Counterexample: Interchange Limit and Integral Given continuous functions $f_n:[0,1]\to\mathbb R$ uniformly converging to 0 and $p>1$ such that $t^{-p}f_n(t)\in L_1[0,1]$, is then 
\begin{align*}
\lim_{n\to\infty}\int_0^1\frac{f_n(t)}{t^p}dt=0
\end{align*}
always true? What if each $f_n$ is increasing? First I tried to prove it using the Dominated Convergence Theorem, but to me it is not clear that $t^{-p}f_n(t)$ is dominated by anything integrable (especially because $t^{-p}$ is not integrable!). Now I think that it is not true in general, but I was not able to find a counterexample. Do you find one?
Any help is highly appreciated. Thank you very much in advance!
 A: I will give counterexample for $p > 2:$
Let 
$$
g_{n}(x) = \begin{cases} 0 \quad \text{if} \quad x < \frac{1}{n}, \\ 
 x - \frac{1}{n} \quad \text{otherwise}; \end{cases}                                   
$$
Then integrals 
$$
I_n = \int^1_0 \frac{g_n(x)}{x^{p}} dx < \infty  
$$
as each $g_n(x)$ is $0$ in some neighbourhood of $0$, and so integrands are bounded.
Hence, $g_n(x)x^{-p} \in L_1[0,1]$. But its clear that
$$
\lim_{n \to \infty} I_n = \lim_{n \to \infty} \frac{n^{p-2}}{(p-1)(p-2)} +  \Theta(1) = +\infty
$$
 as $2 - p  < 0$ and by integration
$$
I_n = \int^1_{1/n} \frac{x - \frac{1}{n}}{x^p}dx =
\frac{1}{2-p}x^{2 - p}|^1_{1/n} - \frac{1}{(1- p)n}x^{1 - p} |^1_{1/n}  = \\=
\frac{n^{p-2}}{(p-1)(p-2)} + \frac{1}{2-p} - \frac{1}{n(1-p)}
$$
So define for $n > 1$
$$ f_n = \frac{g_n}{I_n}. $$
$g_n < 1$  and $I_n \to \infty$ provide uniform convergence  $\lim_{n \to \infty}  f_n = 0$.
Moreover,
$$
 \lim_{n \to \infty} \int^1_0 \frac{f_n(x)}{x^p}dx =
  \lim_{n \to \infty} \int^1_0 \frac{g_n(x)}{I_n x^p}dx  =
 \lim_{n \to \infty} \frac{I_n}{I_n} = 1 \neq 0.
$$
I think you can refine this counerexample for $p \in (1,2)$ by replasing linear function in definition of $g_n$ by suitable sublinear continuous function. But this is still some extra work.
