Let $H$ be a separable Hilbert space and $L^{1}(H)$ be the space of trace class operators on $H$. 


Let $f:\mathbb{R}\to L^{1}(H)$ be a measurable function that is $t\to \operatorname{tr}(f(t))$ is Borel measurable. 



> 
> Q. Suppose that $\int \operatorname{tr}(f(t))d\mu$ is finite. Can we conclude that $\int \operatorname{tr}(|f(t)|)d\mu$ is finite too?
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Compatibility of the absolute value with the integration
What if $H$ is two dimensional, $\mu$ = Lebesgue measure and
$$
f(t) = \begin{bmatrix} 1 & 0\\0 &-1\end{bmatrix}
$$
so that $\mathrm{tr}(f(t))=0$ 


Then I guess
$$
|f(t)| = \begin{bmatrix} 1 & 0\\0 &1\end{bmatrix}
$$
so that $\mathrm{tr}\big(|f(t)|\big) = 2$ is not integrable.


