State and the commutation with the support Let $\varphi$ be a state on the von Neumann algebra $\mathfrak{M}$ with the support $s(\varphi)$ and $M\in \mathfrak{M}$ be positive operator such that $\varphi(MA)=0$ for all $A\in \mathfrak{M}.$

Question is $s(\varphi)M=0$?

I think on this because I know how to prove it for semifinite von Neumann algebras. I do not know if this is true in the general case.
 A: Write $P=s(\varphi)$. We have, since $\varphi$ and $M$ are positive, that $\varphi(AM)=0$ for all $A$. Then, via Cauchy-Schwarz,
$$
|\varphi(PM(I-P))|=|\varphi((MP)^*(I-P))|\leq\varphi(PM^2P)^{1/2}\varphi(I-P)^{1/2}=0,
$$
since $\varphi(I-P)=0$. If follows that $\varphi(PM(I-P)=0$, and so $$\varphi(PMP)=\varphi(PM)-\varphi(PM(I-P))=\varphi(PM)=0.$$ 
But $\varphi$ is faithful on $P\mathfrak MP$ (see below), so $PMP=0$. Then
$$
0=PMP=PM^{1/2}M^{1/2}P=(M^{1/2}P)M^{1/2}P,
$$
and $M^{1/2}P=0$. Multiplying by $M^{1/2}$ on the left, $MP=0$. 

Proof that $\varphi$ is faithful on $P\mathfrak MP$. I've had this fact in my mind for many years, but I couldn't find a reference on a quick check.
Assume that $\varphi(PTP)=0$, with $T$ positive. Let $\varepsilon>0$ and put $Q=1_{(\varepsilon,\|PTP\|]}(PTP)$. Then $\varepsilon Q\leq QT=QTQ$. Thus $$\varphi(Q)\leq\frac1\varepsilon\,\varphi(QTQ)=\frac1\varepsilon\,\varphi(T^{1/2}QT^{1/2})=\frac1\varepsilon\,\varphi(PT^{1/2}QT^{1/2}P)
\leq\frac1\varepsilon\,\varphi(PTP)=0. $$
It follows that $\varphi(Q)=0$, and so $Q\leq I-P$ (because $P$ is the support of $\varphi$). But, as $Q\leq P$, we get $Q=0$. AS $\varepsilon$ was arbitrary, we deduce that $PTP=0$. 
