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$$(\forall n\in\mathbb{N}):\int_{\mathbb R}f_n=1.$$