[proofplan] The dyadic increments of Brownian motion are independent centered Gaussian random variables with common variance $t2^{-n}$. We compute the expectation and variance of the quadratic sum exactly. The variance tends to zero, giving convergence in mean square; the variances are summable in $n$, so [Chebyshev's inequality](/theorems/1126) and Borel-Cantelli upgrade the convergence to almost sure convergence for each fixed time. [/proofplan] [step:Compute the mean of the dyadic quadratic sum] Fix $t\geq 0$ and $n\geq 1$. For $1\leq k\leq 2^n$, define the Brownian increment \begin{align*} \Delta_{n,k}W &= W_{k t2^{-n}}-W_{(k-1)t2^{-n}}. \end{align*} By the definition of standard Brownian motion, the random variables $(\Delta_{n,k}W)_{k=1}^{2^n}$ are independent and each has Gaussian distribution with mean $0$ and variance $t2^{-n}$. Therefore \begin{align*} \mathbb E[Q_n(t)] &= \sum_{k=1}^{2^n}\mathbb E[(\Delta_{n,k}W)^2] \\ &= \sum_{k=1}^{2^n}t2^{-n} \\ &= t. \end{align*} [/step] [step:Compute the variance and prove convergence in $L^2$] Let $\sigma_n^2=t2^{-n}$. If $G$ is a centered Gaussian random variable with variance $\sigma_n^2$, then $\mathbb E[G^4]=3\sigma_n^4$. Hence \begin{align*} \operatorname{Var}(G^2) &= \mathbb E[G^4]-\mathbb E[G^2]^2 \\ &= 3\sigma_n^4-\sigma_n^4 \\ &= 2\sigma_n^4. \end{align*} Using independence of the increments, \begin{align*} \operatorname{Var}(Q_n(t)) &= \sum_{k=1}^{2^n}\operatorname{Var}\left((\Delta_{n,k}W)^2\right) \\ &= 2^n\cdot 2(t2^{-n})^2 \\ &= 2t^2\,2^{-n}. \end{align*} Since $\mathbb E[Q_n(t)]=t$, this gives \begin{align*} \mathbb E\left[\left|Q_n(t)-t\right|^2\right] &= 2t^2\,2^{-n}\to 0. \end{align*} Thus $Q_n(t)\to t$ in $L^2(\Omega,\mathcal F,\mathbb P)$. [/step] [step:Use Borel-Cantelli to obtain almost sure convergence] Let $\varepsilon>0$. [Chebyshev's inequality](/theorems/1126) and the variance computation give \begin{align*} \mathbb P\left(|Q_n(t)-t|>\varepsilon\right) &\leq \frac{\mathbb E[|Q_n(t)-t|^2]}{\varepsilon^2} \\ &= \frac{2t^2}{\varepsilon^2}\,2^{-n}. \end{align*} The series of upper bounds is summable: \begin{align*} \sum_{n=1}^{\infty}\frac{2t^2}{\varepsilon^2}\,2^{-n} &<\infty. \end{align*} By the [Borel Cantelli Lemma I](/theorems/507), the event $|Q_n(t)-t|>\varepsilon$ occurs only finitely many times almost surely. Applying this conclusion to $\varepsilon=1/m$ for each integer $m\geq 1$ and intersecting the resulting probability-one events gives $Q_n(t)\to t$ almost surely. [/step] [step:Identify the limit as quadratic variation] The dyadic partitions of $[0,t]$ have mesh size $t2^{-n}\to 0$. By definition, the quadratic variation $[W]_t$ along these partitions is the limit in probability, when it exists, of the quadratic sums $Q_n(t)$. Since $Q_n(t)\to t$ in $L^2$ and almost surely, it also converges in probability to $t$. Hence \begin{align*} [W]_t &= t \end{align*} for every fixed $t\geq 0$. [/step]