# Why the set of stochastic process Ito Integrable has to be square integrable w.r.t time as well?

Ito Integral

1. Consider a set of stochastic process $f(t)$ mainly such that

a) $$E\left(\int_0^{+\infty}f(t)^2 \,dt\right) < \infty.$$

Denote this set of stochastic process as $M^2$.

Question:

a) For each $w$, $f(t,$$w$$)$ is a continuous funtion, so we all know a bounded continuous funtion will be Riemann integrable, but why it is square integrable too, i.e. why $$\int_0^{\infty}f(t)^2 \,dt$$ has to exist

i.e. why the set $$M^2$$ will for sure be in $$L^2$$ when integrating over time $t$ on the time line $(0, \infty)$?

Thank you

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If a random variable is integrable then it is finite almost surely, in fact it is a basic result from measure theory that: $$\int |g| d\mu < \infty \Rightarrow |g| < \infty \quad \text{almost everywhere }$$
So in this case $\mu = \mathbb{P}$ and $g = \int_0^\infty f(t)^2 dt$
When constructing the Ito integral, you are working on a Hilbert space, $\mathcal{L}^2$, and so for a sequence of functions $f_n$ to converge to a function $f$, you need the $\mathcal{L}^2$ norm to converge to zero, in particular, you want your functions to be square integrable. When constructing the Ito integral, you usually consider the set $M_0^2$ of square integrable simple functions, which are dense in $M^2$. Hope this helps a bit