Proof. Assume without loss of generality that $p, q \in (1, \infty)$.

(Simple): Suppose that $x$ and $y$ are simple and positive. Write $x = \sum_{j = 1}^{m} \lambda_{j} p_{j}$ and $y = \sum_{k = 1}^{n} \mu_{k} q_{k}$, where $\bracsn{p_j}_{1}^{m}, \bracsn{q_k}_{1}^{n} \subset \text{Proj}(M)$ are families of mutually orthogonal projections in $M$, with $1_{M} = \sum_{j = 1}^{m} p_{j} = \sum_{k = 1}^{n} q_{k}$.

Let $u, v \in M$ with $\norm{u}_{M}, \norm{v}_{M} \le 1$, then by the classical Hölder’s Inequality,

(Bounded): By Theorem 43.7.2, there exists positive simple operators $\seq{x_n}, \seq{y_n}\subset M$ such that $x_{n} \to x$ and $y_{n} \to y$ in $M$ as $n \to \infty$. By Definition 44.4.6, $\norm{z}_{L^p(M, \tau)}\le \norm{z}_{M}$ for all $z \in M$.

Let $u, v \in M$ with $\norm{u}_{M}, \norm{v}_{M} \le 1$, then by the simple case,

(Arbitrary): Supopse that $x \in L^{p}(M, \tau)$ and $y \in L^{q}(M, \tau)$ are arbitrary. Since $\norm{x}_{L^p(M, \tau)}= \norm{|x|}_{L^p(M, \tau)}$ and $\norm{y}_{L^q(M, \tau)}= \norm{|y|}_{L^q(M, \tau)}$, assume without loss of generality that $x$ and $y$ are positive.

By Lemma 44.3.4, there exists non-decreasing sequences $\seq{p_n}, \seq{q_n}\subset \text{Proj}(M)$ such that:

1. (i)

   $p_{n}L^{2}(M, \tau) \subset D(x)$ and $q_{n}L^{2}(M, \tau) \subset D(y)$ for all $n \in \nat$.

2. (ii)

   $xp_{n}, yq_{n} \in M$ for all $n \in \nat$.

3. (iii)

   $\tau(p_{n}), \tau(q_{n}) \upto 1$ as $n \to \infty$.

By Lemma 44.3.6, there exists a non-decreasing sequence $\seq{r_n}\subset \text{Proj}(M)$ such that:

1. (iv)

   $r_{n} L^{2}(M, \tau) \subset D(xy) \subset D(|xy|)$ and $yr_{n}L^{2}(M, \tau) \subset p_{n}L^{2}(M, \tau)$ for all $n \in \natp$.

2. (v)

   $r_{n} \le p_{n} \wedge q_{n}$ for all $n \in \natp$.

3. (vi)

   $\tau(r_{n}) \upto 1$ as $n \to \infty$.

Let $xy = u|xy|$ be the polar decomposition of $xy$, then by Lemma 44.4.2 and the bounded case,

$\square$