Let $f_n\to f$ on $[0,1]$, where $f_n$ and $f$ be continuous with $f_n$ bounded. Then, which of the following are true:
1) $\lim_{n\to\infty} \int_{[0,1]}f_n=\int_{[0,1]}f$
2)$\lim_{n\to\infty}\int_{[0,1]}|f_n(t)-f(t)|dt\to0\implies\lim_{n\to\infty}\int_{[0,1]}f_n=\int_{[0,1]}f$?
I think the first should be true by dominated convergence, since $f_n$ are bounded and $f$ is continuous. Any counterexamples? I have no idea about second one. Any hints. Thanks beforehand.
$\def\d{\mathrm{d}}$For the first proposition, take$$ f_n(x) = \begin{cases} nπ \sin(nπx); &\displaystyle 0 \le x \le \frac{1}{n}\\ 0; &\displaystyle \frac{1}{n} < x \le 1 \end{cases}, $$ and $f = 0$. Since $f_n(0) = 0 \ (\forall n \in \mathbb{N}_+)$, then $\lim\limits_{n \to \infty} f_n(0) = 0 = f(0)$. And if $0 < x \leqslant 1$, for $\displaystyle n > \frac{1}{x}$, there is $f_n(x) = 0$, thus $\lim\limits_{n \to \infty} f_n(x) = 0 = f(0)$. So $f_n \to f$. However,\begin{gather*} \int_0^1 f_n(x) \,\d x = \int_0^π \sin u \,\d u = 2, \quad \forall n \in \mathbb{N}_+\\ \int_0^1 f(x) \,\d x = 0, \end{gather*} thus$$ \lim_{n \to \infty} \int_0^1 f_n(x) \,\d x = 2 \neq 0 = \int_0^1 f(x) \,\d x. $$
For the second proposition, because\begin{align*} \varlimsup_{n \to \infty} \left| \int_0^1 f_n(x) \,\d x - \int_0^1 f(x) \,\d x \right| \leqslant \int_0^1 |f_n(x) - f(x)| \,\d x = 0, \end{align*} then$$ \lim_{n \to \infty} \left| \int_0^1 f_n(x) \,\d x - \int_0^1 f(x) \,\d x \right| = 0, $$ which implies$$ \lim_{n \to \infty} \int_0^1 f_n(x) \,\d x = \int_0^1 f(x) \,\d x. $$