# Did there exists a counterexample says that the Dominated convergence Theorem must has a control function rather than the limits function?

Recall the Dominated Convergence Theorem is:

Let $\{f_n\}$ be a sequence of real-valued Lebesgue measurable functions on some open (in fact measurable will sufficient) subset of $E\subset R^d$. $\{f_n\}$ is dominated by $g\in L^1(E)$, i.e., $g$ is Lebesgue integrable on $E$, and $$|f_n(x)|\leq g(x),\quad \forall n,\,\forall x\in E,$$ Then $$\lim_{n\to\infty}\int_{E}f_n d\mu=\int_{E}\lim_{n\to\infty}f_nd\mu.$$

In application is quite tricky to find the dominant function $g$, I just wonder whether we can just take $g$ as the limit function? If not, can you show it by some simple counterexamples? How about $E$ is bounded?

Basically, any of the many examples where the conclusion does not hold will be a counterexample. My favourite is $$f_n(x)=n^2x^n(1-x),\qquad 0\le x\le 1.$$ The pointwise limit is the zero function, yet $\int_0^1 f_n(x)\,dx=1$.