This question comes from Folland's A Course in Abstract Harmonic Analysis , 2ed. Let $(\Omega,M)$ be a measurable space, and $\mathcal{H}$ a Hilbert space over $\mathbb{C}$. We define a projection-valued measure on $(\Omega,M)$ to be a map $P:M\rightarrow \mathcal{L}(\mathcal{H})$ satisfying
$(i)$ $\forall E\in M$, $P(E)$ is an orthogonal projection (i.e. $P^2=P=P^\ast$)
$(ii)$ $P(\varnothing)=0$, $P(\Omega)=I$
$(iii)$ $P(E\cap F)=P(E)P(F)$
$(iv)$ $E_i$ disjoint implies $P(\cup_i E_i)=\sum_i P(E_i)$
Notice $\forall x,y\in\mathcal{H}, E\in M$; $P_{x,y}(E):=\langle P(E)x,y\rangle$ is a complex-valued measure on $\Omega$. If $\Omega$ is locally compact Hausdorff space, Regularity of $P$ is defined via regularity of $P_{x,y}$ for each $x,y$. We want to define an operator called $\int f dP$ for each $f\in B(\Omega)$, where $B(\Omega)$ are bounded Borel measurable functions on $\Omega$. By a polarization argument, we may establish
$$ \lvert \int f\hspace{0.1cm}dP_{x,y}\rvert\leq 4||f||_{sup}\cdot||x||\cdot||y|| $$
so that for each $x,y\in\mathcal{H}$, $f\mapsto \int f\hspace{0.1cm}dP_{x,y}\in\mathbb{C}$ is a bounded linear functional on $B(\Omega)$. Folland then claims there exists $T\in\mathcal{L}(\mathcal{H})$ so that $$\int f \hspace{0.2cm} dP_{x,y}=\langle Tx,y\rangle$$ for each $x,y$, and calls this $T:=\int f dP$. What duality result is being used here? I know from Riesz-Markov that if we started with an operator $T$, we may go backwards and identify a measure $\mu_{x,y}$ so that $$\int f \hspace{0.2cm} d\mu_{x,y}=\langle Tx,y\rangle$$ for each $x,y$, but I am not understaning how we can start with some complex measure and represent it as an inner product. This looks like Riesz representation in $\dim<\infty$ but I'm not sure why it holds here.
Theorem: If $b(x,y)$ is a sesquilinear form on $\mathcal{H}\times\mathcal{H}$ for a complex Hilbert space $\mathcal{H}$, and if $|b(x,y)| \le M\|x\|\|y\|$ for some constant $M$ and all $x,y\in\mathcal{H}$, then there is a unique $B\in\mathcal{L}(\mathcal{H})$ such that $b(x,y)=\langle Bx,y\rangle$. Furthermore, $\|B\|=\sup_{\|x\|,\|y\|\le 1}|b(x,y)| \le M$.
This is proved using the Riesz Representation Theorem, as you suspected. If you apply directly to $b(x,y)$, you end up with $b(x,y)=\langle x,Cy\rangle$ instead. So, instead consider $T_b(y)=\overline{b(x,y)}$, which is a linear functional in $y$ that has a unique representation $$ \overline{b(x,y)} = \langle y,Bx\rangle, $$ where $Bx$ is the unique representing vector that depends on $x$ only. Then, for a fixed $x\in\mathcal{H}$, $$ b(x,y) = \langle Bx,y\rangle,\;\;\; y\in\mathcal{H}. $$ By uniqueness of the Riesz vector $Bx$, you can show that $B$ is linear in $x$ because of the linearity of $b(x,y)$ in $x$. And $\|Bx\|\le M\|x\|$ holds for all $x$ because $$ \|Bx\|^2 = \langle Bx,Bx\rangle = b(x,Bx) = |b(x,Bx)| \le M\|x\|\|Bx\|. $$ So $B\in\mathcal{L}(\mathcal{H})$ is unique.