Let $M$ be a von Neumann algebra and $a,b\in M$. Suppose $a\geq b$, then we have $R_a\geq R_b$, where $R_a$ is the range projection of $a$ and $R_b$ is the range projection of $b$.
According to the definition of $R_b$, $R_b$ is the smallest projection such that $R_b b=b$. If we prove that $R_ab=b$, then we can conclude that $R_a\geq R_b$.
My question is that how to deduce that $R_ab=b$?
Assuming that $a$ and $b$ are positive elements with $a\geq b$, then, for every $n\in {\mathbb N}$, we have that $$ c_n:= (1-a^{1/n})b^{1/2} \quad {\buildrel n\to 0 \over \longrightarrow} \quad 0, $$ because $$ c_nc_n^* = (1-a^{1/n})b(1-a^{1/n}) \leq (1-a^{1/n})a(1-a^{1/n}) \to 0. $$ Therefore $$ a^{1/n}b^{1/2} \to b^{1/2}. $$ Denoting by $\bar R(x)$ the closure of the range of an operator $x$, and observing that $\bar R(a^{1/n}) = \bar R(a)$, we deduce from the above that $$ \bar R(b^{1/2}) \subseteq \bar R(a), $$ but since $\bar R(b^{1/2}) = \bar R(b)$, we get $\bar R(b)\subseteq \bar R(a)$, whence the conclusion.