Siu decomposition hello.
I have the following question:
Let $T$ be a positive closed current of bidimension $(p,p)$ then one has the Siu decomposition
$T=R+S$ where $R$ is a positive closed current such that its Lelong number is zero along analytic sets of dimension greater or equal $p$. Now my question is why this current $R$ is the bigger with this property?
$R$ is called the residual part while $S$ is the singualr one and $S$ is also positive.
thank you
 A: R is the biggest with such property since S is the sum of currents of integration on varieties of dimension p. Moreover, a dimension (p,p) current can not have positive Lelong number along varieties of dimension greater than p. 
A: I am not sure if I understand correctly your question, but anyway I'll try to answer: the point is that Siu's decomposition is in fact unique, but let me explain better.
Let $X$ be a complex manifold and $T$ a closed positive current of bidimension $(p,p)$. Then, there is a unique decomposition of $T$ as a (possibly ﬁnite) weakly convergent series
$$
T=\sum_{j\ge 1}\lambda_j[A_j]+R,\quad\lambda_j\ge 0,
$$ 
where $[A_j]$ is the current of integration over an irreducible $p$-dimensional analytic set 
$A_j\subset X$ and where $R$ is a closed positive current with the property that $\dim E_c(R)< p$ for every $c>0$. Here $E_c(R)$ is the set of point $x\in X$ such that the Lelong number $\nu(R,x)$ of $R$ at $x$ is greater than or equal to $c$.
To prove the uniqueness, assume that $T$ admits such a decomposition. Then, the $p$-dimensional components of $E_c(T)$ are $(A_j)_{\lambda_j\ge c}$, for $\nu(T,x)=\sum\lambda_j\nu([A_j],x)+\nu(R,x)$ is non zero only on $\bigcup A_j\cup\bigcup E_c(R)$, and is equal to $\lambda_j$ generically on $A_j$. In particular $A_j$ and $\lambda_j$ are unique.
Thus $R$ is neither the biggest nor the littlest... It is just unique!  
A: In my opinion, the point is not that $R$ is the biggest in whatever sense you may give to this, but that the decomposition $T=R+S$ with $S=\sum \lambda_j [A_j]$ ($A_j$ being a $p$-dimensional analytic set), and $R$ having zero Lelong number along any $p$-dimensional analytic set. 
The uniqueness is clear because for $x$ generic in $A_j$, $\nu(R+S,x)=\lambda_j \nu([A_j],x)=\lambda_j$, which determines thus uniquely $\lambda_j$, and therefore $S$ and $R$.
Now, if you really want to see that $R$ is the biggest current (in the sense of positivity of currents) such that $R$ has zero Lelong number along any $p$-dimensional analytic set, then you can proceed this way: assume that $T=S'+R'$ is another such decomposition. Then for $x$ generic in $A_j$, $\nu(T,A_j)=\nu(T,x)=\nu(S',x)=\nu(S',A_j)$. But it is a classical fact (see e.g Demailly, Complex analytic and differential Geometry, Proposition 8.16) that $S'-\nu(S', A_j) [A_j]$ is a closed positive current, so that $S' \geqslant S$, and therefore $R' \leqslant R$, which concludes.
