Decompose a positive contraction in a continuous masa in $L(H)$ Let $A$ be a continuous masa in $L(H)$ and $T$ be a positive contraction in $A$.
Then we can assume that $0＜\|Th\|＜1$ for all unit vectors $h\in H$. Otherwise decompose $T$ as $P+T_0$ for some projection $P\in A$ and consider $T_0$ in place of $T$.
Suppose there exists a unit vector $h_0\in H$ such that $\|Th_0\|=1$, how to construct the projection $P\in A$ and $T_0$ such that $T=P+T_0$?
If we have the above decomposition, we have $\|T_0h_0\|=\|(T-P)h_0\|$, how to check that $\|T_0h_0\|＜1$?
 A: The result works for any masa, not necessarily continuous.
If $0≤T≤1$ and $\|Th\|=\|h\|$, then $Th=h$. Indeed, you have
$$
0≤\|(1-T^2)^{1/2}h\|^2=\|h\|^2-\|Th\|^2=0.
$$
It follows that $(1-T^2)h=0$. This we can write as $$(1+T)(1-T)h=0.$$ As $1+T$ is invertible, this means that $(1-T)h=0$. That is, $Th=h$.
Now it becomes evident that $$L=\{h:\ \|Th\|=\|h\|\}=\{h:\ Th=h\}$$ is a subspace. Take $P$ to be the projection onto $L$. We have $P\in A$ because if $S\in A^{\rm sa}$ and $h\in L$ then $TSh=STh=Sh$, so $SL\subset L$ which means that $SP=PSP$; as the right-hand-side is selfadjoint, we get $PS=SP$, and since the selfadjoint elements span $A'$, we have shown that $P\in A'=A$.
We also have $TP=P$. Let $T_0=T(1-P)=T-P$. Then $T=P+T_0$. If $h\in H$ with $\|h\|=1$, then $(1-P)h\not\in L$, so $\|T(1-P)h\|<1$. That is, $\|T_0h\|<1$.
A: Martin Argerami's answer above is probably the one written in "The Book" but let me nevertheless give another proof, presenting
a different perspective, although it relies on a much more sophisticated result.
It is well known that every maximal abelian subalgebra of operators on a (henceforth assumed) separable Hilbert space is unitarily conjugate to $L^\infty(X)$
acting on $L^2(X)$, where $X$ is a measure space (this is the sophisticated result: Theorem 9.4.1 in Kadison & Ringrose).
We may then assume that $T$ is the multiplication operator by a measurable function $\varphi $, with $0\le \varphi (x)\le 1$.
We then let $$Y=\{x\in X: \varphi (x)=1\},$$ and we claim that if $h$ lies in  $L^2(X\setminus Y)$, with $\|h\|\leq 1$,
then $\|\varphi h\|<1$.  Indeed, supposing by contradiction that $\|\varphi h\|=1$, we have that
$$
  0=1-\int_X|\varphi (x)h(x)|^2\, dx \ge \int_X|h(x)|^2\, dx - \int_X|\varphi (x)h(x)|^2\, dx = $$$$ =
  \int_{X}\big (1-\varphi (x)^2\big )|h(x)|^2\, dx \geq  0.
  $$
Consequently this integral vanishes and, since the integrand is nonegative, we must have that
$$
  (1-\varphi (x)^2\big )|h(x)|^2=0,
  $$
for almost all $x$ in $X$.  This shows that $\varphi (x)=1$, for almost all $x$ in the set of points where  $h$ is nonzero, hence $\varphi (x)=1$
on a set of
positive measure disjoint from $Y$, a contradiction.
Considering the projection $P$ on $L^2(Y)$, notice that clearly $TP=P$,  and we then let  $T_0=T-P$.
Clearly both $P$ and $T_0$ lie in
$L^\infty(X)$, and  for every $g$ in $L^2(X)$, with $\|g\|\leq 1$,  we have that $g':=g-Pg$ lies in $L^2(X\setminus Y)$ and
$$
  \|T_0(g)\| = \|T(g) - P(g)\| = \|T(I-P)(g)\| = \|T(g')\| < 1.
  $$
