About an exercise in Rudin's book In the book of Walter Rudin Real And Complex Analysis page 31, exercise number 10 said: 
Suppose $\mu(X) < \infty$, $\{f_n\}$ a sequence of bounded complexes measurables functions on $X$ , and $f_n \rightarrow f $ uniformly on $X$. Prove that $$ \lim_{n \rightarrow \infty} \int_X f_n d \mu = \int_X f d \mu  $$
And the answer is the following 
Let $\epsilon > 0$. Since $f_n \rightarrow f$ uniformly, therefore there exists $n_0 \in  N$
such that
$$|f_n (x) − f (x)| < \epsilon \quad  ∀ n > n_0$$
Therefore $|f (x)| < |f_{n_0} (x)| + \epsilon$. Also $|f_n (x)| < |f (x)| + \epsilon$. Combining both
equations, we get
$$|f_n (x)| < |f_{n_0}| + 2\epsilon  \quad ∀ n > n_0$$
Define $g(x) = max(|f_1 (x)|, · · · , |f_{n_0 −1} (x)|, |f_{n_0} (x)| + 2\epsilon)$, then $f_n (x) \leq g(x)$
for all $n$. Also $g$ is bounded. Since $\mu(X)< \infty$, therefore $g \in \mathcal{L}^1(\mu)$. Now
apply DCT to get
$$ \lim_{n \rightarrow \infty} \int_X f_n d \mu = \int_X f d \mu  $$
What didn't I understand is why the condition $f_n \rightarrow f $ uniformly on $X$ is necessary? I proceed like the following 
For every $x \in X$ we have 
$$ |f_n(x)| \leq h(x)= \max_i (|f_i(x)|)$$
such as every $f_i$ is bounded so $h$ is, and in the other hand we  have $\mu(X) < \infty$, therefore $h \in \mathcal{L}^1(\mu)$, then we  can apply DCT.
Where I am wrong please? are there any counter examples? 

Your proof assumes that $h$ is integrable. It doesn't have to be. Suppose that $X=(0,1]$ (with the Lebesgue measure) and that $f_n=\frac1{x^2}\chi_{\left(\frac1{n+1},\frac1n\right]}$. Then $(f_n)_{n\in\mathbb N}$ converges pointwise to the null function, but $h$ is not integrable, since $h(x)=\frac1{x^2}$. And we don't have$$\lim_{n\in\mathbb N}\int_{\mathbb R}f_n=\int_{\mathbb R}0,$$since$$(\forall n\in\mathbb{N}):\int_{\mathbb R}f_n=1.$$
