Show that $\int_S\left|f_n - f\right|d\mu \to 0$ as $n\to\infty$. Theorem: (Dominated Convergence Theorem (DCT)). Let $(f_n)_{n\geq 1}$ be a sequence of integrable functions such that $\lim\limits_{n\to\infty}f_n = f$ pointwise. If there exists an integrable function $g:S\to[0,\infty]$, such that $\left|f_n\right|\leq g$ for all $n\geq 1$, then $f$ is integrable and $$\lim\limits_{n\to\infty}\int_Sf_nd\mu = \int_S fd\mu.$$
Exercise: Let $(f_n)_{n\geq 1}$ be a sequence of measurable functions with values in $\mathbb{R}$ or $\mathbb{C}$ and $f_n\to f$ pointwise. Assume that there exists an integrable function $g:S\to [0,\infty])$ such that $\left|f_n\right|\leq g$. Show that $\int_S\left|f_n - f\right|d\mu\to 0$ as $n\to\infty$. Hint: apply the DCT in a suitable way.
What I've tried: I think I have to use the DCT on another function like $h_n = f_n -f$. In that case we would have that $\lim\limits_{n\to\infty}h_n = 0$. However, the absolute value signs tell me that just using $h_n$ is not enough. Furthermore, in order to use the DCT, I need to show that the functions $f_n$ are integrable. As $f_n$ takes on values in $\mathbb{R}$ and in $\mathbb{C}$, we have $f_n = u_n + iv_n$. Therefore, $f_n$ is called integrable if $u_n$ and $v_n$ are integrable. I think that it's possible to choose $h_n$ in such a way that it solves the problem of complex values in $f_n$ and $f$, but I don't know how.
Questions: 


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*How do I solve this exercise?

*What does it mean that $\left|f_n\right|\leq g$? I'm a bit confused because $f_n$ does take on complex values while $g$ does not. Should I interpret the absolute value signs as the norm of $f_n$?
Thanks in advance! 

Let $h_n =f_n-f\to 0$ and $$|h_n|\le |f_n|+|f|\le g+|f|$$
and $ g+|f|$ is integrable then conclude yourself using DCT
Addendum 
the expression $|f|$ either mean absolute value of $f$ if $f$ is real-valued or complex modulus of $f$ if $f$ is complex-valued function.
