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Completed squared increments form a nondecreasing step process. The continuous sum that includes the last partial increment differs by at most the square of the path's modulus of continuity over one mesh interval. Therefore both have the same limit in uniform convergence on compacts in probability. This proves monotonicity of the limiting quadratic variation without incorrectly asserting monotonicity of each continuous partial-increment sum.

For continuous local martingales $M,N$ and a sequence in which each term is a partition of an interval whose mesh tends to zero, the sumsconverge in the sense of uniform convergence on compacts in probability to a continuous increasing process. To identify the limit, put $C=[M]+[N]$ and choose Radon-Nikodym derivativesIf $(U,V)$ is a centered bivariate normal distribution with covariance matrix $\left(\begin{array}{cc}a& c\\ c& b\end{array}\right)$, thenLocalizing, representing the pair as stochastic integrals against a two-dimensional Brownian motion, and approximating the integrands by bounded predictable step processes proves the convergence. The step-process case follows from the weak law of large numbers for independent Gaussian increments; the Burkholder-Davis-Gundy inequality controls the approximation error. Since $|\mathbb{E}[UV]|\le \mathbb{E}|UV|\le \sqrt{\mathbb{E}{U}^2\mathbb{E}{V}^2}$,

Uniform convergence in probability means that the supremum of the absolute discrepancy on the entire index set tends to zero in probability. It controls evaluation at random indices in that set, unlike pointwise convergence in probability. On a fixed compact index set it is a special case of uniform convergence on compacts in probability.

If $Q\ll P$, uniform convergence on compacts in probability under $P$ implies the same convergence under $Q$. Indeed, let $Z=dQ/dP$ be the Radon-Nikodym derivative. For every event $E$,Integrability makes the second term arbitrarily small for large $K$. Thus $P(E_n)\to 0$ implies $Q(E_n)\to 0$; apply this to each event where the supremum on $[0,T]$ exceeds a fixed tolerance. Only absolute continuity of measures is needed, not equivalence.