proving $ f\left( x\right)= x $ is riemann integrable over $ \left[ 0,1\right] $ [closed]

I want to prove that $ f\left( x\right)= x $ is riemann integrable over $ \left[ 0,1\right] $ using the Squeeze Theorem
here is my approach i want to know if it is right ?
let $\epsilon $ be given
Let $P = \{I_k\}_{k=1}^n $ be a partition such that $( \lVert P \rVert < \delta_\epsilon=\epsilon/2 $, with $I_k= \left[ x_{k-1},x_{k}\right] $
we define $u_k = min_{x \in I_k } f(x)$
and $v_k =max_{x \in I_k } f(x) $
and we define the function $\alpha_\epsilon= $$\left\{\begin{array}{l}u_k \;\; x\in (x_{k-1},x_{k}) \\x \;\;elswere\end{array}\right.$
and we define the function $\omega_\epsilon= $$\left\{\begin{array}{l}v_k \;\; x\in x\in (x_{k-1},x_{k}) \\x \;\;elswere\end{array}\right.$
we have for exery $x \in \left[ 0,1\right] $$\, \,$$ \alpha_\epsilon \leq f\left( x\right)\leq \omega_\epsilon $
and we have $ \,$$\omega_\epsilon $$\, $ and $\, $$\alpha_\epsilon $ are Riemann integrable over $ \left[ 0,1\right] $ because they are step functions
also $\int_0^1 \omega_\epsilon - \alpha_\epsilon =\sum_{k=1}^{n} \left( x_{k} - x_{k-1}\right) \times \left( v_{k}-u_{k}\right) $
and we khow that $\, v_{k}-u_{k}\leq \lVert P \rVert < \epsilon/2 \, \,$ (because $ f\left( x\right)= x $ )
so $\int_0^1 \omega_\epsilon - \alpha_\epsilon < \epsilon/2 < \epsilon \, \, $ (because $\sum_{k=1}^{n} \left( x_{k} - x_{k-1}\right)=1 $
So this is true for every epsilon
So by squeeze theorem f is reiman integrable over $ \left[ 0,1\right] $