Let g(Y) be a function of the random variable Y, with $E[|g(Y)|]<\infty$. Show that for every $k>0$, $$ P[|(g(Y)| \leq k] \geq 1 - \frac {E[|g(Y)|]}{k}$$
This problem is similar to Markov Theorem.
My attempt: Let $$A=\{y:|(g(Y)| \leq k \}$$ and let $f(y)$ be the pdf of X. $$E[|g(Y)|] = \int_{-\infty}^{\infty} |g(y)|f(y)dy = \int_A g(y)dy + \int_{A^c} g(y)dy \geq \int_A g(y)dy \geq k \int_A f(y)dy$$
Am I on the right path?
No, there are some problems with your argument and it does not give the inequality you are asked to prove. Here is a correct proof: By Chebycheff's inequality $P\{|g(Y)| >k \} \leq \frac 1 k E|g(Y)|$. Hence $P\{|g(Y)| \leq k \}=1-P\{|g(Y)| >k \} \geq 1- \frac 1 k E|g(Y)|$. Some errors in your argument: it is not given that $Y$ has a pdf; absolute value sign in missing in some places.