decomposition of measure Could anyone tell me how to prove the following?
Let $K$ denotes the set of all of the non-negative finite measures on some polish space $X$, let $P$ be a map on $K=K_1\oplus K_2$ which is linear and $K_1$ is invariant under $P$, $P\mu (X)=\mu(X)\quad \forall \mu\in K$. Then if $P\mu=\mu$ where $\mu=\mu_1+\mu_2$ where $\mu_i\in K_i, i=1,2$ then $P\mu_i=\mu_i$. 
 A: Here is an approximation to a proof.  Approximation because I don't know the context of where you are getting this stuff, and have to guess at what $\oplus$ means, and so on.
I assume $K=K_1\oplus K_2$ means for each $\mu\in K$ there exist unique $\mu_i\in K_i$ such that $\mu=\mu_1+\mu_2$.  We are told that $K_1$ is $P$-invariant, which I suppose means $P(\mu_1)\in K_1$ whenever $\mu_1\in K_1$.  And that $(P\mu)(X)=\mu(X)$ for all $\mu\in K$.
We are given $\mu=P\mu$.  Decompose $\mu=\mu_1+\mu_2$;   by linearity, $P\mu= P\mu_1+P\mu_2$.  We know  $P\mu_1\in K_1$.  Decompose $P\mu_2=\nu_1+\nu_2$, where $\nu_i\in K_i$.  So we have the decomposition $P\mu=(P\mu_1+\nu_1) + \nu_2$, where $P\mu_1+\nu_1\in K_1$ and $\nu_2\in K_2$.  By uniqueness of decomposition and by $P\mu=\mu$, we have $P\mu_1+\nu_1=\mu_1$.
But we also know that $(P\mu_1+\nu_1)(X)=(P\mu_1)(X)+\nu_1(X)=\mu_1(X)+\nu_1(X)$.
Since $(P\mu_1)(X)=\mu_1(X)$ we have $\nu_1(X)=0$, that is, $\nu_1=0$.
So $P\mu_1=\mu_1$, and finally by linearity, $P\mu_2=\mu_2$.
Is there some paper you are reading through, with puzzling passages?  It might help if you gave us a reference.
