a point on trace class operators Assume $H$ is separable Hilbert space and fix an orthonormal basis $\{e_n\}_1^{\infty}$. Let us denote $p_n$ by the projection onto the subspace generated by 
$\{e_1\cdots,e_n\}$. 
Let $a$ be a positive trace class operator on $H$. I do not know the following is correct or not! 
$$||a-p_nap_n||_1=\text{Tr}(|a-p_nap_n|)\to0$$

Yes. One can prove $\|a-ap_n\|_1\to0$ (see below), and then
$$
\|a-p_nap_n\|_1\leq\|a-ap_n\|_1+\|ap_n-p_nap_n\|_1\leq\|a-ap_n\|_1+\|a-p_na\|_1
$$
(using that $\|b^*\|_1=\|b\|_1$ and that $\|ab\|_1\leq\|a\|_1\,\|b\|$). 
$$
\ 
$$
Assume first that $a$ is a rank-one projection. Then $a=\langle\cdot,x\rangle\,x$ for some $x\in H$ with $\|x\|=1$. We have, using that $\|bc\|_1=\|cb\|_1$,
\begin{align}
\|a-ap_n\|_1&=\|a(1-p_n)\|_1=\|a^2(1-p_n)\|_1=\|a(1-p_n)a\|_1=\text{Tr}(a(1-p_n)a)\\ \ \\
&=\text{Tr}(\langle(1-p_n)x,x\rangle\,a)=\langle (1-p_n)x,x\rangle=\sum_{j>n}|x_j|^2\to0.
\end{align}
Now we immediately get $\|a-ap_n\|_1\to0$ when $a$ is a linear combination of rank-one projections. For a selfadjoint trace class operators, we have $a=\sum_j\alpha_j q_j$, where the $q_j$ are pairwise orthogonal rank-one projections. Then
\begin{align}
\|a-ap_n\|_1&\leq\|a-\sum_{j=1}^n\alpha_jq_j\|_1+\|\sum_{j=1}^n\alpha_jq_j(1-p_n)\|_1+\|(a-\sum_{j=1}^n\alpha_jq_j)p_n\|_1\\ \ \\
&\leq2\|a-\sum_{j=1}^n\alpha_jq_j\|_1+\|\sum_{j=1}^n\alpha_jq_j(1-p_n)\|_1\to0
\end{align}
(the first term, because $a$ is trace-class, and the second one by the previous paragraph). 
Finally, an arbitrary trace-class operator $a$ can be written as $a=b+i\,c$, where $b,c$ are selfadjoint trace-class operators. 
