$$\int_0^1 B_t dt=\lim_{\omega \to\infty}{1 \over {\omega}}{\int_0^{\omega}{Y_0+}X_t dt}$$ Where $B_t$ is simple brownian motion, and $X_t$ is a discrete random variable that can be 1 or -1 with equal probability. $X_t $ is the random variable given at t in the integral or summation. In other words $X_t=X_t$ wheras $X \not =X$. $Y_0$ is where the Brownian motion starts.
Possible Proof:
$$(1) \quad Y_{t+\epsilon}=Y_t+\epsilon \cdot X_t$$ As $\epsilon$ approaches 0 from the right the recurrance relation approaches $B_t$ and the area of $B_t$ is given by... $$(2) \quad A_{T+\epsilon}=A_{T}+\epsilon \cdot Y_T$$ Solve (1) with $Y_0=C \, ...$ $$(3) \quad Y_t=B_t=\sum_{n=1}^{t}{\epsilon \cdot X_n}$$ we'll sub $\epsilon={1 \over {\omega}}$ $$(4) \quad Y_t=B_t=Y_0+{1 \over {\omega}}\sum_{n=1}^{t}{X_t}$$ Now we solve (2) in a very similar manner... $$(5) \quad A_T={1 \over {\omega}}\sum_{m=1}^{T}{\left({1 \over {\omega}}\sum_{n=1}^{m}{(Y_0+X_n)}\right)}={1 \over {\omega^2}} \sum_{m=1}^{T} \left( \sum_{n=1}^{m}(Y_0+X_n) \right)$$ Where T is distinct from t. Here's the part I'm wondering about. Each inner summation will add $X_1$, every summation with $m \ge 2$ will add $X_2$, every summation with $m \ge 3$ will add $X_3$. So in general the nth $X_n$ will be added $(T-n+1)$ times. Also you can break $Y_0$ apart and sum it. This means you can rewrite the summation as... $$(6) \quad A_T={1 \over {\omega^2}}\sum_{k=1}^T{(k \cdot Y_0+k \cdot X_k)}$$ This means this is just a sum of the integers. You can immediatly evaluate the $Y_0$ part. Instead lets write the integral. Using the definition of the definite integral... $$\int_0^{1}{f(\omega \cdot x)dx}={1 \over {\omega}}\sum_{k=1}^{\omega}\left(f\left({k} \right)\right)$$ We are done. Have $f(k)=k \cdot Y_0 +k \cdot X_k$, let $T=\omega$, interchange the summation for integration, and let $\omega$ approach $\infty$.
Seems like an acceptable proof method. I don't really see how that identity simplifies anything, but it is correct as far as I can tell.