[proofplan] We establish the Kunita--Watanabe inequality by lifting the classical Cauchy--Schwarz inequality from finite-dimensional inner products to stochastic integrals against the covariation measure. The argument proceeds in four stages: first, we prove a pointwise Cauchy--Schwarz bound on increments $|\langle M, N \rangle^t_s| \leq \sqrt{\langle M \rangle^t_s}\sqrt{\langle N \rangle^t_s}$ using the Riemann-sum characterisation of covariation; second, we extend this to a total variation bound on intervals via a telescoping-and-Cauchy--Schwarz argument; third, we pass to general Borel sets by a monotone class argument; finally, we extend from indicator integrands to simple previsible processes and then to general previsible processes by monotone convergence. [/proofplan]

[step:Establish the pointwise increment bound $|\langle M, N \rangle^t_s| \leq \sqrt{\langle M \rangle^t_s}\sqrt{\langle N \rangle^t_s}$]Write $\langle M, N \rangle^t_s := \langle M, N \rangle_t - \langle M, N \rangle_s$ for $0 \leq s \leq t$, and similarly for $\langle M \rangle^t_s$ and $\langle N \rangle^t_s$. [claim:Pointwise increment bound] For all $0 \leq s \leq t$, almost surely, \begin{align*} |\langle M, N \rangle^t_s| \leq \sqrt{\langle M \rangle^t_s} \cdot \sqrt{\langle N \rangle^t_s}. \end{align*} [/claim] [proof] By part (iii) of the [Properties of Covariation](/theorems/2086), the Riemann-sum approximations satisfy \begin{align*} \langle M, N \rangle^{(n)}_t - \langle M, N \rangle^{(n)}_s = \sum_{i : t_{i-1}^{(n)} \geq s}^{t_i^{(n)} \leq t} (M_{t_i^{(n)}} - M_{t_{i-1}^{(n)}})(N_{t_i^{(n)}} - N_{t_{i-1}^{(n)}}) \xrightarrow{\text{u.c.p.}} \langle M, N \rangle^t_s. \end{align*} For each fixed $n$, define the vectors $a, b \in \mathbb{R}^k$ (where $k$ is the number of subintervals of $[s,t]$ in the $n$-th dyadic partition) by \begin{align*} a_i &:= M_{t_i^{(n)}} - M_{t_{i-1}^{(n)}}, \qquad b_i := N_{t_i^{(n)}} - N_{t_{i-1}^{(n)}}. \end{align*} The Cauchy--Schwarz inequality in $\mathbb{R}^k$ gives \begin{align*} \left|\sum_{i=1}^k a_i b_i\right| \leq \left(\sum_{i=1}^k a_i^2\right)^{1/2} \left(\sum_{i=1}^k b_i^2\right)^{1/2}. \end{align*} The left-hand side is $|\langle M, N \rangle^{(n)}_t - \langle M, N \rangle^{(n)}_s|$. The two factors on the right are $(\langle M \rangle^{(n)}_t - \langle M \rangle^{(n)}_s)^{1/2}$ and $(\langle N \rangle^{(n)}_t - \langle N \rangle^{(n)}_s)^{1/2}$ respectively, since the quadratic variation approximations are sums of squared increments. By part (iii) of the [Properties of Covariation](/theorems/2086), each of these converges u.c.p. to $(\langle M \rangle^t_s)^{1/2}$ and $(\langle N \rangle^t_s)^{1/2}$ respectively. Passing to the u.c.p. limit preserves the inequality, giving the claimed bound almost surely. [/proof][/step]

[guided]The idea is to reduce to the classical Cauchy--Schwarz inequality in $\mathbb{R}^k$. The covariation $\langle M, N \rangle$ is defined as the limit of Riemann-sum approximations, and each Riemann sum is a finite inner product $\sum_i a_i b_i$ in $\mathbb{R}^k$. The Cauchy--Schwarz inequality $|\sum a_i b_i| \leq (\sum a_i^2)^{1/2}(\sum b_i^2)^{1/2}$ holds at every finite level. The key point is that u.c.p. convergence preserves weak inequalities: if $|X_n| \leq Y_n Z_n$ and $X_n \to X$, $Y_n \to Y$, $Z_n \to Z$ all in u.c.p., then $|X| \leq YZ$ almost surely. This is because u.c.p. convergence implies convergence along a subsequence uniformly on compacts almost surely, and pointwise limits preserve non-strict inequalities. Concretely, define $a, b \in \mathbb{R}^k$ with $a_i = M_{t_i^{(n)}} - M_{t_{i-1}^{(n)}}$ and $b_i = N_{t_i^{(n)}} - N_{t_{i-1}^{(n)}}$. The Cauchy--Schwarz inequality in $\mathbb{R}^k$ gives \begin{align*} |\langle M, N \rangle^{(n)}_t - \langle M, N \rangle^{(n)}_s| \leq \left(\langle M \rangle^{(n)}_t - \langle M \rangle^{(n)}_s\right)^{1/2} \left(\langle N \rangle^{(n)}_t - \langle N \rangle^{(n)}_s\right)^{1/2}. \end{align*} Passing to the u.c.p. limit via part (iii) of the [Properties of Covariation](/theorems/2086) yields $|\langle M, N \rangle^t_s| \leq \sqrt{\langle M \rangle^t_s} \cdot \sqrt{\langle N \rangle^t_s}$ almost surely.[/guided]

