I am looking at an approximation procedure to define the stochastic integral for $f \in \mathcal{V}$ where
Let $\mathcal{V}=\mathcal{V}(S, T)$ be the class of functions
$$ f(t, \omega):[0, \infty) \times \Omega \rightarrow \mathbf{R} $$
such that $(\mathrm{i}) \quad(t, \omega) \rightarrow f(t, \omega)$ is $\mathcal{B} \times \mathcal{F}$-measurable, where $\mathcal{B}$ denotes the Borel $\sigma-$ algebra on $[0, \infty)$ (ii) $f(t, \omega)$ is $\mathcal{F}_{t}$ -adapted. $(\text { iii }) E\left[\int_{S}^{T} f(t, \omega)^{2} d t\right]<\infty$
On the final step of the procedure the author makes the following claim:
Let $f \in \mathcal{V} .$ Then there exists a sequence $\left\{h_{n}\right\} \subset \mathcal{V}$ such that $h_{n}$ is bounded for each $n$ and
$$ E\left[\int_{S}^{T}\left(f-h_{n}\right)^{2} d t\right] \rightarrow 0 \text { as } n \rightarrow \infty $$
For the proof it is used:
$$h_{n}(t, \omega)=\left\{\begin{array}{ccc} -n & \text { if } & f(t, \omega)<-n \\ f(t, \omega) & \text { if } & -n \leq f(t, \omega) \leq n \\ n & \text { if } & f(t, \omega)>n \end{array}\right. $$
and the claim is that the result follows from the dominated convergence theorem. I am having trouble, however, seeing what is the dominating function. Hints are more appreciated than full answers.
Thanks.
We can decompose the square of difference into three regions:
$$(f - h_n)^2 = \begin{cases} (f-(-n))^2 & f < -n \\ 0 & f \in [-n,n] \\ ( f - n )^2 & f > n \end{cases}$$
where the first one is $(f-(-n))^2 = (-|f|+n)^2 = (|f| - n)^2$
now for $f > n$ $(f - n)^2 \leq f^2$