Prove that if $f_n\to f$ uniformly and $\int_0^\infty|f_n|\le M$, then $\int_0^\infty|f|<\infty$

Question

Again while preparing to calculus I found another interesting question:

Prove or give counterexample that if $f_n\to f$ uniformally $[0,\infty]$ and $\forall n\in\mathbb {N}\int_0^\infty|f_n|dx\le M$, then $\int_0^\infty|f(x)|dx<\infty$

It seems incorrect. I'm looking for a function that tends to a constant $c$ uniformly and its integrals are bounded. If the convergence was in $[2,b] (b\in\mathbb R)$, I could have taken $f_n(x)=\frac {x^n}{x^n+1}$ which indeed is a counterexample. How can I find a counterexample?

Answer 1

Fix $A>0$. Then $$\int_0^A|f(x)|\mathrm dx=\lim_{n\to +\infty}\int_0^A|f_n(x)|\mathrm dx\leqslant \liminf_{n\to +\infty}\int_0^{+\infty}|f_n(x)|\mathrm dx\leqslant M.$$ As $A$ was arbitrary, $\int_0^{+\infty}|f(x)|\mathrm dx\leqslant M$.

It actually works when we only have pointwise convergence, and it is called Fatou's lemma, but it involves more advanced tools.