[step:Upgrade the increment bound to a total variation bound on intervals][claim:Total variation bound on intervals] For all $0 \leq s < t$, almost surely, \begin{align*} \int_s^t |d\langle M, N \rangle_u| \leq \sqrt{\langle M \rangle^t_s} \cdot \sqrt{\langle N \rangle^t_s}. \end{align*} [/claim] [proof] Let $s = u_0 < u_1 < \cdots < u_m = t$ be any partition of $[s, t]$. By the triangle inequality for signed measures and the pointwise increment bound from the previous step applied to each subinterval $[u_{j-1}, u_j]$, \begin{align*} \sum_{j=1}^m |\langle M, N \rangle^{u_j}_{u_{j-1}}| \leq \sum_{j=1}^m \sqrt{\langle M \rangle^{u_j}_{u_{j-1}}} \cdot \sqrt{\langle N \rangle^{u_j}_{u_{j-1}}}. \end{align*} Now apply the Cauchy--Schwarz inequality in $\mathbb{R}^m$ to the vectors $(\sqrt{\langle M \rangle^{u_j}_{u_{j-1}}})_{j=1}^m$ and $(\sqrt{\langle N \rangle^{u_j}_{u_{j-1}}})_{j=1}^m$: \begin{align*} \sum_{j=1}^m \sqrt{\langle M \rangle^{u_j}_{u_{j-1}}} \cdot \sqrt{\langle N \rangle^{u_j}_{u_{j-1}}} \leq \left(\sum_{j=1}^m \langle M \rangle^{u_j}_{u_{j-1}}\right)^{1/2} \left(\sum_{j=1}^m \langle N \rangle^{u_j}_{u_{j-1}}\right)^{1/2} = \sqrt{\langle M \rangle^t_s} \cdot \sqrt{\langle N \rangle^t_s}, \end{align*} where the last equality uses the telescoping property $\sum_{j=1}^m \langle M \rangle^{u_j}_{u_{j-1}} = \langle M \rangle_t - \langle M \rangle_s = \langle M \rangle^t_s$ (and likewise for $N$), which holds since $\langle M \rangle$ is additive on intervals. Taking the supremum over all partitions of $[s,t]$ gives the total variation bound. [/proof][/step]

[guided]The total variation of the signed measure $d\langle M, N \rangle$ on $[s,t]$ is the supremum over all partitions of $\sum_j |\langle M, N \rangle^{u_j}_{u_{j-1}}|$. We need to bound each such sum uniformly in the partition. The strategy is a two-level application of Cauchy--Schwarz. At the first level, we use the pointwise increment bound to replace each $|\langle M, N \rangle^{u_j}_{u_{j-1}}|$ with $\sqrt{\langle M \rangle^{u_j}_{u_{j-1}}} \cdot \sqrt{\langle N \rangle^{u_j}_{u_{j-1}}}$. At the second level, we apply Cauchy--Schwarz in $\mathbb{R}^m$ to collect these into a product of two sums: \begin{align*} \sum_{j=1}^m \sqrt{\langle M \rangle^{u_j}_{u_{j-1}}} \cdot \sqrt{\langle N \rangle^{u_j}_{u_{j-1}}} \leq \left(\sum_{j=1}^m \langle M \rangle^{u_j}_{u_{j-1}}\right)^{1/2} \left(\sum_{j=1}^m \langle N \rangle^{u_j}_{u_{j-1}}\right)^{1/2}. \end{align*} The sums telescope: $\sum_j \langle M \rangle^{u_j}_{u_{j-1}} = \langle M \rangle_t - \langle M \rangle_s = \langle M \rangle^t_s$, because $\langle M \rangle$ is a continuous increasing process and its increments are additive. The bound $\sqrt{\langle M \rangle^t_s} \cdot \sqrt{\langle N \rangle^t_s}$ is independent of the partition, so the supremum over all partitions is bounded by the same quantity.[/guided]

