$f_n,f:\mathbb R\to\mathbb R$ non negative. $f_n$ converge pointwise to $f$ and $\int_\mathbb R f_n = 1 = \int_\mathbb R f$. Show $$\lim_n\int_\mathbb R |f_n-f|=0$$
I realize this just means showing $f_n$ converge to $f$ uniformly but the condition I know for uniform convergence can't be applied here. We don't know if $f_n(x)$ is monotonic in $n$ for all $x$ or that $f_n-f$ is bounded. How can I prove this?
Let $$\begin{align*}f^{+} &= \max\{f,0\}\\ f^{-} &= \max\{-f,0\}\end{align*}$$ then it holds $$\begin{align*}f &= f^+ - f^- \\ |f| &= f^+ + f^-\end{align*}$$
And we get: $$\begin{align*}|f-f_n| &= (f-f_n)^+ + (f-f_n)^- \\ &= 2(f-f_n)^+ - (f-f_n)\\ &\le 2f^+ -(f-f_n)\end{align*} $$
Setting $$\begin{align*}\tilde{f}_n &:= |f-f_n| \\ \tilde{g}_n &:= 2f^+ -(f-f_n)\end{align*}$$ we get by the generalized dominated convegence theorem for $\tilde{f}_n,\tilde{g}_n$ that $$\lim_{n\to\infty} \int |f-f_n| = \lim_{n\to\infty} \int \tilde{f}_n = 0$$
The question has been edited.
Answer for the earlier version:
This is false. Let $f_n=I_{(0,1)}+nI_{(0,\frac 1 n)}-nI_{(-\frac 1 n ,0)}$ and $f=I_{(0,1)}$. Then $\int |f_n-f|=2$.
Adding answer from the deleted answer.
First note that $|f_n-f|= (f_n-f)+2\max(f-f_n,0)$
The integral of the first term in RHS is $0$, the integral of the second term converges to $0$ using DCT with the dominator of $2\max(f-f_n,0)$ being $2f$ (Since all the functions are non negative)