Proving $E\left[\left|\frac1n\sum_{m=1}^n Y_m-Y_m1_{[0,R)}(|Y_m|)\right|\right]\leq E\left[|Y_1|, \,|Y_1|\geq R\right]$ I'm doing an exercise from a book about Markov Processes, where I need to prove that
$$E\left[\left|\frac1n\sum_{m=1}^nC_m\right|\right]\leq E\left[|Y_1|, \,|Y_1|\geq R\right],$$
where $C_m=Y_m-Y_m1_{[0,R)}(|Y_m|),\; R>0$ and all the $Y_i, i=1,...,m$ are i.i.d random variables. 
I'm not 100% sure if my approach is correct but this is what I tried: 
$$E\left[\left|\frac1n\sum_{m=1}^nC_m\right|\right]\underset{tri.eq.}{\leq}\frac1n\sum_{m=1}^n E[|C_m|] \underset{i.i.d.}{=}\frac1n\cdot n E[|C_1|]=E[|C_1|].$$
Now, I was thinking could it be proved like this. Let me denote: 
$$\mu(|C_1|)_a^b=\frac{1}{b-a}\int_{a}^b|C_1|\;dy=\frac{1}{b-a}\left[\int_{a}^{-R}|C_1| \;dy+\underbrace{\int_{-R}^R|C_1|\;dy}_{=0}+\int_{R}^b|C_1|\;dy\right]$$
$$=\frac{1}{b-a}\left[\int_{a}^{-R}|Y_1|\;dy+\int_{R}^b|Y_1|\;dy\right].$$
Next, I denote: 
$$\mu(|Y_1|, \,|Y_1|\geq R)_a^b=\frac{1}{b-a-2R}\left[\int_{a}^{-R}|Y_1| \;dy+\int_{R}^b|Y_1| \;dy\right].$$
Now, since $\frac{1}{b-a}<\frac{1}{b-a-2R}\;$ I have $\;\mu(|C_1|)_a^b \leq \mu(|Y_1|, \,|Y_1|\geq R)_a^b$. By setting the integration domain to $a=-\infty$ and $b=\infty$ I get:
$$E\left[\left|\frac1n\sum_{m=1}^nC_m\right|\right]\leq E[|C_1|]=\mu(|C_1|)_{-\infty}^{\infty} \leq \mu(|Y_1|, \,|Y_1|\geq R)_{-\infty}^{\infty}=E\left[|Y_1|, \,|Y_1|\geq R\right].$$
I'm a bit uncertain about this if I made something illegal in my reasoning. I'm very unsure about the last part, when I equate the expectations and my $\mu$-functions. Is this valid? I'd appreciate if I could get some comments on this and maybe even an alternative better solution. Thank you!
P.S. sorry for posting this question two times. In the first post I noticed a flaw and decided to delete, edit and re-post. 
UPDATE: 
To make my question more clear I will restate it: 
A) Is my solution to the problem correct, and B) if not, how should I solve this problem? 
Problem in reference book (page 16, problem 1.3.2 part b))
 A: I will post my own answer here, thank you everybody for your help!
I need to prove that: 
$$E\left[\left|\frac1n\sum_{m=1}^nC_m\right|\right]\underset{tri.eq.}{\leq}\frac1n\sum_{m=1}^n E[|C_m|] \underset{i.i.d.}{=}\frac1n\cdot n E[|C_1|]=E[|C_1|].$$
Now, I was thinking could it be proved like this. Let me denote: 
$$\mu(|C_1|)_a^b=\frac{1}{b-a}\int_{a}^b|C_1|\;dy=\frac{1}{b-a}\left[\int_{a}^{-R}|C_1| \;dy+\underbrace{\int_{-R}^R|C_1|\;dy}_{=0}+\int_{R}^b|C_1|\;dy\right]$$
$$=\frac{1}{b-a}\left[\int_{a}^{-R}|Y_1|\;dy+\int_{R}^b|Y_1|\;dy\right].$$
Next, I denote: 
$$\mu(|Y_1|, \,|Y_1|\geq R)_a^b=\frac{1}{b-a-2R}\left[\int_{a}^{-R}|Y_1| \;dy+\int_{R}^b|Y_1| \;dy\right].$$
Now, since $\frac{1}{b-a}<\frac{1}{b-a-2R}\;$ I have $\;\mu(|C_1|)_a^b \leq \mu(|Y_1|, \,|Y_1|\geq R)_a^b$. By setting the integration domain to $a=-\infty$ and $b=\infty$ I get:
$$E\left[\left|\frac1n\sum_{m=1}^nC_m\right|\right]\leq E[|C_1|]=\mu(|C_1|)_{-\infty}^{\infty} \leq \mu(|Y_1|, \,|Y_1|\geq R)_{-\infty}^{\infty}=E\left[|Y_1|, \,|Y_1|\geq R\right].$$