[step:Extend the total variation bound to arbitrary Borel sets via a monotone class argument] [claim:Borel set bound] For any bounded Borel set $B \subseteq [0, \infty)$, almost surely, \begin{align*} \int_B |d\langle M, N \rangle_u| \leq \sqrt{\int_B d\langle M \rangle_u} \cdot \sqrt{\int_B d\langle N \rangle_u}. \end{align*} [/claim] [proof] Define \begin{align*} \mathcal{C} := \left\{ B \in \mathcal{B}([0,\infty)) : B \text{ bounded and the Cauchy--Schwarz bound holds a.s. for } B \right\}. \end{align*} The previous step shows that $\mathcal{C}$ contains all half-open intervals $(s, t]$ with $0 \leq s < t$. We verify $\mathcal{C}$ is closed under finite disjoint unions. Let $B_1, \ldots, B_m$ be disjoint bounded Borel sets in $\mathcal{C}$ and set $B = \bigsqcup_{\ell=1}^m B_\ell$. Then \begin{align*} \int_B |d\langle M, N \rangle_u| = \sum_{\ell=1}^m \int_{B_\ell} |d\langle M, N \rangle_u| \leq \sum_{\ell=1}^m \sqrt{\int_{B_\ell} d\langle M \rangle_u} \cdot \sqrt{\int_{B_\ell} d\langle N \rangle_u}. \end{align*} Applying the Cauchy--Schwarz inequality in $\mathbb{R}^m$ to the vectors $\left(\sqrt{\int_{B_\ell} d\langle M \rangle_u}\right)_{\ell=1}^m$ and $\left(\sqrt{\int_{B_\ell} d\langle N \rangle_u}\right)_{\ell=1}^m$, \begin{align*} \sum_{\ell=1}^m \sqrt{\int_{B_\ell} d\langle M \rangle_u} \cdot \sqrt{\int_{B_\ell} d\langle N \rangle_u} \leq \sqrt{\sum_{\ell=1}^m \int_{B_\ell} d\langle M \rangle_u} \cdot \sqrt{\sum_{\ell=1}^m \int_{B_\ell} d\langle N \rangle_u} = \sqrt{\int_B d\langle M \rangle_u} \cdot \sqrt{\int_B d\langle N \rangle_u}. \end{align*} The collection of finite disjoint unions of half-open intervals $(s_\ell, t_\ell]$ is a $\pi$-system (closed under finite intersection) that generates $\mathcal{B}([0,\infty))$ restricted to bounded sets. To extend to all bounded Borel sets, we use a monotone limit argument: if $B_n \nearrow B$ with each $B_n \in \mathcal{C}$ and $B$ bounded, then by monotone convergence for the measures $|d\langle M, N \rangle|$, $d\langle M \rangle$, and $d\langle N \rangle$, \begin{align*} \int_B |d\langle M, N \rangle_u| = \lim_{n \to \infty} \int_{B_n} |d\langle M, N \rangle_u| \leq \lim_{n \to \infty} \sqrt{\int_{B_n} d\langle M \rangle_u} \cdot \sqrt{\int_{B_n} d\langle N \rangle_u} = \sqrt{\int_B d\langle M \rangle_u} \cdot \sqrt{\int_B d\langle N \rangle_u}, \end{align*} where the limit passes through the square roots by continuity. Similarly for decreasing limits. By the monotone class theorem, $\mathcal{C}$ contains all bounded Borel sets. [/proof] [/step]

[step:Extend from indicator integrands to simple previsible processes]Let $H = \sum_{\ell=1}^m h_\ell \, \mathbb{1}_{B_\ell}$ and $K = \sum_{\ell=1}^m k_\ell \, \mathbb{1}_{B_\ell}$ be simple previsible processes, where $B_1, \ldots, B_m$ are disjoint bounded Borel sets and $h_\ell, k_\ell \geq 0$ are non-negative constants (the general case follows by replacing $H, K$ with $|H|, |K|$). Then \begin{align*} \int_0^\infty |H_s| |K_s| \, |d\langle M, N \rangle_s| = \sum_{\ell=1}^m h_\ell k_\ell \int_{B_\ell} |d\langle M, N \rangle_s|. \end{align*} Applying the Borel set bound from the previous step to each $B_\ell$: \begin{align*} \sum_{\ell=1}^m h_\ell k_\ell \int_{B_\ell} |d\langle M, N \rangle_s| \leq \sum_{\ell=1}^m h_\ell k_\ell \sqrt{\int_{B_\ell} d\langle M \rangle_s} \cdot \sqrt{\int_{B_\ell} d\langle N \rangle_s}. \end{align*} Applying the Cauchy--Schwarz inequality in $\mathbb{R}^m$ to the vectors $(h_\ell \sqrt{\int_{B_\ell} d\langle M \rangle_s})_{\ell=1}^m$ and $(k_\ell \sqrt{\int_{B_\ell} d\langle N \rangle_s})_{\ell=1}^m$: \begin{align*} \sum_{\ell=1}^m h_\ell k_\ell \sqrt{\int_{B_\ell} d\langle M \rangle_s} \cdot \sqrt{\int_{B_\ell} d\langle N \rangle_s} &\leq \left(\sum_{\ell=1}^m h_\ell^2 \int_{B_\ell} d\langle M \rangle_s\right)^{1/2} \left(\sum_{\ell=1}^m k_\ell^2 \int_{B_\ell} d\langle N \rangle_s\right)^{1/2} \\ &= \left(\int_0^\infty H_s^2 \, d\langle M \rangle_s\right)^{1/2} \left(\int_0^\infty K_s^2 \, d\langle N \rangle_s\right)^{1/2}. \end{align*}[/step]

