This is the excercise 2.11 in Chapter IX of A Course in Functional Analysis by John B. Conway:
Suppose $H$ is a Hilbert space,$\,N$ is a bounded normal operator with associated spectral measure $E.$ Let $\sigma(N)$ be the spectrum of $N$ and $\sigma_p(N)$ be the set of all eigenvalues of $N.$
I need to prove that if $\sigma_p(N)$ is Borel measurable, then $N$ is diagonalizable iff $E(\sigma(N)\backslash\sigma_p(N))=0.$
This is easy when $\sigma_p(N)$ is at most countable since $$E(\sigma_p(N))=E(\cup_{\lambda\in\sigma_p(N)}\{\lambda\})=\Sigma_{\lambda\in\sigma_p(N)}E(\{\lambda\}).$$
But I have no idea to dealt with the case $\sigma_p(N)$ is not countable. Could someone help me?
If $N$ is diagonalizable, then $$ N=\sum_j\lambda_jE(\{\lambda_j\}), $$ where the index set might be uncountable. Let $K=\sigma(N)\setminus \sigma_p(N)$, which is measurable by hypothesis. By functional calculus, $$ E(K)=1_K(N)=\sum_j1_K(\lambda_j)\,E(\{\lambda_j\})=0. $$ Conversely, if again we write $K=\sigma(N)\setminus\sigma_p(N)$ and $E(K)=0$, we have $$ N=\int_{\sigma(N)}\lambda\,dE(\lambda)=\int_K+\int_{\sigma_p(N)}=\int_{\sigma_p(N)}\lambda\,dE(\lambda). $$ Enumerate $\sigma_p(N)$ as $\{\lambda_j\}$, counting multiplicities (so we allow repetitions but we get that each eigenspace is one-dimensional, and we choose eigenvectors for the same eigenvalue to be pairwise orthogonal). Since $$NE(\{\lambda_j\})=\lambda_jE(\{\lambda_j\})$$ (again by functional calculus), for a unit eigenvector $v_j$ with $Nv_j=\lambda_jv_j$ we have $E(\{\lambda_j\})v_j=v_j$, which shows that $E(\{\lambda_j\})\ne0$.
From normality we get that $\{v_j\}$ is orthonormal. Let $L$ be the orthogonal complement of $\{v_j\}$. Then $N|_L$ cannot have eigenvalues (all eigenspaces have already been accounted by the $\{v_j\}$) nor it can have non-eigenvalue elements of its spectrum because then $N$ would have them. So $L=0$, and $\{v_j\}$ is an orthonormal basis. Which means that $N$ is diagonalizable.
Edit: some more detail.
Note that $N^*v_j=\overline{\lambda_j}\,v_j$. This follows from the fact that if $T$ is normal then $$ \|Tx\|^2=\langle Tx,Tx\rangle=\langle T^*Tx,x\rangle=\langle TT^*x,x\rangle=\langle T^*x,T^*x\rangle=\|T^*x\|^2. $$
Because $N$ is normal, $L$ is invariant for $N$. Indeed, for any $x\in L$ we have $$ \langle Nx,v_j\rangle=\langle x,Nv_j\rangle=\lambda_j\langle x,v_j\rangle=0, $$ so $Nx\in L$. By the above we also have that $N^*x\in L$. This means that $L$ is reducing for $N$, and this can be expressed by $NP_L=P_LN$ where $P_L$ is the orthogonal projection onto $L$.
Let $\lambda\in\rho(N)$. That is, $N-\lambda I$ is invertible: there exists $S$ with $S(N-\lambda I)=(N-\lambda I)S=I$. Then we have $$ (P_LS|_L)(N|_L-\lambda I_L)=P_L\big(S(N-\lambda I)\big)|_L=P_LIP_L=P_L|_L=I_L. $$ Thus $\sigma(N|_L)\subset \sigma(N)$.