Does uniform convergence imply convergence of the integrals? 
Let $(X,A,\mu)$ be a measure space. Take a sequence $(f_n)_{n \in
 \mathbb N}$ of real-valued measurable bounded functions. Suppose $f_n \to f$
  uniformly on $X$ and suppose that $\mu(X)<+\infty$. Then $$ \int_X f_n
 d\mu \to \int_Xf d\mu $$ when $n \to +\infty$.

This exercise is taken from Rudin, Real and Complex Analysis, chapter 1. I do not understand where I should use the hypothesis of boundedness of the functions. Indeed, 
$$
\left \vert \int_X f-f_n d\mu \right\vert \le \int_X \vert f_n - f \vert d\mu \le \int_X \varepsilon d\mu = \varepsilon \mu(X)   
$$
for $n$ sufficiently big. 
Where do I use the boundedness of the functions? Do I need it in order to say $\vert \int_X f_n-f d\mu\vert \le \int \vert f-f_n \vert d\mu$?  

Boundedness of the maps and finiteness of the measure space are used to be sure that $\int_Xf_nd\mu$ and $\int_Xfd\mu$ are real numbers. 
