I was thinking of this problem: If $f_n\rightarrow f$ pointwise a.e. and $\lvert f_n\lvert \leq g$ for some $g\in L^p$, then prove that $f_n \rightarrow f$ in $L^p$.
To use the dominated convergence theorem we need: 1. $\lvert f_n -f\lvert^p\rightarrow 0$ a.e. 2. $\lvert f_n -f\lvert^p \leq G(x)\in L^1$.
The first condition we can directly get from the pointwise convergence, and to get the second condition, we let $G(x)=2^pg(x)^p\in L^1$, and then we have $\lvert f_n-f \lvert^p\leq [2g(x)]^2=G(x)\in L^1$. So we can get the desired result.
However, I don’t understand why we cannot go directly from $\lvert f_n -f\lvert^p \rightarrow 0 $ a.e to $\int \lvert f_n-f\lvert^p\rightarrow 0$?
Any help? Thanks!
Because it is not true in general that point-wise convergence implies convergence in norm. You can have continuous functions $f_n$ tending to zero but the integrals do not. For example, take as $f_n$ a function whose graph is a triangle whose vertices are the points $(n,0),(n+1,0),(n+\frac{1}{2},1)$, and zero elsewhere. Then $f_n(x)\to 0$ for every $x\in [0,\infty)$, but $\int_0^{\infty}f_n(x)\,dx=\frac{1}{2}$ for every $n$, so you get a counterexample in $L^p(0,\infty)$.