Integral of Brownian motion against itself

For each $n\ge 1$, set $L_0^n=B_0$ and $L_s^n=B_{k{2}^{-n}}$ on $(k{2}^{-n},(k+1)2^{-n}]$, $k=0,1,\dots$. Each $L^n$ is predictable by the countable interval generators in [F1]; the coefficients need not be bounded to give measurability. Define ${\beta }_s={\lim }_n{L}_s^n$ wherever this limit exists as a finite real number, and zero elsewhere. The convergence set is measurable by the countable Cauchy criterion, hence $\beta$ is predictable by [F1]. On the single full event of continuous Brownian paths the left grid points tend to $s$ for every $s>0$, so ${\beta }_s=B_s$ simultaneously for all $s\ge 0$. No membership of that full event in ${\mathcal{F}}_0$ is needed, and the limiting map is never defined by multiplying B by that event.

In particular for every fixed $s$, $E{\beta }_s^2=E{B}_s^2=s$. Tonelli in [F4] applies to the measurable nonnegative map ${\beta }^2$ and gives $E{\int }_0^t{\beta }_s^{2}ds=t^2\mathrm{/}2$. For the dyadic grid $t_k=kt\mathrm{/}{2}^n$ put $H_s^n={\sum }_{k<2^n}{B}_{t_k}{1}_{(t_k,t_{k+1}]}(s)$. This is predictable, and its finite energy follows from $E{B}_{t_k}^2=t_k$. By deterministic-time equality of ${\beta }_s$ and $B_s$, [F2] and Tonelli give $E{\int }_0^t\mathrm{\mid }{H}_s^n-{\beta }_s{\mathrm{\mid }}^{2}ds={\sum }_k{\int }_{t_k}^{t_{k+1}}(s-t_k)ds=\frac{t^2}{2^{n+1}}\mathrm{.}$ Thus the integrals of $H^n$ converge in $L^2(P)$ to the integral of $\beta$ by [F3].

Fix $n$ and truncate the coefficient $B_{t_k}$ to $c_r(B_{t_k})$, where $c_r(x)=\max (-r,\min (x,r))$, to obtain bounded elementary $H^{n,r}$. Dominated convergence applies to each coefficient error squared, bounded by $B_{t_k}^2$ and tending to zero. Hence $H^{n,r}\to H^n$ in predictable $L^2$. For each increment ${\mathrm{\Delta }}_{k}B=B_{t_{k+1}}-B_{t_k}$, independence in [F2] gives $E\mathrm{\mid }(c_r(B_{t_k})-B_{t_k}){\mathrm{\Delta }}_{k}B{\mathrm{\mid }}^2=(t_{k+1}-t_k)E\mathrm{\mid }{c}_r(B_{t_k})-B_{t_k}{\mathrm{\mid }}^2⟶0.$ The finite sum therefore converges in $L^2$ by the triangle inequality. Comparing this with the isometric convergence of the elementary integrals proves ${\int }_0^t{H}_s^{n}d{B}_s=S_n:={\sum }_k{B}_{t_k}{\mathrm{\Delta }}_{k}B$ in $L^2(P)$. This explicitly licenses unbounded step coefficients without calling them elementary.

Finite telescoping gives $2S_n=B_t^2-B_0^2-{\sum }_k({\mathrm{\Delta }}_{k}B{)}^2$. Since $B_0=0$ almost surely, [F5] implies $S_n\to Y_t=(B_t^2-t)\mathrm{/}2$ almost surely. Write $I_t$ for the integral class of $\beta$. Steps 2.1 and 3.1 give $E\mathrm{\mid }{I}_t-S_n{\mathrm{\mid }}^2\to 0$, whereas Fatou in [F4] gives $E\mathrm{\mid }{I}_t-Y_t{\mathrm{\mid }}^2\le {\mathrm{lim\hspace{0.17em}inf}}_{n}E\mathrm{\mid }{I}_t-S_n{\mathrm{\mid }}^2=0$. Thus $I_t=Y_t$ almost surely.

Choose the continuous adapted integral version supplied by [F3]. Intersect its continuity event, the common Brownian continuity and zero-start event, and the equality events of step 4.1 for all positive rational t. This is a measurable full event; continuity of both sides extends the equality from rational to all nonnegative real times. At time zero the integral is zero and $B_0=0$ on this event. No claim is made that the identity holds on every exceptional constant Brownian path, or that the entire all-time equality set must itself be measurable in an incomplete space.

By [F3] the mean is zero. By [F2], $E{Y}_t^2=\frac{1}{4}(E{B}_t^4-2tE{B}_t^2+t^2)=t^2\mathrm{/}2,$ in agreement with the isometry and step 2.1. For $t>0$ this variance is positive, while $Y_t\ge -t\mathrm{/}2$. Every centered normal with positive variance gives positive probability to an interval below $-t\mathrm{/}2$, because its density there is positive; a zero-variance normal has zero variance. Thus [F6] rules out a deterministic-integrand law for $t>0$. At $t=0$ both sides vanish and there is no such non-Gaussian claim. Finite grids include both endpoints, and n can start at 1 without changing any limit. Full AC covers [F6]; the predictable representative and grids are explicit and no additional choice of paths is made. No later Ito formula is used.