Bochner Integral of trace class operator-valued function Suppose I have a separable Hilbert space $\mathcal{H}$, a measure space $(M,\mu)$ and a strongly measurable operator-valued function $f:M \rightarrow \mathcal{S_{1}}(\mathcal{H})$ whose values are trace class operators on this Hilbert space. Suppose further that its Bochner integral $\int_{M}f(s)d\mu(s)$ exists. 
1) Is it also a trace class operator, maybe with additional data? 
2) Does something like $\|\int_{M}f(s)d\mu(s)\|_{\mathcal{S_{1}}}\leq \int_{M}\|f(s)\|_{\mathcal{S_{1}}}d\mu(s)$ (where $\| \cdot \|_{\mathcal{S_{1}}}$ is the trace class norm) hold, maybe with additional data? 
3) If these questions both have a positive answer, do their appropriate modifications also have a positive answer for other $\mathcal{S_{p}}$ (Schatten p-class) operator-valued functions?
I know the appropriate modification of the first question has a positive answer for compact operator-valued functions and various papers point out to both the first two questions having a positive answer (e.g. page 13 of https://arxiv.org/pdf/1402.0763.pdf). I also suspect the third should be an easy generalisation.
I think this should be easy but can't seem to approach it well. I would be very grateful for every hint.
 A: The properties listed on the wiki page for Bochner integration are useful here.
The answer to 1 is yes.  In particular: $Tr$ is a continuous linear operator with respect to the trace-norm, which means that for any trace class operator $f$, we have
$$
Tr\left[\int f \,d\mu\right] = \int Tr(f)\,d\mu
$$
The tricky bit here is to say that
$$
\left| \int f \,d\mu\right| \overset{!}{\leq} \int |f| \, d\mu
$$
Which is to say that $ \int |f| \, d\mu - \left| \int f \,d\mu\right|$ is a positive operator.  In particular, we need to show that the map $f \mapsto f^*f$ is convex, then apply Jensen's inequality.
With that, we would have
$$
\left\| \int f \,d\mu\right\| = 
Tr\left| \int f \,d\mu\right| \overset{!}{\leq} Tr\int |f| \, d\mu = 
\int Tr |f|\,d\mu = \int \|f\|\,d\mu
$$
The answer to 2 is yes; no additional data required.
The answer to 3 is also yes.

With regards to the inequality: note that $f \leq g$ if and only if for every $x \in \mathcal H$, we have $(x,fx) \leq (x,gx)$.  Now, note by linearity that
$$
\left( x,\left[\int f \,du\right]x
\right) = \int (x,f(x))\,d\mu
$$
Thus: for convexity, we note that for any fixed $x$, operators $f_1,f_2$, and $t \in [0,1]$ we have
$$
(x,[(1-t)f_1 + tf_2]^*[(1-t)f_1 + tf_2] x) = \\
\|[(1-t)f_1 + tf_2]x\|^2 = \\
\|(1-t)f_1x + tf_2x\|^2 \leq\\
((1-t)\|f_1x\| + t\|f_2x\|)^2 \leq\\
(1-t)\|f_1x\|^2 + t\|f_2x\|^2 = \\
(1-t)(x,f_1^*f_1 x) + t(x,f_2^*f_2 x)
$$
since this holds for all $x$, we have 
$$
[(1-t)f_1 + tf_2]^*[(1-t)f_1 + tf_2] \leq 
(1-t)f_1^*f_1 + t f_2^*f_2
$$
as desired.
A: I realised that this has a very simple answer and I will share it here for future reference, hoping it is correct. By the Bochner Integrability theorem, a strongly measurable function $f:M→\mathcal{S}_{p}(\mathcal{H})$ that satisfies $ \int_{M}\|f(s)\|_{\mathcal{S_{p}}(\mathcal{H})}d\mu(s) < \infty$ converges by definition in the appropriate norm, and furthermore satisfies 2) above. So if this absolute integrability condition is satisfied, the integral is in $\mathcal{S_{p}}(\mathcal{H})$. Fortunately, for the concrete case I want to use this, this holds, so it solves my question.
