[proofplan] We construct $R$ and $U$ as Borel functions of $T$ via the [Borel Functional Calculus](/theorems/2696). Set $K = \sigma(T) \subseteq \mathbb{C}$ and let $r(\lambda) = |\lambda|$ and $u(\lambda) = \lambda/|\lambda|$ (with $u(0) := 1$); then $R := r(T)$ is positive because $r$ is real and non-negative, and $U := u(T)$ is unitary because $|u| \equiv 1$ on $K$. The factorisation $T = RU$ comes from the pointwise identity $r u = \mathrm{id}$ on $K$, and pairwise commutation follows from the fact that the BFC is a commutative algebra homomorphism out of $L^\infty(K)$. [/proofplan] [step:Set up the spectrum and the two Borel functions $r$ and $u$] Let $K = \sigma(T) \subseteq \mathbb{C}$. By the [Spectrum is Compact](/theorems/2649) (and indeed the more refined statement that the spectrum of a Hilbert-space operator is a non-empty compact subset of $\mathbb{C}$), $K$ is a non-empty compact subset of $\mathbb{C}$, hence a Polish space, and Borel measurable functions $K \to \mathbb{C}$ with values in a bounded set lie in $L^\infty(K, \mathcal{B}(K))$. Define \begin{align*} r: K &\to [0, \infty) \\ \lambda &\mapsto |\lambda|, \end{align*} and \begin{align*} u: K &\to \mathbb{C} \\ \lambda &\mapsto \begin{cases} \lambda / |\lambda| & \text{if } \lambda \ne 0, \\ 1 & \text{if } \lambda = 0. \end{cases} \end{align*} *$r$ is bounded and Borel.* The map $\lambda \mapsto |\lambda|$ is continuous on $\mathbb{C}$, hence Borel. Since $K$ is compact, $r$ is bounded with $\|r\|_{L^\infty(K)} = \max_{\lambda \in K} |\lambda| = \rho(T)$ (the spectral radius). *$u$ is bounded and Borel.* On $K \setminus \{0\}$, $u(\lambda) = \lambda/|\lambda|$ is continuous (the denominator is bounded away from zero on each set $\{|\lambda| \ge \varepsilon\}$); on the single point $\{0\}$ (if $0 \in K$), $u$ takes the value $1$. Both $K \setminus \{0\}$ and $\{0\}$ are Borel sets, and $u$ is continuous (hence Borel) on each. Thus $u$ is Borel measurable on $K$. Since $|u(\lambda)| = 1$ for all $\lambda \in K$, $\|u\|_{L^\infty(K)} = 1$. Hence $r, u \in L^\infty(K, \mathcal{B}(K))$. [/step] [step:Define $R := r(T)$ and $U := u(T)$ via the Borel Functional Calculus] By the [Borel Functional Calculus](/theorems/2696), the map \begin{align*} \Psi_T: L^\infty(K, \mathcal{B}(K)) &\to \mathcal{L}(H) \\ f &\mapsto f(T) \end{align*} is a unital $*$-homomorphism. Set \begin{align*} R := \Psi_T(r) = r(T), \qquad U := \Psi_T(u) = u(T). \end{align*} Both $R$ and $U$ are bounded operators on $H$, with operator norms bounded by $\|r\|_{L^\infty(K)} = \rho(T)$ and $\|u\|_{L^\infty(K)} = 1$ respectively, by property (ii) of the BFC. [/step] [step:Verify that $R$ is positive] We show $R = R^*$ and $\sigma(R) \subseteq [0, \infty)$, which is the operator-theoretic definition of a positive operator (see [Positive Element](/theorems/2654) for the $C^*$-algebra characterisation that coincides on $\mathcal{L}(H)$). *Self-adjointness.* Since $r$ is real-valued, $\bar{r} = r$ pointwise on $K$. As $\Psi_T$ is a $*$-homomorphism, \begin{align*} R^* = \Psi_T(r)^* = \Psi_T(\bar{r}) = \Psi_T(r) = R. \end{align*} *Spectrum in $[0, \infty)$.* By property (iv) of the [Borel Functional Calculus](/theorems/2696), $\sigma(R) = \sigma(\Psi_T(r)) \subseteq \overline{r(K)}$. Since $r$ is continuous on the compact set $K$, $r(K)$ is compact, hence closed, so $\overline{r(K)} = r(K) \subseteq [0, \infty)$. Therefore $\sigma(R) \subseteq [0, \infty)$. Hence $R$ is a positive operator. [guided] We need to verify that $R = r(T)$ is what the theorem calls a "positive operator" — the standard definition for a Hilbert-space operator is: self-adjoint with non-negative spectrum (equivalently $(Rx, x)_H \ge 0$ for all $x \in H$). We check both criteria via the BFC. *Step A: self-adjointness.