On a lecture notes, there is a following arguement: To make $\int_0^T \pi_t dW_t$ well-defined, (maybe it means to make $\int_0^T \pi_t dW_t<\infty \ \ a.s.$) we only need $\int_0^T \pi_t^2 dt<\infty$. Where $W_t$ is a standard one dimention Brownian Motion, $\pi_t$ is a previsible process.
Question: I don't understand how we can use condition $\int_0^T \pi_t^2 dt<\infty$ to make the stochastic integral well-defined. Do we need to add some other conditions??
Thanks
I think that you might have missed the fact that you cannot define a path-by-path Stochastic Integral with respect to Brownian Motion. That's because the paths of BM are of infinite variations on compacts almost surely.
On the other hand the quadratic variations of the paths of BM tends are equal to $t$ over the interval $[0,t]$ which is why this kind of conditions (among others like predictability which is also mandatory) makes it possible to define a stochastic integral with respect to a BM.
Best regards