An alternate proof of Fuglede's theorem To prove Fuglede's Theorem for normal operators on a separable Hilbert space, why does it suffice to show that $E(S_1)T E(S_2)=0$ for all disjoint Borel sets $S_1$ and $S_2$, where $E$ is the spectral measure associated to the given normal operator $N$?
Further, why is this true for $S_1$ and $S_2$ at a positive distance from one another, and how can an approximation by compact sets be used to prove it in the general case?
 A: I'm not exactly sure where your questions are coming from (are you reading some proof?). But here's a way to prove the theorem using ideas related to those that you mention. 
The hypothesis is that $TN=NT$. By the Spectral Theorem,
$$
N=\int_{\sigma(N)}\,\lambda\,dE(\lambda)
$$
for a certain spectral measure $E$, and 
$$
N^*=\int_{\sigma(N)}\,\overline\lambda\,dE(\lambda).
$$
By definition, we have that $N$ is a norm-limit of operators of the form $\sum_j\lambda_jE(S_j)$ (i.e. "simple functions"), where $S_1,\ldots,S_n$ is a partition of $\sigma(N)$, and 
$$
E(S_j)=\int_{\sigma(N)}\,1_{S_j}(\lambda)\,dE(\lambda).
$$
This last integral, in turn, is a wot limit of integrals of polynomials. 
Now, since $TN=NT$, $TN^2=N^2T$, and $TN^k=N^kT$ for all $k\in\mathbb N$. So $Tp(N)=p(N)T$ for all polynomials $p$. Then $TE(S_j)=E(S_j)T$ (see the second edit) for Borel sets as above, and so $T$ commutes with all operators of the form $\sum_j\overline\lambda_j\,E(S_j)$. As $N^*$ is a norm limit of such operators, $T$ commutes with $N^*$. 
Edit: I'll leave the answer above since it is now mentioned in your question. Regarding the condition you care about, you can use it the following way: Let $S_1,\ldots,S_n$ be a partition of the spectrum of $N$. Then
$$
TE(S_j)=ITE(s_j)=\sum_kE(S_k)TE(S_j)=E(S_j)TE(s_j).
$$
So $TE(S_j)=E(S_j)TE(S_j)$. Doing the same process on the right side we get $E(S_j)T=E(S_j)TE(S_j)$. So $TE(S_j)=E(S_j)T$. That is, $T$ commutes with the spectral projections of $N$, and thus it commutes with $N^*$.
