb{N}} \{A \in \mathbb{B}(H) \mid \|Av_n\| < 1\}, $$ hence the norm unit ball is strongly measurable. The same is true for scalings and translates of the unit ball. Since these sets generate the norm topology, norm-open sets are strongly measurable. In total, we get that the Borel $\sigma$-algebra for the norm topology coincides with the $\sigma$-algebra for the strong topology.

Q: Is a similar statement true for the $\sigma$-algebra of sets with the Baire property?

Here a subset $S$ of a topological space has the Baire property if there exists an open set $U \subseteq X$ such that $S \Delta U$ is meagre; these sets form a $\sigma$-algebra.

So I wonder: How to meagre sets in the strong, respectively, norm topology coincide?

Let $H$ be a separable, infinite-dimensional Hilbert space and let $\mathbb{B}(H)$ be the algebra of bounded operators on $H$. The norm topolology on $\mathbb{B}(H)$ is stricly finer, hence the identity map from the strong to the norm topology is not continuous.

However, it is Borel measurable [This claim is false; see the comments. This relates to the fact that $\mathbb{B}(H)$ is not separable with respect to the norm topology]. To see this, let $v_1, v_2, \dots$ be a countable dense subset of the unit sphere of $H$. Then $$\{A \in \mathbb{B}(H) \mid \|A\|<1\} = \bigcap_{n \in \mathbb{N}} \{A \in \mathbb{B}(H) \mid \|Av_n\| < 1\}, $$ hence the norm unit ball is strongly measurable. The same is true for scalings and translates of the unit ball. Since these sets generate the norm topology, norm-open sets are strongly measurable. In total, we get that the Borel $\sigma$-algebra for the norm topology coincides with the $\sigma$-algebra for the strong topology.

Q: Is a similar statement true for the $\sigma$-algebra of sets with the Baire property?

Here a subset $S$ of a topological space has the Baire property if there exists an open set $U \subseteq X$ such that $S \Delta U$ is meagre; these sets form a $\sigma$-algebra.

So I wonder: Do meagre sets in the strong, respectively, norm topology coincide?