Convergent functions and integrals Let $V$ be real normed vector space of Riemann integrable functions $f : [0,1]\to\mathbb{R}$ with norm
$||f||:=\sup\{|f(x)|:x\in[0,1]\}$.
Let $f_n:[0,1]\to\mathbb{R}$ be continuous for $n\in\mathbb{N}$ and let there be a function $f\in V$ such that
$\lim_{n\to\infty} ||f-f_n|| = 0$.
I have to show that $\lim_{n\to\infty} \int_0^1f_n(x)dx=\int^1_0f(x)dx$.
I have already shown that $f$ is continous.
My strategy:
Prove that $\lim_{n\to\infty} \int_0^1f_n(x)dx -\int^1_0f(x)dx = 0$ 
so $\lim_{n\to\infty} \int_0^1(f_n(x) - f(x))dx = 0$.
I then think I have to use the squeeze theorem:
$\lim_{n\to\infty}\inf_{[0,1]}(f_n(x) - f(x)) \leq\lim_{n\to\infty} \int_0^1(f_n(x) - f(x))dx \leq \lim_{n\to\infty}\sup_{[0,1]}(f_n(x) - f(x))$
Is it then allowed to put the limit inside the supremum and infimum, which makes them both equal to 0?
 A: You are almost done, just modify the last line as 
$$0\leq \left|\int_0^1(f_n(x) - f(x))dx\right|\leq \int_0^1\left|f_n(x) - f(x)\right|dx \leq \sup_{[0,1]}\left|f_n(x) - f(x)\right|$$
then use $\sup_{[0,1]}\left|f_n(x) - f(x)\right|=\|f-f_n\|\to 0$
and apply the squeeze theorem.
P.S. Following your approach, you need
$$-\sup_{[0,1]}\left|f_n(x) - f(x)\right|\leq\inf_{[0,1]}(f_n(x) - f(x)) \leq \int_0^1(f_n(x) - f(x))dx\\ \leq \sup_{[0,1]}(f_n(x) - f(x))\leq\sup_{[0,1]}\left|f_n(x) - f(x)\right|$$
and again use $\sup_{[0,1]}\left|f_n(x) - f(x)\right|=\|f-f_n\|\to 0$.
A: 
Is it then allowed to put the limit inside the supremum and infimum, which makes them both equal to 0?

You can't switch infimum/supremum and the limit in general.
Consider $f_n(x)=\max\{0,1-|x-n|\}$. Then $\sup_{x\in\mathbb R}|f_n(x)|=1$ while $\lim_{n\to\infty} |f_n(x)|=0$. Hence,
$$
\lim_{n\to\infty}\sup_{x\in\mathbb R}|f_n(x)|=1\neq 0=\sup_{x\in\mathbb R}\lim_{n\to\infty} |f_n(x)|.
$$
But your case is a bit more special because $[0,1]$ is compact and $f_n$ and $f$ are continuous. Hence, the infimum and supremum are achieved at some points. Then you can apply the limit as your last step.
A: The map $f \mapsto \int_0^1 f(x)\,dx$ is continuous w.r.t your norm $\|\cdot\|$.
Indeed, it is linear and bounded:
$$\left|\int_0^1 f(x)\,dx\right| \le \int_0^1 |f(x)|\,dx \le \int_0^1 \|f\|\,dx  = \|f\|$$
Therefore $f_n \xrightarrow{\|\cdot\|} f$ implies $ \int_0^1 f_n(x)\,dx \to  \int_0^1 f(x)\,dx$
