Scheffé lemma says : $f_n\to f$ a.e. and $\int f_n\to \int f$ (and $f_n,f\in L^1(\mathbb R^d)$) then $f_n\to f$ in $L^1(\mathbb R^d)$
In wikipedia they do as follow : $$\int (f_n-f)^++\int (f_n-f)^-=\int|f_n-f|\tag{1}$$ $$\int (f_n-f)^+-\int(f_n-f)^-=\int(f_n-f)\tag{2}.$$
Since $0\leq (f_n-f)^+\leq f$ and $(f_n-f)^+\to 0$, we have by DCT,$$\lim_{n\to \infty }\int(f_n-f)^+=0.$$ Using $(2)$ and $\int f_n\to \int f$, we get $$\int (f_n-f)^-\to 0,$$ therefore, using $(1)$, the claim follow.
Question
Why do we need $(2)$ ? (of course it's not important, but we can make the argument easier). Don't we have $0\leq (f_n-f)^\pm\leq f$, and thus, by DCT $$\int (f_n-f)^\pm\to 0,$$ and thus, using $(1)$ the claim follow. This is not correct ?
$(f-f_n)^{+}$ is either $0$ or $f-f_n$ so $(f-f_n)^{+} \leq f$. This allows us to use DCT. It is not true that $0 \leq (f_n-f)^{\pm} \leq f$. In fact the proof above is also wrong. DCT applies to $\int(f-f_n)^{+}$ but not to $\int (f_n-f)^{+}$. Once you see that $\int(f-f_n)^{+} \to 0$ you can use the facts that $\int(f-f_n)^{-}=\int(f-f_n)^{+}-\int (f-f_n) $ and $|f-f_n|=(f-f_n)^{+}+(f-f_n)^{-}$.