Let $H$ be a $\mathbb R$-Hilbert space and $A$ be a bounded linear self-adjoint operator on $H$. Let $\mathcal M_b(\sigma(A))$ denote the $\mathbb R$-vector space of bounded Borel measurable functions $\sigma(A)\to\mathbb R$. We know that there is a unique continuous algebra homomorphism $\Phi:\mathcal M_b(\sigma(A))\to\mathfrak L(H)$ with $\Phi(1)=\operatorname{id}_H$ and $\Phi\left(\operatorname{id}_{\sigma(A)}\right)=A$. The spectral measure on $\mathcal B(\mathbb R)$ associated with $A$ is given by $$E:\mathcal B(\mathbb R)\to\mathfrak L(H)\;,\;\;\;B\mapsto\Phi\left(1_{\sigma(A)\:\cap\:B}\right).$$
Now assume $H=\mathbb R^d$, $d\in\mathbb N$ and let $\lambda_1,\ldots,\lambda_k$ denote the pairwise distinct eigenvalues of $A$, $k\in\{1,\ldots,d\}$. We know that $$A=\sum_{i=1}^k\lambda_i\pi_i,\tag1$$ where $\pi_i$ denotes the orthogonal projection from $\mathbb R^d$ onto $\mathcal N(\lambda_i-A)$ for $i\in\{1,\ldots,k\}$. Let $B\in\mathcal B(\mathbb R)$. Since $\sigma(A)=\{\lambda_1,\ldots,\lambda_k\}$, $$E(B)=\sum_{i=1}^k\Phi\left(1_{\{\lambda_i\}\:\cap\:B}\right)\tag2.$$
While it intuitively should hold, I can't figure out how we see that $$E(B)=\sum_{\substack{1\le i\le k\\\lambda_i\in B}}\pi_i.\tag3$$
Maybe we need to note that, by $(1)$, $A$ is the sum of bouned linear self-adjoint operators with closed range, but I don't see how the defining properties of $\Phi$ yied $(3)$.