Lebesgue Integral of $f^{\alpha}$ I have a homework problem:
Let $m$ denote the lebesgue measure. Let $A \subset \mathbb{R}$ be a measureable set with $m(A) < \infty$. Let $f$ be a non-negative measurable function with finite lebesgue integral. Decide whether $$\lim_{\alpha \rightarrow 1^{+}} \int_{A}f^{\alpha} dm = \int_{A}f dm$$
I previously proved this statement is true as $\alpha \rightarrow 1^-$, but I think it's false from the right since if $f \geq 1$ $, \{f^{\alpha}\}$ is a pointwise non-increasing collection (for decreasing $\alpha$), which I thought I could exploit somehow in the vein of an example of a strict inequality in Fatou's lemma. 
But I can't seem to get anywhere. Any hints or advice would be greatly appreciated. 
Edit: I'm starting to think this is true. I have an argument in mind, but I need to find some $\alpha>1$ such that $f^{\alpha}$ is Lebesgue integrable.
 A: If there is an $\alpha_0 > 1$ such that
$$\int_A f^{\alpha_0}\,dm < \infty,$$
then we can dominate $f^\alpha$ by $1 + f^{\alpha_0}$ for $1 < \alpha < \alpha_0$, and the dominated convergence theorem asserts that then indeed
$$\lim_{\alpha \to 1^+} \int_A f^\alpha\,dm = \int_A f\,dm.$$
But, for $f\in L^1(A)$, there need not exist such an $\alpha_0$. For example, let's take $A = \left(0,\frac12\right)$, and
$$f(x) = \frac{1}{x(\log x)^2}.$$
Then $f(x) > 0$ on $A$, and
$$\int_A f\,dm = \int_0^{1/2} \frac{dx}{x(\log x)^2} = \int_{-\infty}^{-\log 2} \frac{du}{u^2} = \frac{1}{\log 2}$$
is finite, but $f^\alpha$ has a non-integrable singularity at $0$ for all $\alpha > 1$, since
$$\frac{1}{x^\alpha(\log x)^{2\alpha}} > \frac{1}{x}$$
for all small enough $x$, so
$$\lim_{\alpha\to 1^+} \int_A f^\alpha\,dm = +\infty \neq \int_A f\,dm.$$
A: If $f^\alpha$ is integrable and $f\ge 1$ almost everywhere, then you can use Lebesgue dominated convergence theorem.
A: This answer was motivated by the comment relating to the previous version of this post. 
Let $f = \sum_{n=1}^\infty \frac{1}{2^n} x^{-1+1/n}$, $A=(0,1]$, and $m = \lambda$ (Lebesgue measure). Then 
$$
  \int_A f dm
= \sum_{n=1}^\infty \frac{1}{2^n} \int_0^1 x^{-1+1/n}dx
= \sum_{n=1}^\infty \frac{1}{2^n} \left. nx^{1/n} \right|^1_0
= \sum_{n=1}^\infty \frac{n}{2^n} 
< \infty.
$$
However, for every $\alpha > 1$ there is an $n$ for which $\frac{n}{n-1} < \alpha$, in which case $-\alpha + \alpha/n < -1$. Then 
$$
     \int_A f^\alpha dm
\geq \int_0^1 (x^{-1+1/n})^\alpha
=    \infty.
$$
I may have made another error here, but it seems not to be true.
