Let $\{f_{n}(x)\}$ be a sequence of integrable functions defined on the closed interval $[a,b]$ that is convergent pointwise to $f(x)$ and uniformly bounded.
I am wondering if this condition alone is sufficient to guarantee that the following limit exists:
$$\lim\limits_{n\to\infty}\int_{a}^{b} f_{n}(x)\,\text{d}x$$
On one hand, I tried proving this by using the Cauchy's convergence test which rquires that for any given $\varepsilon>0$, there should exist a positive integer $N$ such that for all integers $m,n\ge N$, the following inequality holds:
$$\left|\int_a^bf_n(x)-f_m(x)\,\text{d}x\right|\le\int_a^b|f_n(x)-f_m(x)|\,\text{d}x\overset{?}<\varepsilon $$
However, I could not establish this bound since the sequence is not uniformly convergent.
On the other hand, I attempted to find a conterexample to show that the integral limit might not always exist. For example, I considered a sequence like:
$$h_n(x)=\left\{\begin{aligned}&(-2)^n\quad0\le x\le2^{-n}\\&\quad1\qquad2^{-n}<x\le1\end{aligned}\right.$$
which converges pointwise and whose integrals diverges. However, this example fails to be uniformly bounded.
Is there a counterexample where the sequence is uniformly bounded, but the integral limit does not exist, or is the initial statement correct under the given conditions?
Importantly, I do not assume any additional properties about $f$.
Thank you for any insights!