Pointwise convergence implies convergence in $L^p$?

Question

In this question a user asks if pointwise convergence implies convergence in $L^p$. I would have thought that the answer is yes. I am not experienced with measure theory, which is how that question is framed. The following statement seems to assert that p.w. convergence implies convergence in $L^p$: $$ \lim_{n\to \infty} ||f_n - f||_{L^p(\Omega)}^p = \lim_{n\to \infty} \int_\Omega |f_n(x)-f(x)|^p dx = \int_\Omega |\lim_{n\to \infty} f_n(x)-f(x)|^p dx = \int_\Omega |0|^p dx = 0. $$ But the answers to the other post say that p.w. convergence does not imply convergence in $L^p$, so what am I missing?

Answer 1

For $\Omega=\Bbb R$ and Lebesgue measure I like this example. $$f_n(x)=e^{-(x-n)^2}.$$ Then $f_n\to0$ pointwise, but $$\|f_n\|_p=\|f_1\|_p>0.$$

Answer 2

In general $ \lim_{n\to \infty} \int_\Omega |f_n(x)-f(x)|^p dx = \int_\Omega |\lim_{n\to \infty} f_n(x)-f(x)|^p dx$ is false !

Example: Let $p=1$ , $ \Omega =[0,1]$ and let $f_n$ be defined as follows $( n \ge 3)$:

$f_n(x)=n^2x$ , if $0 \le x \le 1/n$, $f_n(x)=-n^2x+2n$, if $1/n \le x \le 2/n$ and $f_(x)=0$, if $2/n \le x \le1$.