Let $\lambda$ be the Lebesgue-measure on $\Omega =[0,1]$. Given a sequence of non-negative measurable functions $$f_n:\Omega\to\Bbb R: x \mapsto ne^{-nx},$$ how can I show that $f_n$ converges $\lambda$-almost everywhere to a measurable function $f$, but $$\int f d\lambda \neq \lim_{n\to\infty} \int f_nd\lambda $$
The theorem of dominated convergence seems to fail here..
Any hints for the proof?
Edit: Ok what I have done so far: $(f_n)$ converges almost everywhere to $f(x)=0$. Obviously $f_n(0)=n$ for all $n \in \Bbb N_{\gt 0}$. Hence, $$\int f d\lambda =0 \neq 1 = \lim_{n\to\infty} \int f_nd\lambda$$
Now, how can I show that there is no integrable function that dominates $(f_n)$ without utilizing the theorem of dominated convergence?
For $x\in(0,1]$ we have \begin{align} \lim_{n\to\infty} ne^{-nx}&=e^{\lim_{n\to\infty}\left(\log n-nx\right)}=0, \end{align} and for $x=0$ $$\lim_{n\to\infty}ne^{-n\cdot 0}=\infty. $$ Therefore $f_n\to 0$ a.e. and hence $$\int_{[0,1]}\lim_{n\to\infty} f_n\ \mathsf d\lambda=0. $$ However, $$\int_{[0,1]}f_n\ \mathsf d\lambda = \int_0^1 ne^{-nx}\ \mathsf dx=-e^{-nx}|_0^1 =1-e^{-nx}.$$ For $x\in(0,1]$ we have $$\lim_{n\to\infty} (1-e^{-nx}) = 1, $$ and thus $$\lim_{n\to\infty}\int_{[0,1]}f_n\ \mathsf d\lambda = 1. $$