Equivalent definition of range projection of positive elements in a von Neumann algebra Let $M$ be a von Nuemann algebra and $A$ be a positive element in $A$. There are two different ways of defining the range projection of $A$.
(1) the range projection $R_A$ of $A$ is  $\chi_A(0,\|A\|]$, where $\chi_A$ is the spectral measure of $A$.
(2) Consider the polar decomposition of $A$, we have $A=V|A|$, then $VV^*$ is the renge projection of $A$.
How to prove that the two definitions are equivalent?
 A: Things are fairly simple because $A$ is positive, and so one can show that $V$ is already a projection. But with basically the same effort one gets the result for $A$ normal.
We need to  have $R_A=1_{(0,\|A\|]}$ (the interval has to be closed on $\|A\|$ as otherwise you might be excluding part or all of the range of $A$).
There are three projections involved, which we want to show they are the same.

*

*The range projection $R$ of $A$. This is the smallest orthogonal projection $R$ such that $RA=A$.


*The projection $P=1_{\sigma(A)\setminus\{0\}}(A)$.


*The projection $Q=VV^*$, where $A=V|A|$ is the minimal polar decomposition of $A$.
We have
$$
PA=A,\qquad\qquad QA=A\qquad\qquad RA=A. 
$$
The first one, because $t\,1_{\sigma(A)\setminus\{0\}}(t)=t$ on $\sigma(A)$. The second one, because $QA=VV^*A=V|A|=A$. And the third one by definition of $R$. From $QA=A$ we get $Qf(A)=f(A)$ for every continuous function $f$, and then for every Borel $f$; thus $QP=P$, which gives us $P\leq Q$. So
$$
R\leq P\leq Q. 
$$
Now, from $RA=A$ we get $RV|A|=V|A|$. So, on the range of $|A|$, $RV=V$. The range projection of $|A|$ is $V^*V$. Thus $RVV^*V=VV^*V$. Multiplying on the right by $V^*$ we get $RQ=Q$. Thus $Q\leq R$.
Then
$$
R=Q=P. 
$$
