Ultra weakly continuous linear functional on von neuman algebras

I started to learn von Neumann algebras and I wonder about the proof of the following. Can anyone outline it? this is theorem 4.2.10 in Murphy's book:

A linear functional $\tau$ on von Neuman algebra $A$ is ultra-weakly continuous iff there exist a trace class operator $v\in L^1(H)$ such that $\tau(a)=\text{tr}(av)$ for all $a\in A$.

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• I tried to clarify your question, but as stated you still need to specify what H is supposed to be. There are issues here, to do with the "standard form" of a von Neumann algebra. Perhaps it would help if you gave a quote of a precise statement you have seen somewhere and are trying to understand? – Yemon Choi Dec 20 '15 at 1:32
• this is theorem 4.2.10 in murphy's book and H is hilbert space – 89701407 Dec 20 '15 at 6:48
• A is our von neuman algebra and T is our linear functional which is ultra weakly continiuos on A – 89701407 Dec 20 '15 at 18:13

In Murphy's context, Theorem 4.2.9 shows that $A$ is the dual of $A_*$. This implies directly that $$a_j\xrightarrow{w^*} a \ \ \ \text { if and only if }\ \ \ \text{tr}(v(a_j-a))\to0 \text{ for all }v\in L^1(H).$$ Thus the weak$^*$ topology of $A$ agrees with the restriction of the $\sigma$-weak topology to $A$. Now, if $\tau$ is $\sigma$-weak continuous, then it is weak$^*$-continuous. By Theorems A.2 and 4.2.9, there exists $v\in L^1(H)$ such that $$\tau(a)=\theta(a)(v+A^\perp)=\text{tr}(av),\ \ \ \ a \in A.$$