[guided]We now want to move from integrals of indicator functions (the Borel set bound) to integrals with non-trivial integrands $H$ and $K$. For simple processes of the form $H = \sum h_\ell \mathbb{1}_{B_\ell}$, the integral decomposes into a finite sum, and we can apply the Borel set bound to each term. The computation involves three applications of Cauchy--Schwarz at different levels. The first is the Borel set bound (already proved). The second is Cauchy--Schwarz in $\mathbb{R}^m$, which collects the $m$ terms into a single product. The key is to pair $h_\ell \sqrt{\int_{B_\ell} d\langle M \rangle}$ with $k_\ell \sqrt{\int_{B_\ell} d\langle N \rangle}$. Cauchy--Schwarz gives \begin{align*} \sum_{\ell} h_\ell k_\ell \sqrt{\int_{B_\ell} d\langle M \rangle} \sqrt{\int_{B_\ell} d\langle N \rangle} &\leq \left(\sum_\ell h_\ell^2 \int_{B_\ell} d\langle M \rangle\right)^{1/2} \left(\sum_\ell k_\ell^2 \int_{B_\ell} d\langle N \rangle\right)^{1/2}. \end{align*} But $\sum_\ell h_\ell^2 \int_{B_\ell} d\langle M \rangle = \int_0^\infty H_s^2 \, d\langle M \rangle_s$ since $H^2 = \sum_\ell h_\ell^2 \mathbb{1}_{B_\ell}$ on the disjoint sets $B_\ell$. This completes the inequality for simple processes.[/guided]

[step:Pass to general previsible processes by monotone convergence] For general previsible processes $H, K \geq 0$, choose sequences of non-negative simple previsible processes $H^n \nearrow |H|$ and $K^n \nearrow |K|$ pointwise. By the monotone convergence theorem applied to the measure $|d\langle M, N \rangle|$ (which is a positive measure on $[0,\infty)$ for almost every $\omega$), \begin{align*} \int_0^\infty |H_s| |K_s| \, |d\langle M, N \rangle_s| = \lim_{n \to \infty} \int_0^\infty H^n_s K^n_s \, |d\langle M, N \rangle_s|. \end{align*} For each $n$, the inequality for simple processes gives \begin{align*} \int_0^\infty H^n_s K^n_s \, |d\langle M, N \rangle_s| \leq \left(\int_0^\infty (H^n_s)^2 \, d\langle M \rangle_s\right)^{1/2} \left(\int_0^\infty (K^n_s)^2 \, d\langle N \rangle_s\right)^{1/2}. \end{align*} By the monotone convergence theorem applied to the positive measures $d\langle M \rangle$ and $d\langle N \rangle$, \begin{align*} \lim_{n \to \infty} \int_0^\infty (H^n_s)^2 \, d\langle M \rangle_s = \int_0^\infty H_s^2 \, d\langle M \rangle_s, \qquad \lim_{n \to \infty} \int_0^\infty (K^n_s)^2 \, d\langle N \rangle_s = \int_0^\infty K_s^2 \, d\langle N \rangle_s. \end{align*} The monotone convergence theorem applies because $\langle M \rangle$ and $\langle N \rangle$ are increasing processes, so $d\langle M \rangle$ and $d\langle N \rangle$ are positive measures, and $(H^n)^2 \nearrow H^2$ and $(K^n)^2 \nearrow K^2$ are non-negative and increasing. Taking the limit in the inequality for each $n$ yields the Kunita--Watanabe inequality for general previsible $H$ and $K$. [/step]