# One stochastic integrability problem

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

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The construction of the Ito integral starts with simple (=piecewise constant) functions for which the formula is just  –  Ilya Jan 31 at 17:13
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.
Thank you for your answer. I think we need condition like $E \int (...)^2dt<\infty$ to make its integrable well-defined. I don't understand why we don't need to take expectation in the condition. –  XXX11235 Jan 26 at 16:37