Let $f_n \to f$ pointwise on $[0,1]$. If $f_n, f$ are continuous, is it true that $\int_0^1 f_n(x) dx \to \int_0^1 f(x)dx$?

Question

Let $\{f_n\}$ be a sequence of continuous functions on on $[0,1]$. Let $f_n \to f$ pointwise. If $f$ is continuous on $[0,1]$, is it true that $$\int_0^1 f_n(x) dx \to \int_0^1 f(x)dx?$$

I couldn't think of a counter-example, so my inclination is that it is true. If I can show that $f_n \to f$ uniformly, then I would be done, since I can choose an $N$ such that for all $n > N$ and $x\in [0,1]$ we have $|f_n(x) - f(x)| < \varepsilon$, which gives $$ \left| \int_0^1 f_n(x) - f(x) dx \right| \leq \int_0^1 |f_n(x) - f(x)|dx < \varepsilon$$

Can it be shown that $f_n \to f$ uniformly since we're working on a compact set and $f$ is continuous? Or can a counter-example be constructed from here?

Answer 1

No. Take$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}n\sin(n\pi x)&\text{ if }x\leqslant\frac1n\\0&\text{ otherwise.}\end{cases}\end{array}$$Then $\lim_{n\to\infty}f_n(x)=0$ for each $x\in[0,1]$, but, for each $n\in\mathbb N$, $\int_0^1f_n(x)\,\mathrm dx=\frac2\pi$.

Answer 2

There is an easy and well-known counter-example. Consider $f_n=(n-n^2x)X_{[0,1/n]}$.

Answer 3

Let $f_n(x) = n^2x^n(1-x).$ Then $f_n\to 0$ pointwise everywhere in $[0,1],$ but $\int_0^1 f_n(x)\,dx \to 1.$