real analysis - 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?