* The function $r(\lambda) = |\lambda|$ takes real values, so $\bar{r} = r$ pointwise. The [Borel Functional Calculus](/theorems/2696) preserves the involution: property (i) states that $\Psi_T$ is a $*$-homomorphism, meaning $\Psi_T(\bar{f}) = \Psi_T(f)^*$ for every $f \in L^\infty(K)$. Applying this to $f = r$: \begin{align*} R^* = \Psi_T(r)^* = \Psi_T(\bar{r}) = \Psi_T(r) = R. \end{align*} So $R$ is self-adjoint. *Step B: non-negative spectrum.* We use property (iv) of the BFC, which states $\sigma(\Psi_T(f)) \subseteq \overline{f(K)}$ for every $f \in L^\infty(K, \mathcal{B}(K))$. The closure is needed in general for Borel $f$ (the image $f(K)$ may not be closed), but here $r$ is continuous on the compact set $K$, so $r(K)$ is the continuous image of a compact set, hence itself compact, hence closed: $\overline{r(K)} = r(K)$. Since $r$ takes values in $[0, \infty)$ by definition, $r(K) \subseteq [0, \infty)$, and hence \begin{align*} \sigma(R) \subseteq r(K) \subseteq [0, \infty). \end{align*} Combining: $R$ is self-adjoint with $\sigma(R) \subseteq [0, \infty)$, so $R$ is positive. [/guided] [/step] [step:Verify that $U$ is unitary] We show $U U^* = U^* U = I$ using the BFC and the pointwise identity $u \bar{u} = 1$ on $K$. For every $\lambda \in K$, \begin{align*} u(\lambda) \overline{u(\lambda)} = |u(\lambda)|^2 = 1, \end{align*} so $u \bar{u} = 1$ as elements of $L^\infty(K, \mathcal{B}(K))$. Apply $\Psi_T$: \begin{align*} U U^* = \Psi_T(u) \cdot \Psi_T(u)^* = \Psi_T(u) \cdot \Psi_T(\bar{u}) = \Psi_T(u \bar{u}) = \Psi_T(1) = I, \end{align*} using that $\Psi_T$ is a $*$-homomorphism (so $\Psi_T(\bar{u}) = \Psi_T(u)^* = U^*$ and $\Psi_T(u) \Psi_T(\bar{u}) = \Psi_T(u\bar{u})$) and that $\Psi_T$ is unital ($\Psi_T(1) = I$). The same argument with $\bar{u} u = 1$ gives $U^* U = I$. Hence $U$ is unitary. [/step] [step:Verify $T = RU$ via the pointwise identity $r u = \mathrm{id}$ on $K$] We compute $r(\lambda) u(\lambda)$ for $\lambda \in K$ in two cases. *Case $\lambda \ne 0$.* Then $r(\lambda) u(\lambda) = |\lambda| \cdot \lambda/|\lambda| = \lambda$. *Case $\lambda = 0$.* Then $r(0) u(0) = 0 \cdot 1 = 0 = \lambda$. So $r(\lambda) u(\lambda) = \lambda = \mathrm{id}(\lambda)$ for every $\lambda \in K$, i.e., $r u = \mathrm{id}$ in $L^\infty(K, \mathcal{B}(K))$. Apply $\Psi_T$: \begin{align*} R U = \Psi_T(r) \Psi_T(u) = \Psi_T(r u) = \Psi_T(\mathrm{id}) = \mathrm{id}(T) = T, \end{align*} using property (i) of the [Borel Functional Calculus](/theorems/2696), which states that $\mathrm{id}(T) = T$. Hence $T = RU$. [/step] [step:Verify pairwise commutation of $R$, $U$, $T$] The functions $r$, $u$, $\mathrm{id}$ all lie in the commutative algebra $L^\infty(K, \mathcal{B}(K))$ (which is commutative under pointwise multiplication of Borel functions). For any pair $f, g \in L^\infty(K, \mathcal{B}(K))$: \begin{align*} fg = gf \quad \text{pointwise on } K. \end{align*} Since $\Psi_T$ is a homomorphism, \begin{align*} \Psi_T(f) \Psi_T(g) = \Psi_T(fg) = \Psi_T(gf) = \Psi_T(g) \Psi_T(f). \end{align*} Apply this to the three pairs $(r, u)$, $(r, \mathrm{id})$, $(u, \mathrm{id})$ to obtain \begin{align*} RU = UR, \qquad RT = TR, \qquad UT = TU. \end{align*} Hence $R$, $U$, and $T$ pairwise commute. [/step] [step:Wrap up] We have constructed $R = r(T)$ and $U = u(T)$ via the [Borel Functional Calculus](/theorems/2696) such that: (a) $R$ is a positive operator (Step 3), (b) $U$ is a unitary operator (Step 4), (c) $T = RU$ (Step 5), (d) $R$, $U$, and $T$ pairwise commute (Step 6). This establishes the polar decomposition of the normal operator $T$. [/step]