The $L^{2}-$extension construction of Ito integral in several literature I am learning the $L^{2}-$extension of Ito integral, and in the construction of extension, Karatzas and Shreve (and some online notes) seemed to analyze such a class of functions:

Let the "probability space" be $([0,T],\mathcal{B}, Leb)$, so we are just analyzing the usual lebesgue integral, etc. 
Let $f\in L^{2}([0,T],\mathcal{B}, Leb)$ and for al $h>0$, we define a kind of averaging as follows: 
  $$f_{h}(t)=\left\{
             \begin{array}{ll}
             0, \ \ \ 0\leq t<h;\\
             \\
             \dfrac{1}{h}\int_{(k-1)h}^{kh}f(s)ds,\ \ \ kh\leq t<(k+1)h\wedge T.
             \end{array}
\right.
$$
Then, the literature claims that $\|f_{h}\|_{L^{2}}\leq\|f\|_{L^{2}}$ and $f_{h}$ converges to $f$ in $L^{2}$ as $h\rightarrow 0$.

The only thing I can derive from this definition is using Jensen's inequality to obtain:$$|f_{h}(t)|^{2}\leq\dfrac{1}{h^{2}}\Big|\int_{(k-1)h}^{kh\wedge T}f(s)ds\Big|^{2}\leq\dfrac{1}{h^{2}}\int_{(k-1)h}^{kh\wedge T}f(s)^{2}ds\leq\dfrac{1}{h^{2}}\|f\|_{L^{2}}.$$ 
What is exactly the $L^{2}$ norm of $f_{h}(t)$? When I integrate $f_{h}(t)^{2}$ over $t\in [0,T]$, then value of $f_{h}(t)$ changes given different $kh\leq t<(k+1)h\wedge T$....
What can I do? Thank you!
 A: I will show the bound : For given $h>0$, let $\tilde{f}(s):=f(sh)$. Then
$$||f_h||_{L^2}^2=\int_0^Tf_h^2\leq\sum_{k=1}^{\lfloor T/h\rfloor}h\bigg(\int_{(k-1)h}^{kh}f(s)h^{-1}ds\bigg)^2=\sum_{k=1}^{\lfloor T/h\rfloor}h\bigg(\int_{k-1}^k\tilde{f}\bigg)^2\leq\sum_{k=1}^{\lfloor T/h\rfloor}h\int_{k-1}^k\tilde{f}^2$$$$=h\int_0^{\lfloor T/h\rfloor}\tilde{f}^2=h\int_0^{\lfloor T/h\rfloor h}f^2h^{-1}=\int_0^{\lfloor T/h\rfloor h}f^2\leq ||f||_{L^2}^2.$$
EDIT : I add the $L^2$ convergence result. First, suppose $f$ is continuous. If we fix $t\in [0,T]$, then $\forall h>0$, $\exists k=k(h)$ s.t. $kh\leq t<(k+1)h$.  For such $k$, $$|f_h(t)-f(t)|\leq h^{-1}\int_{(k-1)h}^{kh}|f(s)-f(t)|ds\leq\max_{(k-1)h\leq s\leq kh}|f(s)-f(t)|\xrightarrow{h\rightarrow 0}0$$ by continuity of $f$. Therefore $f_h\xrightarrow{h\rightarrow 0} f$, $\forall f\in C([0,T])$.  Now let $f\in L^2([0,T])$. Take any $\epsilon>0$, and since $C^\infty$ is dense in $L^2$, we have $g\in C^\infty([0,T])$ such that $||g-f||_{L^2}<\epsilon/3$. Therefore $$||f_h-f||_{L^2}\leq ||f_h-g_h||_{L^2}+||g_h-g||_{L^2}+||g-f||_{L^2}\leq||g_h-g||_{L^2}+2||g-f||_{L^2}$$ $$\leq \frac{\epsilon}{3}+\frac{2\epsilon}{3}=\epsilon,$$ by taking $h$ small enough. (used $||g_h-f_h||_{L^2}=||(g-f)_h||_{L^2}\leq||g-h||_{L^2}$)
EDIT2 : I will show how did I obtain the first inequality of the first part, as detail as possible. Note that  $T\leq(\lfloor T/h\rfloor +1)h$, hence $k=\lfloor T/h\rfloor$ is last $k$ possible for given $h>0$. Then, 
$$\int_0^Tf_h^2=\bigg(\sum_{k=1}^{\lfloor T/h\rfloor-1}\int_{kh}^{(k+1)h}f^2_h\bigg)+\underbrace{\int_{\lfloor T/h\rfloor h}^{T}f^2_h}_{\text{term for }k=\lfloor T/h\rfloor,\text{ but the integral ends at }T}$$ 
$$=\bigg(\sum_{k=1}^{\lfloor T/h\rfloor-1}\int_{kh}^{(k+1)h}\bigg(\int_{(k-1)h}^{kh}f(s)h^{-1}ds\bigg)^2\bigg)+\int_{\lfloor T/h\rfloor h}^{T}\bigg(\int_{(\lfloor T/h\rfloor-1)h}^{\lfloor T/h\rfloor h}f(s)h^{-1}ds\bigg)^2$$ 
$$=\bigg(\sum_{k=1}^{\lfloor T/h\rfloor-1}\bigg(\int_{(k-1)h}^{kh}f(s)h^{-1}ds\bigg)^2\int_{kh}^{(k+1)h}dt\bigg)+\bigg(\int_{(\lfloor T/h\rfloor-1)h}^{\lfloor T/h\rfloor h}f(s)h^{-1}ds\bigg)^2\int_{\lfloor T/h\rfloor h}^{T}dt$$ 
$$=\bigg(\sum_{k=1}^{\lfloor T/h\rfloor-1}\bigg(\int_{(k-1)h}^{kh}f(s)h^{-1}ds\bigg)^2h\bigg)+\bigg(\int_{(\lfloor T/h\rfloor-1)h}^{\lfloor T/h\rfloor h}f(s)h^{-1}ds\bigg)^2(T-\lfloor T/h\rfloor h).$$
And since 
$$T-\lfloor T/h\rfloor h\leq h,$$ 
$$\int_0^Tf_h^2\leq\bigg(\sum_{k=1}^{\lfloor T/h\rfloor-1}\bigg(\int_{(k-1)h}^{kh}f(s)h^{-1}ds\bigg)^2h\bigg)+\bigg(\int_{(\lfloor T/h\rfloor-1)h}^{\lfloor T/h\rfloor h}f(s)h^{-1}ds\bigg)^2h$$ 
$$=\sum_{k=1}^{\lfloor T/h\rfloor}\bigg(\int_{(k-1)h}^{kh}f(s)h^{-1}ds\bigg)^2h.$$
Maybe this was not so trivial.
