Suppose $(f_{n})_{n\in \mathbb {N} }$ is a sequence of continuous functions on the closed interval $[0,1]$ converging to $0$ point-wise. Then the integral
$$\int_{0}^{1} f_n(x) dx$$
converges to $0$.
I have to check that the whether the above statement is TRUE or FALSE.
My attempt:
If the sequence $(f_{n})_{n\in \mathbb {N} }$ is uniformly convergent on $[0,1]$ with limit $0$. Then for every $\epsilon >0$, there exists a natural number $N$ such that for all $n\geq N$ and $x\in [0,1]$
$$|f_{n}(x)|<\epsilon.$$
This implies
$$\Big{|} \int_{0}^{1} f_n(x) dx \Big{|} \leq \int_{0}^{1} |f_n(x)| dx <\epsilon$$
for all $n \geq N$. Thus $\int_{0}^{1} f_n(x) dx$ converges to 0.
The problem with the point-wise convergence is there is no guarantee that such a $N$ would exist which works for all $x \in [0,1]$.
I think the statement seems to be true nonetheless. It would be helpful if some could help me prove it or provide me with a counter example.
Best Answer
Let $f_n$ be a broken line joining $(0,0), \left(\frac{1}{2n},n\right), \left(\frac{1}{n},0\right)$ and $(1,0)$. Then $f_n\to 0$ pointwise, while all the integrals are $\frac{1}{2}$.